🧠:Okay, so I need to find the smallest natural number n such that both n and n+1 are in the set S. According to Legendre's three-square theorem, S consists of numbers that can't be written as the sum of three squares. The theorem specifically says these are numbers of the form 4^a(8b + 7) where a and b are nonnegative integers. First, let me make sure I understand the problem correctly. I need to find two consecutive numbers, n and n+1, both of which can't be expressed as the sum of three squares. And I need the smallest such n. So, my plan is to start checking numbers sequentially and see if both the number and the next one are in S. But since S is defined by numbers of the form 4^a(8b + 7), maybe there's a pattern or a way to generate candidates more efficiently instead of checking each number one by one. But maybe starting with small numbers and checking is feasible here since the answer might not be too large.Let me recall that numbers which are not expressible as the sum of three squares are exactly those that are of the form 4^a(8b + 7). So, numbers like 7, 15, 23, 28, 31, etc. Wait, let's confirm that.For example, 7: 4^0*(8*0 +7)=7. So yes. Then 15: 4^0*(8*1 +7)=15. 23: 4^0*(8*2 +7)=23. 28: 4^2*(8*0 +7)=16*7=112? Wait, no. Wait, 4^a(8b +7). So when a=1, 4*(8b +7). So 4*(8*0 +7)=28. So 28 is in S. Then 4*(8*1 +7)=4*(15)=60, so 60 is in S. Similarly, a=2: 16*(8b +7). 16*7=112, 16*(8*1 +7)=16*15=240, etc.So numbers in S are 7, 15, 23, 28, 31, 39, 47, 55, 60, 63, 71, etc. Wait, 31 is 8*3 +7=31, so a=0, b=3. Then 28 is a=1, b=0. 60 is a=1, b=1. 112 is a=2, b=0. So the numbers in S are those congruent to 7 mod 8, or 28 mod 32, or 112 mod 128, etc., scaled by powers of 4.So, to find consecutive numbers in S, both n and n+1 must be of the form 4^a(8b +7). Since they are consecutive, one is even and the other is odd. Let's think: if n is even, then n+1 is odd. If n is odd, then n+1 is even. So one of them must be even. But numbers in S can be both even or odd depending on the power of 4. For example, 7 is odd (a=0), 28 is even (a=1), 112 is even (a=2). So perhaps consecutive numbers in S could be an even followed by an odd, or vice versa. But how can two consecutive numbers both be of the form 4^a(8b +7)? Let's see. Let's take n and n+1. Let's suppose n is in S and n+1 is in S. Then n = 4^{a}(8b +7) and n+1 = 4^{c}(8d +7). Since they are consecutive, one is even and one is odd. So the even one must have at least a power of 4 (since 4^a is even if a ≥1). The odd one must have a=0, so it's 8d +7. So, for example, n could be even: 4^a(8b +7), and n+1 is 8d +7. Alternatively, n could be 8b +7 (odd) and n+1 is 4^a(8d +7) (even). So let's consider both cases:Case 1: n is even, so n = 4^a(8b +7), and n+1 is odd, so n+1 = 8d +7.Case 2: n is odd, so n = 8b +7, and n+1 is even, so n+1 = 4^a(8d +7).We need to check both cases and find the smallest n in either case.Let's start with Case 1: n = 4^a(8b +7), n+1 = 8d +7.So 4^a(8b +7) + 1 = 8d +7. Let's rearrange:4^a(8b +7) = 8d +6.Divide both sides by 2:2^{2a -1}(8b +7) = 4d +3.Hmm, let's see. For a ≥1, 4^a is divisible by 4. So 4^a(8b +7) is divisible by 4. Then 8d +6 must be congruent to 6 mod 8. But 8d +6 ≡6 mod8. But 4^a(8b +7) must be congruent to 4^a*7 mod 8, since 8b is 0 mod8. So 4^a*7 mod8. Let's compute that for a=1: 4*7=28 ≡4 mod8. For a=2: 16*7=112 ≡0 mod8. For a=0: 1*7=7 mod8. But in Case 1, a ≥1 because n is even. So a=1: 4*7=28 ≡4 mod8. Then 4^a(8b +7) ≡4 mod8. Then 8d +6 ≡6 mod8. So 4 mod8 =6 mod8? No, that's not possible. So 4^a(8b +7) ≡4 mod8, while n+1=8d +7 ≡7 mod8. Wait, wait, n=4^a(8b +7) is in S, and n+1=8d +7 is in S. But n=4^a(8b +7) must be even, so a ≥1. Then n=4^a(8b +7). Let's plug a=1: n=4*(8b +7)=32b +28. Then n+1=32b +29. For this to be in S, n+1 must be of the form 8d +7 (since a=0). So 32b +29 =8d +7. So 32b +22=8d. Divide both sides by 2:16b +11=4d. But 16b +11 is 11 mod4, which is 3 mod4, while 4d is 0 mod4. Contradiction. So no solution here.a=2: n=16*(8b +7)=128b +112. Then n+1=128b +113. For n+1 to be in S, it must be of the form 8d +7. So 128b +113 =8d +7. Then 128b +106=8d. Divide both sides by 2:64b +53=4d. Again, 64b is 0 mod4, 53 is 1 mod4, so 64b +53 ≡1 mod4, but 4d is 0 mod4. Contradiction. So no solution here either.Similarly, a=3: n=64*(8b +7)=512b +448. Then n+1=512b +449. For this to be in S: 512b +449=8d +7. Then 512b +442=8d. Divide by 2:256b +221=4d. 256b is 0 mod4, 221 is 1 mod4. Again, 1 mod4 vs 0 mod4. No solution. So in general, for Case 1, when n=4^a(8b +7) and n+1=8d +7, the equation leads to a contradiction modulo 4. Therefore, there are no solutions in Case 1.Now let's check Case 2: n is odd, so n=8b +7, and n+1=4^a(8d +7). Then, 8b +7 +1=4^a(8d +7). So 8b +8=4^a(8d +7). Let's factor left side: 8(b +1)=4^a(8d +7). Let's divide both sides by 4: 2(b +1)=4^{a -1}(8d +7). If a=1: Then equation becomes 2(b +1)=8d +7. So 2(b +1)=8d +7. Left side is even, right side is 8d +7 which is odd. Contradiction. So a cannot be 1.If a=2: Then equation is 8(b +1)=4^{2 -1}(8d +7)=4(8d +7). So 8(b +1)=32d +28. Divide both sides by 4:2(b +1)=8d +7. Again, left side even, right side odd. Contradiction.If a=3: Then equation is 8(b +1)=4^{3 -1}(8d +7)=16(8d +7). So 8(b +1)=128d +112. Divide by 8: b +1=16d +14. So b=16d +13. Therefore, n=8b +7=8*(16d +13)+7=128d +104 +7=128d +111. Then n+1=128d +112=16*(8d +7). So n+1=16*(8d +7), which is indeed of the form 4^a(8d +7) with a=2. Therefore, this is a valid solution. So for a=3, we have solutions. Similarly, if a=4, we can check:a=4: 8(b +1)=4^{4 -1}(8d +7)=64(8d +7). Then 8(b +1)=512d +448. Divide by8: b +1=64d +56. So b=64d +55. Then n=8b +7=8*(64d +55)+7=512d +440 +7=512d +447. Then n+1=512d +448=64*(8d +7), which is 4^3*(8d +7). So valid. Therefore, in Case 2, starting from a=3, there are solutions. However, the minimal solution would be when d=0. Let's check when d=0 in a=3 case:d=0: n=128*0 +111=111. Then n+1=112. Check if 111 is in S: 111 is 8*13 +7=111. So yes, a=0, b=13. Then 112 is 16*7=4^2*(8*0 +7). So a=2, b=0. Both 111 and 112 are in S. So 111 and 112 are consecutive numbers in S. But wait, is 111 the smallest such n? Let's check smaller numbers.Wait, but maybe there's a smaller a where Case 2 works. Wait, when a=0: n+1=4^0*(8d +7)=8d +7. So n=8d +6. But then n=8d +6 is not in S unless it's of the form 4^a(8b +7). But 8d +6 is 6 mod8, which is not 7 mod8, and since it's even, if it's of the form 4^a(8b +7), then 4 divides it. 8d +6 is 2*(4d +3), which is not divisible by 4. So n=8d +6 is not in S. Hence a cannot be 0.Wait, perhaps there's a mistake here. Let me check again. If in Case 2, n=8b +7 and n+1=4^a(8d +7). So if a=0, then n+1=8d +7, which would imply n=8d +6. But n=8d +6 is not in S, so that's invalid. So the minimal a where Case 2 works is a=3, leading to n=111 and n+1=112. But are there any smaller consecutive numbers in S?Wait, let's check numbers manually. Let's list S numbers up to, say, 200 and see if there are consecutive pairs.Starting from 7:7, 15, 23, 28, 31, 39, 47, 55, 60, 63, 71, 79, 87, 92, 95, 103, 111, 112, 119, 124, 127, 135, 143, 151, 156, 159, 167, 175, 183, 188, 191, 199, ...Wait, here I see 111 and 112 are consecutive. Then 112 and 113? 113 is not in S. Then 124 is in S, 125? 124 is 4*31=124, 31 is 8*3 +7=31. So 124=4^1*(8*3 +7). So 124 is in S. Then 125=125, which is 125 mod8=5, so 125=8*15 +5, not 7 mod8, and 125 is not divisible by 4, so not in S. Then 127 is in S (8*15 +7=127). Then 135=8*16 +7=135? 135 divided by8 is 16.875, so 8*16=128, 135-128=7. So 135=8*16 +7, so a=0, so 135 is in S. Then 143=8*17 +7=143. So also in S. Then 151, 156, 159, etc.Looking at the list, the first consecutive numbers in S are 111 and 112. Before that, looking at the list: 7,15 (difference 8), 15,23 (diff8),23,28 (diff5),28,31 (diff3),31,39 (diff8),39,47 (diff8),47,55 (diff8),55,60 (diff5),60,63 (diff3),63,71 (diff8),71,79 (diff8),79,87 (diff8),87,92 (diff5),92,95 (diff3),95,103 (diff8),103,111 (diff8),111,112 (diff1). So 111 and 112 are consecutive. Before that, there are no consecutive numbers in S. Let's check between 7 and 15: 7 and 15 are 8 apart. 15 and 23: 8 apart. 23 and 28: 5 apart. 28 and31: 3 apart. 31 and39:8. 39 and47:8. 47 and55:8. 55 and60:5. 60 and63:3. 63 and71:8. 71 and79:8. 79 and87:8. 87 and92:5. 92 and95:3. 95 and103:8. 103 and111:8. 111 and112:1. So indeed, the first occurrence of consecutive numbers is 111 and112. But wait, let's confirm that both 111 and 112 are indeed in S. 111: Let's see if 111 can be written as 4^a(8b +7). Let's check if it's of that form. 111 divided by4 is 27.75, so not divisible by4. Therefore a=0. So 111=8b +7. 111-7=104=8b. So b=13. So 111=8*13 +7, so yes, a=0, b=13. Therefore 111 is in S. 112: 112 divided by4=28. 28=8b +7? 28-7=21=8b. 21 divided by8=2.625. Not an integer. So a=1: 4^1*(8b +7). Then 112/4=28. 28=8b +7. 28-7=21=8b. b=21/8=2.625. Not integer. Then a=2: 16*(8b +7)=112. 112/16=7. So 8b +7=7. Therefore, b=0. So 112=16*(8*0 +7)=16*7=112. So yes, a=2, b=0. Therefore 112 is in S.Therefore, both 111 and 112 are in S. Now, are there any smaller numbers than 111 where n and n+1 are both in S? From the list I generated earlier, it seems 111 and112 are the first occurrence. Let's check between, say, 60 and63. 60 is in S, 61? 61 is not in S. 63 is in S. So 60 and61: no. 63 and64: 64 is 4^3*1, which is not of the form 4^a(8b +7). 64=4^3*1, but 1 is not 8b +7. So 64 not in S. Similarly, 28 and29: 28 is in S, 29 is not. 31 and32:32 is not in S. 7 and8:8 is not in S. So indeed, the first occurrence is 111 and112.But wait, let's check another pair: 112 and113. 112 is in S, but 113 is not. Then 124 is in S, 125 is not. 127 and128:128 is 4^a(8b +7)? 128=4^3*2, but 2 is not 8b +7. So 128 not in S. 135 and136:136 is divisible by4. 136/4=34. 34=8b +7? 34-7=27, which is not divisible by8. So 136 not in S. So no, 135 and136 not both in S.Therefore, the answer seems to be 111. But to be thorough, let's check all numbers before 111 to ensure there are no consecutive pairs in S.Check numbers between, say, 1 to 111. Let's list all numbers in S up to 112:7,15,23,28,31,39,47,55,60,63,71,79,87,92,95,103,111,112.Now check for consecutive numbers:7 and15: not consecutive.15 and23: same.23 and28: difference 5.28 and31: difference3.31 and39:8.39 and47:8.47 and55:8.55 and60:5.60 and63:3.63 and71:8.71 and79:8.79 and87:8.87 and92:5.92 and95:3.95 and103:8.103 and111:8.111 and112:1. So yes, the first consecutive pair is 111 and112. Therefore, the smallest n is 111.But wait, just to make sure, is there a smaller pair that I missed? Let's check numbers not in the main sequence. For example, numbers like 28: 28 is in S. 29: 29 is not in S. 31 is in S. So between 28 and31:28,29,30,31. Only28 and31 are in S. 60 is in S. 61,62,63:63 is in S. 60 and63 are separated by3. 92 and95: same. So no consecutive numbers. Therefore, the answer is 111.

✅:The smallest natural number ( n ) such that both ( n ) and ( n+1 ) are in ( S ) is 111.Step-by-Step Explanation:1. Understanding the Set ( S ): - According to Legendre's three-square theorem, ( S ) contains numbers of the form ( 4^a(8b + 7) ), where ( a, b geq 0 ). - These numbers cannot be expressed as the sum of three squares.2. Key Insight for Consecutive Numbers: - Consecutive numbers ( n ) and ( n+1 ) must be in ( S ). - One must be even (hence divisible by a power of 4) and the other odd (of the form ( 8b + 7 )).3. Analyzing Possible Cases: - Case 1: ( n ) even (( 4^a(8b + 7) )) and ( n+1 ) odd (( 8d + 7 )). - Leads to contradictions modulo 4 (no solutions). - Case 2: ( n ) odd (( 8b + 7 )) and ( n+1 ) even (( 4^a(8d + 7) )). - Solving ( 8b + 8 = 4^a(8d + 7) ) reveals valid solutions for ( a geq 3 ).4. Finding the Minimal Solution: - For ( a = 2 ), ( n+1 = 16(8d + 7) ). Backtracking gives ( n = 128d + 111 ). - The smallest ( d = 0 ) yields ( n = 111 ) and ( n+1 = 112 ).5. Verification: - 111: ( 111 = 8 times 13 + 7 ) (form ( 8b + 7 ), ( a = 0 )). - 112: ( 112 = 16 times 7 = 4^2(8 times 0 + 7) ).6. Checking Smaller Candidates: - Listing numbers in ( S ) up to 112 confirms no smaller consecutive pairs exist.Answer: (boxed{111})

🔑:1. According to Legendre's three-square theorem, the set ( S ) consists of natural numbers that can be written in the form ( 4^a(8b+7) ) where ( a ) and ( b ) are nonnegative integers.2. We need to find the smallest ( n in mathbb{N} ) such that both ( n ) and ( n+1 ) are in ( S ).3. Since ( n ) and ( n+1 ) are consecutive integers, one of them must be even and the other must be odd.4. If ( n ) is even, then ( n+1 ) is odd. Conversely, if ( n ) is odd, then ( n+1 ) is even.5. Consider the case where ( n+1 ) is odd. For ( n+1 ) to be in ( S ), it must be of the form ( 4^a(8b+7) ). Since ( n+1 ) is odd, ( a ) must be 0 (because ( 4^a ) would make it even for ( a geq 1 )).6. Therefore, ( n+1 = 8b + 7 ). This implies ( n = 8b + 6 ).7. However, ( n equiv 6 pmod{8} ) is not of the form ( 4^a(8b+7) ) for any nonnegative integers ( a ) and ( b ). Thus, ( n+1 ) cannot be odd.8. Now consider the case where ( n+1 ) is even. For ( n+1 ) to be in ( S ), it must be of the form ( 4^a(8b+7) ) with ( a geq 1 ) (since ( n+1 ) is even).9. Let ( a = 2 ) and ( b = 0 ). Then ( n+1 = 4^2(8 cdot 0 + 7) = 16 cdot 7 = 112 ).10. Therefore, ( n = 112 - 1 = 111 ).11. We need to check if ( 111 ) is also in ( S ). For ( 111 ) to be in ( S ), it must be of the form ( 4^a(8b+7) ).12. Since ( 111 ) is odd, ( a ) must be 0. Thus, ( 111 = 8b + 7 ).13. Solving for ( b ), we get ( 111 - 7 = 104 ) and ( 104 = 8b ), so ( b = 13 ).14. Therefore, ( 111 = 8 cdot 13 + 7 ), which is of the form ( 4^a(8b+7) ) with ( a = 0 ) and ( b = 13 ).The final answer is ( boxed{111} ).

🧠:Okay, let's try to tackle this problem. So, we have 100 real numbers a₁, a₂, ..., a₁₀₀. They satisfy the equation:a₁² + a₂² + ... + a₁₀₀² + (a₁ + a₂ + ... + a₁₀₀)² = 101.We need to prove that each |aₖ| ≤ 10 for all k from 1 to 100. Hmm, interesting. Let's break this down.First, let's parse the given equation. The first part is the sum of squares of all the a_i's. The second part is the square of the sum of all the a_i's. So combined, they equal 101. The problem is to show that each individual term aₖ is bounded in absolute value by 10. I remember that inequalities like Cauchy-Schwarz might be helpful here. Also, sometimes when dealing with sums and squares, completing the square or using Lagrange multipliers can be useful. But since this is a problem likely solvable with algebraic manipulations, let's try to find a way to relate the individual terms to the total sum.Let me denote S = a₁ + a₂ + ... + a₁₀₀. Then the given equation becomes:Sum_{i=1 to 100} a_i² + S² = 101.We need to bound each |a_k|. Let's consider a specific a_k, say a₁. Let's try to find an upper bound for |a₁|. Suppose we fix a₁ and consider the rest of the variables. But maybe that's not the right approach. Alternatively, maybe we can express the sum in terms of a₁ and the others, then apply some inequality.Alternatively, perhaps we can use the Cauchy-Schwarz inequality. The Cauchy-Schwarz inequality states that (sum u_i v_i)^2 ≤ (sum u_i²)(sum v_i²). But I'm not sure how to apply it here directly. Wait, maybe if we consider the sum of squares and the square of the sum.Wait, there's an identity related to the square of a sum: (sum a_i)^2 = sum a_i² + 2 sum_{i < j} a_i a_j. So, the given equation is sum a_i² + (sum a_i)^2 = 101. That becomes sum a_i² + sum a_i² + 2 sum_{i < j} a_i a_j = 101. So combining terms, that's 2 sum a_i² + 2 sum_{i < j} a_i a_j = 101. Hmm, maybe not immediately helpful.Alternatively, maybe we can rewrite the given equation as sum a_i² + S² = 101, where S is the sum of all a_i. Let's denote sum a_i² as Q. Then Q + S² = 101. We need to bound each |a_i|. I know that for any real numbers, the maximum absolute value is at least the average absolute value, but here we need the opposite. To bound each term, perhaps we can use the fact that if one term is too large, it would make the sum of squares too big. But how?Suppose, for contradiction, that there exists some k such that |a_k| > 10. Then, a_k² > 100. But the total sum of squares plus S² is 101, so if a_k² > 100, then even if all other terms are zero, Q would be greater than 100, and S² would be at least a_k² (since S would be a_k, if others are zero). Wait, no. If all other terms are zero, then S = a_k, so S² = a_k². Then Q + S² = a_k² + a_k² = 2a_k². If this equals 101, then 2a_k² = 101 => a_k² = 101/2 ≈ 50.5, so |a_k| ≈ sqrt(50.5) ≈ 7.1. Hmm, but that contradicts the assumption that |a_k| > 10. Wait, but in this case, if |a_k| > 10, then 2a_k² > 200, which is much larger than 101. So this is impossible. Therefore, if all other terms are zero, the maximum possible |a_k| would be sqrt(101/2) ≈ 7.1. So if someone is claiming |a_k| ≤ 10, which is a larger bound, but in the problem, the total sum could be distributed among multiple variables. Wait, but perhaps in the general case, even when other variables are non-zero, the maximum |a_k| is still bounded by 10. So maybe we need to use some inequality.Alternatively, think of each a_k in terms of the sum S. Let's fix some k and consider the variable a_k. Let’s write S = a_k + T, where T is the sum of the other 99 variables. Then, our equation becomes:sum_{i=1}^{100} a_i² + S² = 101.But sum a_i² = a_k² + sum_{i≠k} a_i². Let’s denote sum_{i≠k} a_i² = Q'. Then the equation becomes:a_k² + Q' + (a_k + T)^2 = 101.Expanding (a_k + T)^2 gives a_k² + 2a_k T + T². Therefore, substituting back:a_k² + Q' + a_k² + 2a_k T + T² = 101.Simplify:2a_k² + 2a_k T + Q' + T² = 101.But Q' + T² is the sum of the squares of the other variables plus the square of their sum. Wait, perhaps that's not directly helpful. Alternatively, note that Q' is the sum of squares of the other variables, and T is their sum.Hmm. Let me try to bound a_k. Let's suppose that a_k is positive (since we can take absolute value). So assume a_k > 10. Then we need to show that this leads to a contradiction.But how? Let's see. If a_k is large, then Q' + T² must be small. But Q' is the sum of squares, which is non-negative. So 2a_k² + 2a_k T must be less than or equal to 101. But Q' and T² are non-negative. Wait, perhaps not. Let me see:From the equation:2a_k² + 2a_k T + Q' + T² = 101.But Q' is the sum of squares of variables other than a_k, so Q' ≥ 0. Similarly, T² is the square of the sum of variables other than a_k, so also non-negative. Therefore:2a_k² + 2a_k T ≤ 101.So 2a_k² + 2a_k T ≤ 101. Now, if we can bound T in terms of a_k, maybe we can get an inequality only involving a_k.But T is the sum of the other variables. How are they related? Let's denote the variables other than a_k as b₁, b₂, ..., b₉₉. Then T = b₁ + b₂ + ... + b₉₉, and Q' = b₁² + b₂² + ... + b₉₉².By Cauchy-Schwarz, we have (b₁ + b₂ + ... + b₉₉)^2 ≤ 99(b₁² + b₂² + ... + b₉₉²). So T² ≤ 99 Q'. Therefore, Q' ≥ T² / 99.So substituting back into the equation:2a_k² + 2a_k T + Q' + T² ≤ 101.But since Q' ≥ T² / 99, then:2a_k² + 2a_k T + (T² / 99) + T² ≤ 101.Simplify:2a_k² + 2a_k T + (100/99) T² ≤ 101.Hmm, this seems complicated. Maybe another approach.Alternatively, think of variables other than a_k. Let’s denote T = sum_{i≠k} a_i, and Q' = sum_{i≠k} a_i². Then, by Cauchy-Schwarz, T² ≤ 99 Q'. So Q' ≥ T² / 99.Our equation is:a_k² + Q' + (a_k + T)^2 = 101.Which becomes:a_k² + Q' + a_k² + 2a_k T + T² = 101.So:2a_k² + 2a_k T + Q' + T² = 101.But since Q' ≥ T² / 99, we can substitute:2a_k² + 2a_k T + (T² / 99) + T² ≤ 101.Which simplifies to:2a_k² + 2a_k T + (100/99) T² ≤ 101.Hmm. Let’s denote x = a_k and y = T. Then the inequality becomes:2x² + 2xy + (100/99) y² ≤ 101.We need to find the maximum possible x given this inequality. Since we want to maximize x, perhaps we can consider this as a quadratic in y. Let's treat y as a variable and x as a constant. Then:(100/99) y² + 2x y + 2x² - 101 ≤ 0.This is a quadratic inequality in y. For real solutions y to exist, the discriminant must be non-negative. The discriminant D is:(2x)^2 - 4*(100/99)*(2x² - 101) ≥ 0.Compute D:4x² - (400/99)(2x² - 101) ≥ 0.Multiply through by 99 to eliminate denominators:4x² * 99 - 400(2x² - 101) ≥ 0Calculate:396x² - 800x² + 400*101 ≥ 0Combine like terms:(396 - 800)x² + 40400 ≥ 0-404x² + 40400 ≥ 0Multiply both sides by -1 (reversing inequality):404x² - 40400 ≤ 0Divide by 404:x² - 100 ≤ 0Thus,x² ≤ 100Therefore,|x| ≤ 10.Which is exactly what we wanted to prove. Therefore, if |a_k| > 10, the discriminant would be negative, meaning there are no real solutions for y, which contradicts the existence of such a_k. Hence, |a_k| ≤ 10 for all k.Wait, let me check that again. So treating the inequality in terms of y, we set up the quadratic in y and required that it has a real solution (since T is real). If the discriminant is negative, then there's no real y, which would mean that our original assumption (that such a_k exists) is impossible. Therefore, the discriminant must be non-negative, leading to x² ≤ 100. Hence, |x| ≤ 10. Therefore, each |a_k| is bounded by 10. That seems to work!Alternatively, maybe there's a more straightforward inequality. Let me think. Since we have sum a_i² + (sum a_i)² = 101. We can use the Cauchy-Schwarz inequality on (sum a_i)² ≤ 100 sum a_i². Then, sum a_i² + (sum a_i)^2 ≤ sum a_i² + 100 sum a_i² = 101 sum a_i². But the left-hand side is given as 101. Therefore, 101 sum a_i² ≥ 101 => sum a_i² ≥ 1. Hmm, but that gives a lower bound on the sum of squares, which might not help directly.Wait, but maybe we can relate the maximum term to the sum. Since sum a_i² ≥ a_k² for any k. So a_k² ≤ sum a_i². But sum a_i² = 101 - (sum a_i)^2. Therefore, a_k² ≤ 101 - (sum a_i)^2. But (sum a_i)^2 is non-negative, so a_k² ≤ 101. Hence, |a_k| ≤ sqrt(101) ≈ 10.05. But sqrt(101) is slightly more than 10. So this approach only gives |a_k| ≤ ~10.05, which is not sufficient. Therefore, this approach is not tight enough. The previous method with the discriminant gives exactly |a_k| ≤ 10.Therefore, the discriminant approach seems necessary here. Let me recap that:For each a_k, express the equation in terms of a_k and the sum of the other variables. Then, use Cauchy-Schwarz on the sum of the other variables to relate their sum and sum of squares. Substitute back into the equation to form a quadratic inequality in the sum of the other variables. For this inequality to have a solution, the discriminant must be non-negative, leading to a bound on a_k.Therefore, this method successfully shows that |a_k| ≤ 10. That's pretty clever. So, even though at first glance the problem might seem like it needs more advanced techniques, breaking it down with Cauchy-Schwarz and quadratic discriminant gives the required bound.Final Answerboxed{10}

✅:Given the real numbers (a_1, a_2, ldots, a_{100}) that satisfy the relationship[a_1^2 + a_2^2 + cdots + a_{100}^2 + (a_1 + a_2 + cdots + a_{100})^2 = 101,]we need to prove that (|a_k| leq 10) for all (k = 1, 2, ldots, 100).First, denote (S = a_1 + a_2 + cdots + a_{100}). The given equation can be rewritten as:[sum_{i=1}^{100} a_i^2 + S^2 = 101.]Consider a specific (a_k) and denote the sum of the other 99 variables as (T = sum_{i neq k} a_i). Let (Q' = sum_{i neq k} a_i^2). The equation becomes:[a_k^2 + Q' + (a_k + T)^2 = 101.]Expanding and simplifying, we get:[2a_k^2 + 2a_k T + Q' + T^2 = 101.]Using the Cauchy-Schwarz inequality on (T) and (Q'), we have (T^2 leq 99 Q'). Substituting (Q' geq frac{T^2}{99}) into the equation, we get:[2a_k^2 + 2a_k T + frac{T^2}{99} + T^2 leq 101.]Simplifying further:[2a_k^2 + 2a_k T + frac{100}{99} T^2 leq 101.]Treating this as a quadratic inequality in (T), we rewrite it as:[frac{100}{99} T^2 + 2a_k T + 2a_k^2 - 101 leq 0.]For this quadratic inequality to have real solutions for (T), the discriminant must be non-negative:[(2a_k)^2 - 4 left(frac{100}{99}right) (2a_k^2 - 101) geq 0.]Simplifying the discriminant:[4a_k^2 - frac{400}{99}(2a_k^2 - 101) geq 0.]Multiplying through by 99:[396a_k^2 - 800a_k^2 + 40400 geq 0,]which simplifies to:[-404a_k^2 + 40400 geq 0.]Solving for (a_k^2):[a_k^2 leq 100 implies |a_k| leq 10.]Thus, we conclude that (|a_k| leq 10) for all (k = 1, 2, ldots, 100).[boxed{10}]

🔑:1. We start with the given equation: [ a_1^2 + a_2^2 + cdots + a_{100}^2 + (a_1 + a_2 + cdots + a_{100})^2 = 101. ]2. Let ( S = a_1 + a_2 + cdots + a_{100} ). Then the equation becomes: [ a_1^2 + a_2^2 + cdots + a_{100}^2 + S^2 = 101. ]3. Without loss of generality, assume ( max{|a_1|, |a_2|, ldots, |a_{100}|} = |a_{100}| ).4. We can rewrite the sum of squares as: [ a_1^2 + a_2^2 + cdots + a_{99}^2 + a_{100}^2. ]5. Using the Cauchy-Schwarz inequality, we have: [ (a_1^2 + a_2^2 + cdots + a_{99}^2)(1^2 + 1^2 + cdots + 1^2) geq (a_1 + a_2 + cdots + a_{99})^2. ] Since there are 99 terms, this simplifies to: [ (a_1^2 + a_2^2 + cdots + a_{99}^2) cdot 99 geq (a_1 + a_2 + cdots + a_{99})^2. ] Let ( T = a_1 + a_2 + cdots + a_{99} ). Then: [ a_1^2 + a_2^2 + cdots + a_{99}^2 geq frac{T^2}{99}. ]6. Substitute this back into the original equation: [ a_1^2 + a_2^2 + cdots + a_{99}^2 + a_{100}^2 + S^2 = 101. ] Using ( S = T + a_{100} ), we get: [ a_1^2 + a_2^2 + cdots + a_{99}^2 + a_{100}^2 + (T + a_{100})^2 = 101. ]7. Substitute the inequality from step 5: [ frac{T^2}{99} + a_{100}^2 + (T + a_{100})^2 = 101. ]8. Expand and simplify: [ frac{T^2}{99} + a_{100}^2 + T^2 + 2Ta_{100} + a_{100}^2 = 101. ] Combine like terms: [ frac{T^2}{99} + T^2 + 2a_{100}T + 2a_{100}^2 = 101. ]9. Let ( x = a_{100} ). Then: [ frac{T^2}{99} + T^2 + 2Tx + 2x^2 = 101. ]10. Combine the terms involving ( T^2 ): [ left(1 + frac{1}{99}right)T^2 + 2Tx + 2x^2 = 101. ] Simplify the coefficient of ( T^2 ): [ frac{100}{99}T^2 + 2Tx + 2x^2 = 101. ]11. This is a quadratic inequality in ( T ): [ frac{100}{99}T^2 + 2Tx + 2x^2 - 101 leq 0. ]12. The discriminant of this quadratic must be non-negative for real solutions to exist: [ Delta = 4x^2 - frac{400}{99}(2x^2 - 101) geq 0. ]13. Simplify the discriminant: [ 4x^2 - frac{800x^2}{99} + frac{40400}{99} geq 0. ] [ 4x^2 - frac{800x^2}{99} + frac{40400}{99} geq 0. ] [ frac{396x^2}{99} + frac{40400}{99} geq 0. ] [ 396x^2 + 40400 geq 0. ] [ 396x^2 leq 40400. ] [ x^2 leq frac{40400}{396}. ] [ x^2 leq 101. ] [ |x| leq sqrt{101}. ]14. Since ( sqrt{101} approx 10 ), we have: [ |a_{100}| leq 10. ]15. Therefore, for all ( k = 1, 2, ldots, 100 ): [ |a_k| leq 10. ]The final answer is ( boxed{ |a_k| leq 10 } ) for all ( k = 1, 2, ldots, 100 ).

🧠:Okay, let's see. So the problem is about Tony tossing 40 nickels, and for each one that lands in the jar, he gets 5 points. If it misses, he loses 2 points. After all 40 tosses, his total score is 88. We need to find the greatest number of nickels he could have successfully tossed into the jar. Hmm, so the question is asking for the maximum possible number of successes (nickels in the jar) that could result in a score of 88, right?First, let me parse the problem step by step. He has 40 nickels, so that's 40 tosses. Each success gives +5, each failure gives -2. Total score is 88. We need to find the maximum number of successes possible. So, intuitively, to maximize the number of successes, we want as many successes as possible, but the total score has to be 88. So if he had all 40 successes, his score would be 40*5 = 200, which is way higher than 88. So obviously, some of those must have been failures. But each failure not only subtracts 2 points but also means that instead of a +5 from a success, he gets -2, so the net difference is -7 per failure compared to a success. Hmm, maybe that's a way to think about it.Alternatively, maybe set up equations. Let me try that. Let x be the number of successes, y the number of failures. Then we know that x + y = 40, because he tossed all 40 nickels. Then the total score is 5x - 2y = 88. So we have two equations:1. x + y = 402. 5x - 2y = 88We can solve this system of equations to find x and y. Let me do that first. From the first equation, y = 40 - x. Substitute into the second equation:5x - 2(40 - x) = 88Simplify:5x - 80 + 2x = 88Combine like terms:7x - 80 = 88Add 80 to both sides:7x = 168Divide by 7:x = 24So according to this, x is 24. Then y would be 16. So he made 24 successes and 16 failures. Then total points would be 24*5 - 16*2 = 120 - 32 = 88, which matches.But wait, the problem says "Find the greatest number of nickels he could have successfully tossed into the jar." So according to this, the answer is 24. But why is the question asking for the greatest number? Because if there's a unique solution, then 24 is the only possible number. But maybe there are multiple solutions? But with two equations and two variables, there's a unique solution. So maybe 24 is the only possible number. Therefore, the maximum is 24.Wait, but that seems contradictory. If the answer is uniquely determined, then the maximum is 24. So why does the question ask for the greatest number? Maybe I'm missing something here. Let me think again.Wait, perhaps there's a different way where you can have more successes but some of them are canceled by more failures. But how? Let me check.Wait, suppose he has x successes and (40 - x) failures. The total score is 5x - 2(40 - x) = 5x - 80 + 2x = 7x - 80. So 7x - 80 = 88, so 7x = 168, so x=24. So this is unique. Therefore, there's only one possible value for x, which is 24. Hence, the maximum number of successes is 24. So the answer should be 24. But why does the question say "the greatest number of nickels he could have successfully tossed"? Is there a scenario where there could be multiple numbers of successes leading to the same score, and we need the maximum possible? But according to the equations, no. So maybe 24 is the only possible number, hence the maximum is 24.Wait, but maybe there's another interpretation. For example, if some of the nickels neither land in the jar nor hit the ground? But the problem states that he tosses them one at a time toward his desk where his change jar sits. He awards himself 5 points for each that lands in the jar, and takes away 2 points for each that hits the ground. So all tosses must result in either a jar landing or hitting the ground. There's no third possibility. So all 40 nickels are either successes or failures, so x + y = 40. Therefore, the equations are correct. Then x must be 24, so 24 is the only possible number. Therefore, the answer is 24.But the problem says "the greatest number of nickels he could have successfully tossed into the jar". So maybe the answer is 24. But the problem seems to be tricking me here. Wait, let me check again.Wait, maybe there's a different way to model the problem. Let me think. Suppose that for each nickel, he either gets 5 points or loses 2 points. So total points is 5x - 2y, and x + y = 40. Then solving gives x=24. So that's unique. Hence, 24 is the only possible number. Therefore, the maximum is 24.Alternatively, maybe the problem is in the interpretation of "the greatest number of nickels he could have successfully tossed into the jar" given that the score is 88. So maybe there are multiple ways to get 88 points with different numbers of successes and failures? For example, if there were some other way to get 88 points with more successes but compensated by something else. But given the way the problem is stated, each toss is either a success (5 points) or a failure (-2 points). So the total score is 5x - 2y, and since x + y =40, the equations are fixed. So solving gives x=24. Hence, there's only one solution.Wait, unless there are some constraints that I'm missing. For example, the number of nickels tossed is 40, but maybe some nickels could be partially in or something? No, the problem says he either lands in the jar or hits the ground. So each nickel is either a success or a failure.Wait, let me check the arithmetic again. If x=24, then y=16. Then 24*5=120, 16*2=32. 120 -32=88. That's correct. If he had more than 24 successes, say 25. Then y=15. Then total score would be 25*5 -15*2=125 -30=95, which is more than 88. If he had 23 successes, y=17, 23*5 -17*2=115 -34=81, which is less than 88. So the score increases by 7 for each additional success (since each success gives +5 and reduces failures by 1, which would have taken away -2, so net change is +5 +2 = +7). Therefore, the score is 7x -80. So for each x, 7x -80. So to get 88, x must be (88 +80)/7=168/7=24. Hence, unique solution.Therefore, the answer is 24. So why does the question ask for the greatest number? Maybe there's a trick here. Wait, maybe if some of the failures could be negative? No, the number of failures can't be negative. Similarly, x and y must be non-negative integers. So x=24 is the only solution.Alternatively, maybe the problem is in the interpretation of the scoring. For example, does each nickel that hits the ground take away 2 points from the total, or does it subtract 2 points per nickel in addition to not getting the 5 points? The problem says "awards himself 5 points for each nickel that lands in the jar, and takes away 2 points from his score for each nickel that hits the ground." So for each success, +5. For each failure, -2. So total score is 5x -2y. Which is how we set it up. So that is correct.Therefore, the answer is 24. So maybe the problem is straightforward, but the initial thought that there might be a different answer is just a red herring. Let me just check once more.Suppose x=24, y=16. 24+16=40. 24*5=120, 16*2=32, 120-32=88. Correct. If x=25, y=15: 25*5=125, 15*2=30, 125-30=95≠88. x=23, y=17: 23*5=115, 17*2=34, 115-34=81≠88. So indeed, 24 is the only solution. Therefore, the greatest number of nickels he could have successfully tossed is 24.Hence, the answer is 24. So maybe the question is trying to test if people might think that there's a different maximum, but in reality, the equations pin it down to exactly 24. Therefore, 24 is the answer.Final AnswerThe greatest number of nickels Tony could have successfully tossed into the jar is boxed{24}.

✅:Tony tosses 40 nickels, earning 5 points for each success and losing 2 points for each failure. His total score is 88. We need to find the greatest number of nickels he could have successfully tossed into the jar.Let ( x ) be the number of successes and ( y ) be the number of failures. We have the following equations:1. ( x + y = 40 )2. ( 5x - 2y = 88 )From the first equation, we get ( y = 40 - x ). Substituting this into the second equation:[5x - 2(40 - x) = 88]Simplifying:[5x - 80 + 2x = 88][7x - 80 = 88][7x = 168][x = 24]Thus, the number of successes ( x ) is 24, and the number of failures ( y ) is ( 40 - 24 = 16 ). Verifying the score:[24 times 5 - 16 times 2 = 120 - 32 = 88]The equations confirm that 24 is the unique solution. Therefore, the greatest number of nickels Tony could have successfully tossed into the jar is (boxed{24}).

🔑:1. Let ( x ) be the number of successful tosses (nickels that land in the jar).2. Let ( y ) be the number of failed tosses (nickels that hit the ground).3. We know that the total number of nickels is 40, so we have the equation: [ x + y = 40 ]4. Tony awards himself 5 points for each successful toss and deducts 2 points for each failed toss. Therefore, the total score can be expressed as: [ 5x - 2y = 88 ]5. We now have a system of linear equations: [ begin{cases} x + y = 40 5x - 2y = 88 end{cases} ]6. To eliminate ( y ), we can multiply the first equation by 2: [ 2x + 2y = 80 ]7. Adding this to the second equation: [ 5x - 2y + 2x + 2y = 88 + 80 ] Simplifying, we get: [ 7x = 168 ]8. Solving for ( x ): [ x = frac{168}{7} = 24 ]9. Substituting ( x = 24 ) back into the first equation to find ( y ): [ 24 + y = 40 implies y = 40 - 24 = 16 ]Thus, the greatest number of nickels Tony could have successfully tossed into the jar is ( boxed{24} ).

🧠:Okay, so I need to prove that the limit of the sequence defined by u₁ = a, u₂ = b, and u_{n+1} = (u_n + u_{n-1})/2 for n ≥ 2 exists, and then find its value in terms of a and b. Hmm, let's start by understanding what this sequence does. First, the recursive formula is u_{n+1} = (u_n + u_{n-1})/2. That means each term is the average of the two previous terms. So, starting from a and b, each subsequent term is kind of smoothing out the difference between the previous two. Maybe the sequence is converging to some value between a and b? But I need to be precise here.I remember that for sequences defined by linear recursions, we can sometimes find a closed-form expression using characteristic equations. Maybe that's the way to go here. Let me recall how that works. If we have a linear recurrence relation, we can assume a solution of the form u_n = r^n and substitute it into the recurrence to find the characteristic equation. But this recurrence is second-order, right? Because u_{n+1} depends on u_n and u_{n-1}, which are two terms back. So the general form for a second-order linear recurrence is u_{n+1} = c*u_n + d*u_{n-1}. In this case, c = 1/2 and d = 1/2. So, let's set up the characteristic equation. Assume a solution u_n = r^n. Then substituting into the recurrence:u_{n+1} = (u_n + u_{n-1})/2r^{n+1} = (r^n + r^{n-1})/2Divide both sides by r^{n-1}:r^2 = (r + 1)/2Multiply both sides by 2:2r^2 = r + 1Bring all terms to one side:2r^2 - r - 1 = 0Now, solving this quadratic equation for r. The quadratic formula gives:r = [1 ± sqrt(1 + 8)] / (2*2) = [1 ± 3]/4So, the roots are (1 + 3)/4 = 4/4 = 1 and (1 - 3)/4 = -2/4 = -1/2. Therefore, the general solution to the recurrence is u_n = A*(1)^n + B*(-1/2)^n, where A and B are constants determined by the initial conditions. So, u_n = A + B*(-1/2)^n. Now, applying the initial conditions. For n = 1: u₁ = a = A + B*(-1/2)^1 = A - B/2.For n = 2: u₂ = b = A + B*(-1/2)^2 = A + B*(1/4).So we have the system of equations:1) A - B/2 = a2) A + B/4 = bLet's solve this system. Subtract equation 1 from equation 2:(A + B/4) - (A - B/2) = b - aA + B/4 - A + B/2 = b - a( B/4 + B/2 ) = b - aConvert to like terms: B/4 + 2B/4 = 3B/4 = b - aTherefore, 3B/4 = b - a => B = (4/3)(b - a)Now substitute back into equation 1:A - ( (4/3)(b - a) ) / 2 = aA - (2/3)(b - a) = aA = a + (2/3)(b - a) = a + (2b/3 - 2a/3) = (3a/3 - 2a/3) + 2b/3 = (a/3 + 2b/3)So, A = (a + 2b)/3Therefore, the general term is u_n = (a + 2b)/3 + (4/3)(b - a)*(-1/2)^nSo, u_n = (a + 2b)/3 + (4(b - a)/3)*(-1/2)^nNow, to find the limit as n approaches infinity, we can analyze the expression. The term (-1/2)^n as n approaches infinity will go to zero because | -1/2 | < 1. Therefore, the limit of u_n as n approaches infinity is just (a + 2b)/3.Therefore, the limit exists and is equal to (a + 2b)/3.Wait, let me verify this with another approach just to be sure. Maybe by considering the difference between consecutive terms and seeing if the sequence is a Cauchy sequence or something.Alternatively, maybe look at the behavior of the sequence. Let's compute the first few terms manually.Given u₁ = a, u₂ = b.u₃ = (u₂ + u₁)/2 = (b + a)/2u₄ = (u₃ + u₂)/2 = ( (b + a)/2 + b ) / 2 = ( (b + a + 2b ) / 2 ) / 2 = (a + 3b)/4u₅ = (u₄ + u₃)/2 = ( (a + 3b)/4 + (a + b)/2 ) / 2 = ( (a + 3b + 2a + 2b ) / 4 ) / 2 = (3a + 5b)/8 / 2 = Wait, no. Wait:Wait, first compute numerator:(a + 3b)/4 + (a + b)/2 = (a + 3b)/4 + 2(a + b)/4 = (a + 3b + 2a + 2b)/4 = (3a + 5b)/4Then divide by 2: (3a + 5b)/8So u₅ = (3a + 5b)/8Similarly, u₆ = (u₅ + u₄)/2 = ( (3a + 5b)/8 + (a + 3b)/4 ) / 2 = ( (3a + 5b + 2a + 6b ) / 8 ) / 2 = (5a + 11b)/16 / 2 = (5a + 11b)/32?Wait, hold on. Wait, let's compute u₆:u₄ is (a + 3b)/4, u₅ is (3a + 5b)/8.So u₆ = (u₅ + u₄)/2 = [ (3a + 5b)/8 + (a + 3b)/4 ] / 2Convert to common denominator:= [ (3a + 5b)/8 + 2(a + 3b)/8 ] / 2= [ (3a + 5b + 2a + 6b)/8 ] / 2= (5a + 11b)/8 / 2= (5a + 11b)/16Hmm, so u₆ = (5a + 11b)/16Looking at the coefficients of a and b:u₁: 1a + 0bu₂: 0a + 1bu₃: (1a + 1b)/2u₄: (1a + 3b)/4u₅: (3a + 5b)/8u₆: (5a + 11b)/16u₇: (11a + 21b)/32?Wait, let's check u₇:u₇ = (u₆ + u₅)/2 = [ (5a + 11b)/16 + (3a + 5b)/8 ] / 2Convert to common denominator:= [ (5a + 11b)/16 + (6a + 10b)/16 ] /2= [ (11a + 21b)/16 ] /2 = (11a + 21b)/32So the coefficients of a and b in each term seem to follow a pattern. Let's list them:n | coefficient of a | coefficient of b1 | 1 | 02 | 0 | 13 | 1/2 | 1/24 | 1/4 | 3/45 | 3/8 | 5/86 | 5/16 | 11/167 | 11/32 | 21/32Looking at the coefficients for a: 1, 0, 1/2, 1/4, 3/8, 5/16, 11/32...Similarly for b: 0, 1, 1/2, 3/4, 5/8, 11/16, 21/32...These coefficients look like they might be following a pattern where each coefficient is (previous coefficient + something). Let's see:For the a coefficients:Term 1: 1Term 3: 1/2 = (Term1 + Term2)/2 = (1 + 0)/2 = 1/2Term 4: 1/4 = (Term3 + Term2)/2 = (1/2 + 0)/2 = 1/4Term 5: 3/8 = (Term4 + Term3)/2 = (1/4 + 1/2)/2 = (3/4)/2 = 3/8Term 6: 5/16 = (Term5 + Term4)/2 = (3/8 + 1/4)/2 = (5/8)/2 = 5/16Term 7: 11/32 = (Term6 + Term5)/2 = (5/16 + 3/8)/2 = (11/16)/2 = 11/32Similarly for the b coefficients:Term 2: 1Term3: 1/2 = (Term2 + Term1)/2 = (1 + 0)/2 = 1/2Term4: 3/4 = (Term3 + Term2)/2 = (1/2 + 1)/2 = 3/4Term5: 5/8 = (Term4 + Term3)/2 = (3/4 + 1/2)/2 = 5/8Term6: 11/16 = (Term5 + Term4)/2 = (5/8 + 3/4)/2 = 11/16Term7: 21/32 = (Term6 + Term5)/2 = (11/16 + 5/8)/2 = 21/32So, the coefficients themselves follow the same recurrence relation as the original sequence. Interesting.But how does this help? Well, maybe if we can find a closed-form for the coefficients, we can then take the limit. However, we already have the closed-form from the characteristic equation approach. So, perhaps the coefficients of a and b in u_n are:For a: [ ( -1/2 )^{n} * something + something ] ?Wait, since the general solution is u_n = A + B*(-1/2)^n, where A = (a + 2b)/3 and B = (4/3)(b - a). Therefore, u_n can be written as:u_n = (a + 2b)/3 + (4/3)(b - a)*(-1/2)^nTherefore, the coefficient of a in u_n is [1/3 + (4/3)(-1/2)^n*(-1)] ?Wait, let's expand the expression:u_n = (a + 2b)/3 + (4/3)(b - a)*(-1/2)^n= (a)/3 + (2b)/3 + (4b/3)(-1/2)^n - (4a/3)(-1/2)^n= a[1/3 - (4/3)(-1/2)^n] + b[2/3 + (4/3)(-1/2)^n]So, coefficient of a is [1/3 - (4/3)(-1/2)^n], coefficient of b is [2/3 + (4/3)(-1/2)^n]Therefore, as n approaches infinity, the terms involving (-1/2)^n go to zero, so the coefficient of a tends to 1/3 and coefficient of b tends to 2/3. Therefore, the limit is (1/3)a + (2/3)b = (a + 2b)/3, which matches our previous result.Therefore, this confirms that the limit is indeed (a + 2b)/3.Alternatively, another approach: since the recurrence is linear and the coefficients form a converging sequence, maybe we can use the concept of a contracting sequence? The recursive formula averages the previous two terms, which might bring the terms closer together each time, leading to convergence.Suppose the limit L exists. Then, taking the limit on both sides of the recurrence:lim_{n→∞} u_{n+1} = lim_{n→∞} (u_n + u_{n-1})/2Assuming the limit L exists, then L = (L + L)/2 = (2L)/2 = L. So, the equation is satisfied for any L, which doesn't help directly. But this tells us that if the limit exists, the equation doesn't contradict it, but we need another way to find L.Alternatively, if the sequence converges, then the limit must satisfy L = (L + L)/2, which is trivial. So, we need another method to find L.But perhaps we can consider the difference between consecutive terms. Let’s define d_n = u_n - u_{n-1}. Then, maybe we can find a relation for d_n.Compute d_{n+1} = u_{n+1} - u_n = (u_n + u_{n-1})/2 - u_n = (-u_n + u_{n-1})/2 = (-1/2)(u_n - u_{n-1}) = (-1/2)d_nSo, d_{n+1} = (-1/2)d_n. Therefore, the differences form a geometric sequence with ratio -1/2.Therefore, d_n = d_2*(-1/2)^{n-2}Since d_2 = u_2 - u_1 = b - a.Therefore, d_n = (b - a)*(-1/2)^{n-2}Then, the difference between terms is decreasing in absolute value by a factor of 1/2 each time and alternating in sign. Since | -1/2 | < 1, the differences go to zero as n approaches infinity. Therefore, the sequence is Cauchy, hence convergent.Now, to find the limit L, we can express u_n as u_1 + sum_{k=2}^n d_k. Wait, because u_n = u_1 + sum_{k=2}^n (u_k - u_{k-1}) ) = a + sum_{k=2}^n d_k.So, as n approaches infinity, u_n = a + sum_{k=2}^∞ d_k.Compute the sum: sum_{k=2}^∞ d_k = sum_{k=2}^∞ (b - a)*(-1/2)^{k-2}Let’s make substitution m = k - 2, so when k=2, m=0. Therefore, sum_{m=0}^∞ (b - a)*(-1/2)^mThis is a geometric series with first term (b - a) and ratio (-1/2). The sum is (b - a)/(1 - (-1/2)) ) = (b - a)/(3/2) = 2(b - a)/3Therefore, the limit L = a + 2(b - a)/3 = (3a + 2b - 2a)/3 = (a + 2b)/3, which again gives the same result.Therefore, regardless of the method used—whether solving the recurrence relation via characteristic equations, analyzing the coefficients, or considering the differences and summing the series—the limit is (a + 2b)/3.Just to be thorough, let's test with specific numbers. Let’s pick a = 0 and b = 3. Then the limit should be (0 + 2*3)/3 = 2.Compute the terms:u1 = 0, u2 = 3u3 = (3 + 0)/2 = 1.5u4 = (1.5 + 3)/2 = 2.25u5 = (2.25 + 1.5)/2 = 1.875u6 = (1.875 + 2.25)/2 = 2.0625u7 = (2.0625 + 1.875)/2 = 1.96875u8 = (1.96875 + 2.0625)/2 = 2.015625u9 = (2.015625 + 1.96875)/2 = 1.9921875u10 = (1.9921875 + 2.015625)/2 = 2.00390625So, oscillating around 2, getting closer each time. The terms are approaching 2, which is (0 + 2*3)/3 = 2. That checks out.Another example: a = 1, b = 4. The limit should be (1 + 8)/3 = 9/3 = 3.Compute terms:u1=1, u2=4u3=(4+1)/2=2.5u4=(2.5 + 4)/2=3.25u5=(3.25 + 2.5)/2=2.875u6=(2.875 + 3.25)/2=3.0625u7=(3.0625 + 2.875)/2=2.96875u8=(2.96875 + 3.0625)/2=3.015625u9=(3.015625 + 2.96875)/2=2.9921875u10=(2.9921875 + 3.015625)/2=3.00390625Again, converging to 3. So, the examples support the result.Therefore, after multiple approaches and verification, it's solid that the limit exists and is (a + 2b)/3.Final AnswerThe limit is boxed{dfrac{a + 2b}{3}}.

✅:To prove that the limit of the sequence defined by ( u_1 = a ), ( u_2 = b ), and ( u_{n+1} = frac{u_n + u_{n-1}}{2} ) for ( n ge 2 ) exists and to find its value in terms of ( a ) and ( b ), we can use the following steps:1. Characteristic Equation Approach: - Assume a solution of the form ( u_n = r^n ). Substituting this into the recurrence relation ( u_{n+1} = frac{u_n + u_{n-1}}{2} ) gives the characteristic equation: [ r^{n+1} = frac{r^n + r^{n-1}}{2} ] - Simplifying, we get the quadratic equation: [ 2r^2 - r - 1 = 0 ] - Solving this quadratic equation using the quadratic formula, we find the roots ( r = 1 ) and ( r = -frac{1}{2} ).2. General Solution: - The general solution to the recurrence relation is: [ u_n = A cdot 1^n + B left( -frac{1}{2} right)^n = A + B left( -frac{1}{2} right)^n ] - Using the initial conditions ( u_1 = a ) and ( u_2 = b ), we set up the system of equations: [ A - frac{B}{2} = a ] [ A + frac{B}{4} = b ] - Solving this system, we find ( A = frac{a + 2b}{3} ) and ( B = frac{4(b - a)}{3} ).3. Limit Analysis: - The general term of the sequence is: [ u_n = frac{a + 2b}{3} + frac{4(b - a)}{3} left( -frac{1}{2} right)^n ] - As ( n to infty ), the term ( left( -frac{1}{2} right)^n ) approaches 0 because ( left| -frac{1}{2} right| < 1 ). Therefore, the limit of ( u_n ) is: [ lim_{n to infty} u_n = frac{a + 2b}{3} ]4. Verification with Differences: - Defining the difference ( d_n = u_n - u_{n-1} ), we find ( d_{n+1} = -frac{1}{2} d_n ), forming a geometric sequence with ratio ( -frac{1}{2} ). The sum of this geometric series confirms the limit.5. Examples and Testing: - Specific examples with different values of ( a ) and ( b ) confirm the limit calculation, showing convergence to ( frac{a + 2b}{3} ).Thus, the limit exists and is given by:[boxed{dfrac{a + 2b}{3}}]

🔑:1. We start with the given sequence defined by ( u_1 = a ), ( u_2 = b ), and ( u_{n+1} = frac{u_n + u_{n-1}}{2} ) for ( n geq 2 ).2. To find the limit of the sequence as ( n to infty ), we assume that the sequence converges to some limit ( L ). If ( lim_{n to infty} u_n = L ), then ( lim_{n to infty} u_{n+1} = L ) and ( lim_{n to infty} u_{n-1} = L ).3. Using the recurrence relation ( u_{n+1} = frac{u_n + u_{n-1}}{2} ), we take the limit on both sides as ( n to infty ): [ L = frac{L + L}{2} ] Simplifying this, we get: [ L = frac{2L}{2} implies L = L ] This confirms that ( L ) is consistent with the recurrence relation.4. Next, we need to find the explicit value of ( L ). We observe the behavior of the sequence: [ u_3 = frac{u_2 + u_1}{2} = frac{b + a}{2} ] [ u_4 = frac{u_3 + u_2}{2} = frac{frac{b + a}{2} + b}{2} = frac{a + 3b}{4} ] [ u_5 = frac{u_4 + u_3}{2} = frac{frac{a + 3b}{4} + frac{a + b}{2}}{2} = frac{a + 3b + 2a + 2b}{8} = frac{3a + 5b}{8} ]5. We notice that the sequence is approaching a value that is a weighted average of ( a ) and ( b ). To find the exact limit, we consider the infinite series approach: [ u_{n+1} = frac{u_n + u_{n-1}}{2} ] This can be interpreted as the sequence taking the midpoint of the previous two terms, which suggests a convergence to a stable value.6. To find the exact limit, we use the fact that the sequence is a linear combination of ( a ) and ( b ). We can express the limit as: [ L = lim_{n to infty} u_n = lim_{n to infty} left( a + (b - a) left( 1 - frac{1}{2} + frac{1}{4} - frac{1}{8} + cdots right) right) ]7. The series inside the parentheses is a geometric series with the first term ( 1 ) and common ratio ( -frac{1}{2} ): [ sum_{k=0}^{infty} left( -frac{1}{2} right)^k = frac{1}{1 - (-frac{1}{2})} = frac{1}{1 + frac{1}{2}} = frac{1}{frac{3}{2}} = frac{2}{3} ]8. Therefore, the limit ( L ) is: [ L = a + (b - a) cdot frac{2}{3} = a + frac{2(b - a)}{3} = a + frac{2b - 2a}{3} = frac{3a + 2b - 2a}{3} = frac{a + 2b}{3} ]The final answer is ( boxed{frac{a + 2b}{3}} )

🧠:Okay, so I need to figure out the greatest value of k for which Alex the T-rex can place positive integers on an infinite chessboard such that no two numbers are equal, and each number m on a square C requires that at least k of the orthogonally adjacent squares (up, down, left, right) have multiples of m. Hmm, let me start by breaking down the problem.First, the chessboard is infinite, so we don't have to worry about edges or boundaries. Each square must have a unique positive integer. The key condition here is that for any number m placed on a square, at least k of its four neighbors must be multiples of m. We need to find the maximum possible k where such an arrangement exists.Let me think about possible values of k. Since each square has four neighbors, k can be 0, 1, 2, 3, or 4. Obviously, k=4 is impossible because every neighbor would need to be a multiple of m, but since all numbers are unique, you can't have four different multiples of m around a square. Wait, but maybe if m is 1? Because every number is a multiple of 1. But hold on, if m is 1, then all four neighbors must be multiples of 1, which they already are. However, the problem states that no two numbers are equal. If we place 1 on a square, then all four adjacent squares need to have multiples of 1 (which is any number), but since numbers must be unique, we can have different numbers. Wait, but m=1 doesn't impose any restrictions except that the neighbors can't be equal to 1 again. But actually, the condition is about the neighbors being multiples of m. So if m=1, then all neighbors are automatically multiples of 1, so that condition is trivially satisfied. But since all numbers must be unique, the neighbors just need to be different from 1. But actually, even if you have 1 on a square, the neighbors can be any numbers as long as they are unique. So maybe using 1 cleverly can help? But 1 can be placed anywhere, but each number must be unique. So only one 1 can exist on the entire chessboard. Wait, no. Wait, the problem states that each square has a positive integer, and no two squares have equal numbers. So each number is unique. Therefore, 1 can only appear once. So if we place 1 on a square, then its four neighbors can be any numbers except 1, but they must be multiples of 1, which they are. Therefore, the 1 doesn't impose any constraints except that 1 can't be adjacent to itself. But since 1 is only placed once, that's not a problem. However, the main challenge is for other numbers. For example, if we place a 2 on a square, then at least k of its neighbors must be multiples of 2. But since all numbers are unique, we can't repeat 2, but we can have 4, 6, 8, etc. So maybe arranging numbers such that each number's multiples are appropriately placed around them. But how?Let me consider the possible maximum k. If k=4, as I thought before, that would require all four neighbors of a square to be multiples of the number on that square. But if the number on the square is m, then all four neighbors have to be distinct multiples of m. However, since the numbers must be unique, each multiple of m can only be used once. But since the chessboard is infinite, maybe we can have an infinite number of multiples? Wait, but each square is assigned a unique number. So for any m, you can have multiples of m assigned to different squares. However, if every time you place an m, you need four unique multiples of m around it. Then, when you place those multiples, each of those multiples would in turn require their own neighbors to satisfy their k conditions. This seems very restrictive, especially if k=4. Let's try to see if k=4 is possible.Suppose we start with a square labeled m. Then all four neighbors must be multiples of m. Let's pick m=1. Then neighbors can be any numbers, but they have to be unique. So if we place 1 somewhere, the four neighbors can be 2,3,4,5. Then, each of those numbers (2,3,4,5) must have at least k=4 multiples in their neighbors. But 2, for example, would need all four of its neighbors to be multiples of 2. However, the square adjacent to 2 that was previously 1 is already 1, which is not a multiple of 2. Therefore, this would violate the condition for 2. Hence, k=4 is impossible even with m=1. Because the number 2 would have a neighbor (1) that isn't a multiple of 2, but k=4 would require all four neighbors to be multiples. Therefore, k=4 is impossible.Therefore, the maximum possible k is less than 4. Let's check k=3. Maybe we can arrange numbers such that each number has at least three multiples around it. But again, similar issues might arise. Let's try to see.Suppose we start with m=1. Place 1 in the center. Then, we need three of its neighbors to be multiples of 1, which is trivial. Wait, but actually, since k=3, we need at least three neighbors to be multiples. But all numbers are multiples of 1, so actually, all four neighbors can be any numbers, but since they must be unique, but the 1 just needs three of them to be multiples of 1, which they all are. But the problem is when we place the neighbors. For example, place 1 at (0,0). Then, the neighbors can be 2,3,4,5. Now, take square (1,0) which is 2. For 2, we need at least three of its neighbors to be multiples of 2. The neighbors of 2 are (0,0)=1, (2,0), (1,1), (1,-1). The square (0,0)=1 is not a multiple of 2, so already one neighbor isn't a multiple. Therefore, the other three neighbors (2,0), (1,1), (1,-1) must all be multiples of 2. But (2,0) is adjacent to 2 and also to (2,1) and (3,0), etc. So we have to assign multiples of 2 to (2,0), (1,1), (1,-1). Let's assign 6, 8, 10. Wait, but each number must be unique. So 6,8,10 are unique. Then, (2,0) is 6. Now, 6 needs at least three of its neighbors to be multiples of 6. The neighbors of 6 (at (2,0)) are 2 (at (1,0)), (3,0), (2,1), (2,-1). The square (1,0)=2 is not a multiple of 6, so already one neighbor is invalid. Therefore, the other three neighbors must be multiples of 6: (3,0), (2,1), (2,-1). Assign 12, 18, 24. Then, each of those numbers (12,18,24) must in turn have three multiples around them. This is getting unwieldy, and we can see that each step requires assigning three new multiples, which may not be feasible indefinitely without conflicts. Moreover, numbers have to be unique across the entire infinite board. Although the board is infinite, the numbers are positive integers, so each number can only be used once. But since we have an infinite number of squares, we need an infinite supply of numbers. However, each number must be a multiple of previous numbers, which may require an exponentially increasing set of numbers. But even if that's possible, the problem is that each time we place a multiple, that multiple's own neighbors have to satisfy the condition. It's a chain reaction that might not be sustainable. For instance, if we place 12 at (3,0), then 12 needs three of its neighbors to be multiples of 12. The neighbors are 6 (at (2,0)), (4,0), (3,1), (3,-1). The 6 is not a multiple of 12, so again, the other three neighbors must be multiples of 12. Assign 24, 36, 48. But 24 was already used at (2,-1). Wait, no, we assigned 24 to (2,-1)? Wait, no, (2,-1) was assigned 24? Wait, actually, earlier, we assigned 6 at (2,0), and then its neighbors (3,0), (2,1), (2,-1) were assigned 12, 18, 24. So (3,0)=12, (2,1)=18, (2,-1)=24. Then, when we go to 12 at (3,0), its neighbors are (2,0)=6, (4,0), (3,1), (3,-1). So (4,0), (3,1), (3,-1) need to be multiples of 12. Assign 36, 48, 60. But each of those will then require three multiples themselves. This seems like an endless chain where each number requires three larger multiples. Since numbers are unique and positive integers, this could be possible in theory, but practically, arranging them without overlap or contradiction is challenging. However, since the chessboard is infinite, maybe this is possible? But we also need to ensure that all numbers across the entire board are unique. For example, 24 is placed at (2,-1), then later when we need to assign multiples of 12, we need to use 24 again, but we can't. Therefore, each multiple must be unique. Therefore, we need an infinite number of multiples for each number. But for each m, there are infinitely many multiples (2m, 3m, 4m, ...), so in theory, we could assign distinct multiples each time. But when building this structure, each time we place a multiple, it needs to have its own set of multiples. This seems like a fractal-like structure where each node branches into three new nodes, each requiring a unique multiple. Since the chessboard is infinite, maybe such a construction is possible. But is there a way to tile the chessboard in such a way that this works? Alternatively, maybe there's a periodic pattern or coloring that can be used.Alternatively, maybe using prime numbers and arranging them such that their multiples are placed appropriately. For example, primes would require their neighbors to have multiples of themselves, but primes can't be adjacent to their multiples except in specific ways. However, primes are only divisible by 1 and themselves, so if a prime p is placed on a square, then at least k of its neighbors must be multiples of p. Since multiples of p are p, 2p, 3p, etc. But since numbers must be unique, we can only use each multiple once. So for each prime p, we need to place multiples of p in k of its neighbors. However, primes are infinite, so each prime would require their own set of multiples. This seems complex.Alternatively, maybe using a spiral pattern where each number is placed in a spiral and each number's multiples are placed in specific positions. But ensuring that each number's neighbors include enough multiples is tricky.Wait, maybe another approach. Let's consider the problem as a graph where each square is a node connected to its four neighbors. We need to assign a unique positive integer to each node such that for every node, at least k of its neighbors are multiples of its number.What is the maximum k possible? For k=3, as we tried earlier, the problem seems challenging because every number requires three multiples around it, leading to an exponential growth in the required multiples, but maybe on an infinite board, it's feasible. However, we have to ensure that when we assign a multiple to a neighbor, it doesn't conflict with another number's requirements. For example, if two different numbers require the same square to be their multiple, but the square can only have one number, which can't be a multiple of both unless one is a multiple of the other. But since all numbers are unique, if two numbers m and n are such that m divides n, then n can be a multiple of m. So if a square is assigned n, which is a multiple of m, then it can satisfy the condition for m if m is adjacent to it. But the problem is that each square has only one number, so it can be a multiple of multiple numbers, but each adjacent number would require that. For example, if square C has number 6, then any adjacent square with number 2 or 3 would have 6 as a multiple, satisfying their conditions if needed. However, square C=6 would itself need that at least k of its neighbors are multiples of 6. Therefore, it's possible for a number to satisfy multiple conditions for its neighbors, but each number's own condition must be met independently.This seems complex but maybe manageable with careful planning. However, constructing such an arrangement is non-trivial. Maybe another way to think about it is through graph coloring, but instead of colors, we have numbers with divisibility conditions. Alternatively, perhaps using layers where each layer is a multiple of the previous one. But how?Alternatively, consider that each number must have k multiples around it. To maximize k, we need numbers that have many small multiples, but since numbers are unique, each multiple can only be used once. Therefore, numbers with many small multiples (like 1) are beneficial, but 1 can only be used once. Wait, but 1 can only be placed once, so it's not helpful beyond that.Alternatively, think of arranging numbers in such a way that each number is a multiple of its neighbor. For example, if we have a chain where each number is a multiple of the previous one, but on a 2D grid, this is more complicated.Wait, maybe using a checkerboard pattern where alternating squares have numbers and their multiples. For example, divide the chessboard into black and white squares like a checkerboard. Assign numbers to black squares such that each white square is a multiple of its adjacent black squares, and vice versa. But since numbers must be unique, this might not work directly. For instance, if black squares have numbers m1, m2, m3, etc., then white squares need to be multiples of their neighbors. However, a white square has four black neighbors, so it would need to be a multiple of all four m's. But unless those m's are 1, which can only be used once, this would require the white square's number to be a common multiple of four different numbers, which would have to be very large. Moreover, each white square's number must be unique. This seems difficult.Alternatively, maybe using a bipartite graph structure where the chessboard is divided into two sets (like black and white squares), and numbers on one set are multiples of numbers on the other set. For example, all black squares have numbers that are factors of the numbers on white squares. Then, each black square's number m would require that the adjacent white squares are multiples of m. If k=4, this would require all four white neighbors to be multiples of m, but since each white square is adjacent to four black squares, it would need to be a multiple of four different m's. However, except for 1, numbers can't be multiples of four different numbers unless those numbers are factors of it. But each white square's number must be unique, so you would need an infinite number of numbers that are multiples of four different numbers each time, which is possible (for example, using products of four primes), but then the numbers would get very large. Also, each white square's number must satisfy its own condition, i.e., having k multiples on its neighbors, which are black squares. But if white squares have numbers that are products of four primes, then the adjacent black squares would need to have factors of that product. But since each black square's number is unique, this becomes a tangled web of dependencies.This approach seems too complicated. Maybe a different strategy. Let's consider smaller k values and see if we can find a pattern.For k=0, it's trivial. You can place any numbers as long as they are unique, since there's no requirement on the neighbors.For k=1, it's manageable. Each number needs at least one neighbor to be a multiple. One possible way is to arrange the numbers in a spiral where each number is followed by a multiple of it. For example, 1, 2, 3, 4, 5, 6, etc., but ensuring that each number has at least one multiple adjacent. However, ensuring uniqueness complicates things. Alternatively, use a diagonal pattern where each row or column alternates between primes and their multiples. But this is vague.For k=2, it's more challenging. Each number needs two neighbors to be multiples. One possible approach is to use a grid where each number is placed such that two of its neighbors are multiples. For example, using rows where even numbers are placed in such a way that each even number has two even neighbors, but since numbers must be unique, this would require a more intricate setup.But the question is about the greatest possible k. Since k=4 is impossible, and k=3 seems very difficult, maybe the answer is k=2? Let's check if k=2 is possible.To arrange the numbers so that each number has at least two multiples adjacent to it. How can we do this? Maybe using a repeating pattern where numbers are arranged in blocks. For example, in a 2x2 block:a bc dWhere a, b, c, d are numbers such that each has two multiples in their neighbors. But since the numbers must be unique, this would require careful assignment.Alternatively, using a diagonal pattern where each number is a multiple of the number two steps away. Hmm, not sure.Alternatively, consider using a binary tree structure embedded in the grid, where each node has two children that are multiples of it. But on a 2D grid, each node has four neighbors, so maybe two parents and two children? Not sure.Wait, another idea: use the fact that every number can have multiple factors. If we place a number m, then two of its neighbors can be multiples of m, and the other two neighbors can be factors of m. Because factors of m would require that m is a multiple of them, so those factors would have m as a multiple neighbor. This way, mutual dependencies can be satisfied.For example, place a number m. Then, two of its neighbors are 2m and 3m (multiples), and the other two neighbors are factors of m, say p and q, where p and q divide m. Then, for p and q, their multiple neighbor is m, satisfying their k=1 condition (if k=2, they need two multiples). Wait, but if p is a factor of m, then to satisfy p's condition, p needs two multiples adjacent to it. But p is adjacent to m and three other squares. If one of those three other squares is a multiple of p, then maybe p has two multiples: m and another multiple. But m is already a multiple of p, so that's one. Then, another neighbor of p can be 2p, but then 2p needs to have two multiples adjacent, and so on. This seems possible, but we need to ensure that all numbers are unique.Let me try to construct such a pattern. Start with m=1. Place 1 at the center. Its neighbors need to have two multiples of 1 (which is any number) and two factors of 1, but 1 has no factors other than itself. Since 1 can't be used again, this approach might not work. Alternatively, maybe start with a composite number.Let's start with m=6. Place 6 at (0,0). Then, two of its neighbors can be multiples of 6: 12 and 18. The other two neighbors can be factors of 6: 2 and 3. Now, 2 and 3 each need two multiples adjacent to them. For 2 at (1,0), its neighbors are 6 (which is a multiple), and three others: (2,0), (1,1), (1,-1). To satisfy k=2, we need another multiple of 2. Let's assign 4 to (2,0). Similarly, for 3 at (0,1), its neighbors are 6 (multiple), and three others: (0,2), (-1,1), (1,1). Assign 9 to (0,2) as a multiple of 3. Now, 4 at (2,0) needs two multiples. Its neighbors are 2 (at (1,0)), (3,0), (2,1), (2,-1). 2 is a factor, so we need two multiples of 4. Assign 8 and 12 to (3,0) and (2,1). But 12 is already used at (0,0)'s neighbor. Wait, 12 was placed at (say) (1,0)? Wait, no. Wait, original neighbors of 6 at (0,0) were 12, 18, 2, 3. So 12 is at (1,0), 18 at (-1,0), 2 at (0,1), and 3 at (0,-1)? Maybe not, I need to track coordinates carefully.Let me define coordinates more precisely. Let's say 6 is at (0,0). Then, assign:- (1,0) = 12 (multiple of 6)- (-1,0) = 18 (multiple of 6)- (0,1) = 2 (factor of 6)- (0,-1) = 3 (factor of 6)Now, for 2 at (0,1), it needs two multiples. Its neighbors are 6 (at (0,0)), (1,1), (-1,1), (0,2). So 6 is a multiple of 2. Assign 4 at (1,1) and 8 at (0,2). Then, 4 at (1,1) needs two multiples. Its neighbors are 2 (at (0,1)), 12 (at (1,0)), (2,1), (1,2). Assign 8 at (2,1) and 16 at (1,2). Wait, but 8 was already assigned to (0,2). No, (0,2) is 8, so (2,1) can be 16. Then 16 needs two multiples, and so on.Similarly, for 3 at (0,-1), neighbors are 6 (at (0,0)), (1,-1), (-1,-1), (0,-2). Assign 9 at (1,-1) and 15 at (0,-2). Then 9 needs two multiples, etc.This seems like a possible way to expand, ensuring that each number has two multiples adjacent. However, we need to ensure that all numbers are unique. For example, 12 at (1,0) has neighbors 6 (at (0,0)), (2,0), (1,1), (1,-1). (1,1) is 4, (1,-1) is 9, so assign 24 to (2,0) and 36 to (1,1). Wait, but (1,1) is already 4. Hmm, conflict. Maybe I need to adjust.This process is getting complicated, but perhaps on an infinite board, such a construction is possible by continually assigning new multiples and factors, ensuring uniqueness. If each time we place a multiple, we use a new number, and similarly for factors, then it might work. However, factors are limited since each number has only finite factors, but since we can choose numbers with multiple factors, we can manage.Alternatively, maybe using prime factors. For example, each time we place a prime number, we need to assign its multiples next to it. Since primes have only 1 and themselves as factors, except 1, but 1 can only be used once. So if we place a prime p, then two of its neighbors must be multiples of p, say 2p and 3p, and the other two neighbors would have to be numbers that have p as a factor, but since those numbers are unique, they would have to be multiples of p. However, that would require more than two multiples, which conflicts. Wait, no. If p is a prime, then its neighbors must include two multiples of p. The other two neighbors can be any numbers, but those numbers must have p not dividing them, unless they are multiples. But since we need uniqueness, we can't repeat p. Therefore, the other two neighbors must be numbers not divisible by p. However, those numbers would then need to satisfy their own k=2 condition with their other neighbors. So if a prime p is placed, two neighbors are multiples of p (like 2p, 3p), and the other two neighbors are numbers q and r, which are not multiples of p. Then, q and r must have two multiples each among their neighbors. However, q is adjacent to p and three other squares. Since p is not a multiple of q (unless q=1), but q's multiples must come from its other three neighbors. Similarly for r. This seems possible if we can arrange the q and r numbers such that their other neighbors provide the required multiples.However, this approach requires careful planning and might lead to a recursive structure where each prime branches into multiples and non-multiples, each requiring their own multiples. Since the board is infinite, this could be feasible, but it's not straightforward. Nonetheless, if such a construction is possible, then k=2 is achievable. On the other hand, if we can't find a concrete example or a systematic way to place the numbers for k=2, maybe the answer is k=1. But I need to check if k=2 is actually possible.Another angle: consider that each number must have two multiples adjacent. To minimize conflicts, we can use numbers with many small multiples. For example, even numbers can have multiples like 4, 6, 8, etc. If we arrange the even numbers in a checkerboard pattern, but since all numbers must be unique, we can't repeat even numbers. Alternatively, use a spiral of numbers where each even number is followed by a multiple. However, uniqueness is still an issue.Wait, here's an idea inspired by the Collatz conjecture or Ulam spiral: arrange numbers in a spiral where each number is placed such that its multiples are positioned in specific directions. For example, starting from 1, spiral outward, placing multiples of each number in two directions. However, ensuring that each number's two neighbors are multiples would require careful placement.Alternatively, use a diagonal grid where each diagonal contains multiples of a base number. For example, the main diagonal has powers of 2, the next diagonal has powers of 3, etc. But this might not ensure that each number has two multiples adjacent.Alternatively, use a bipartition of the grid into two sets: one set containing numbers with even numbers and the other with odd numbers. But even numbers can have multiples in both even and odd sets (since even times odd is even). Not sure.Wait, going back to the initial idea of mutual dependencies. If we pair numbers such that each number has two multiples and two factors adjacent, then maybe this can work. For example, a number m is adjacent to two multiples (am and bm) and two factors (m/c and m/d). However, m/c and m/d must be integers, so m must be divisible by c and d. This implies that m must be a multiple of the least common multiple of c and d. But since m is unique, this requires careful selection.Alternatively, construct a grid where each row is a sequence of numbers, each being a multiple of the previous one. For example, row 0: 1, 2, 4, 8, 16,...; row 1: 3, 6, 12, 24,...; row 2: 5, 10, 20, 40,...; etc. Then, each number in a row is a multiple of the one before it, and each column could be primes or something. However, in this case, the neighbors of a number would be the next number in the row (a multiple) and the number from the next row (not necessarily a multiple). So maybe only one neighbor is a multiple, which would give k=1. Not enough.Alternatively, arrange numbers in a grid where each cell (i,j) contains the number 2^i * 3^j. Then, each number's right and up neighbors are multiples (multiplied by 2 or 3), while left and down neighbors are factors (divided by 2 or 3). In this case, each number would have two multiples (right and up) and two factors (left and down). Therefore, for each number m=2^i*3^j, the right neighbor is 2^{i+1}*3^j (a multiple) and the up neighbor is 2^i*3^{j+1} (a multiple). Similarly, left and down are factors. However, in this arrangement, every number would have two multiples and two factors adjacent. Therefore, for each m, k=2 because two neighbors are multiples. This seems to satisfy the condition. Moreover, all numbers are unique since each (i,j) pair gives a unique product. Yes! This seems like a valid construction. Let's verify:Consider the grid where each cell (i,j) has the number 2^i * 3^j. Then, the four neighbors of (i,j) are:- Right: (i+1,j) => 2^{i+1}*3^j = 2*(2^i*3^j) = 2m, which is a multiple of m.- Left: (i-1,j) => 2^{i-1}*3^j = m/2, which is a factor (if i>0).- Up: (i,j+1) => 2^i*3^{j+1} = 3m, which is a multiple of m.- Down: (i,j-1) => 2^i*3^{j-1} = m/3, which is a factor (if j>0).Therefore, for any cell (i,j) where i and j are non-negative integers, the right and up neighbors are multiples of m, and the left and down neighbors are factors of m. However, this grid is only for non-negative integers, but the original problem states the chessboard is infinite in all directions. So we need to extend this to negative i and j as well.But if we allow negative i and j, then cells can have negative exponents, leading to fractions. However, the problem specifies positive integers. Therefore, we cannot use negative exponents because 2^{-1}*3^{0}=1/2, which is not a positive integer. Therefore, this construction only works for the first quadrant (i,j ≥0), but not for the entire infinite chessboard.However, maybe we can modify it. Instead of using 2 and 3, use primes in such a way that each dimension corresponds to a different prime. For example, assign to each cell (i,j) the number p_i * q_j, where p_i and q_j are primes. Then, the right neighbor would be p_{i+1} * q_j, and the up neighbor would be p_i * q_{j+1}. However, these are not necessarily multiples of the original number. Wait, no. If we use a different prime for each row and column, then the neighbors won't be multiples. So this approach doesn't work.Alternatively, use a pairing of primes where moving in one direction multiplies by a prime, and moving in another direction multiplies by another prime. For example, right neighbor is *2, left is /2, up is *3, down is /3. But this requires that the numbers can be divided by 2 or 3 when moving left or down, which is only possible if the number is a multiple of 2 or 3. Therefore, to cover the entire chessboard (including negative directions), we would need numbers that are products of 2^i * 3^j for all integers i and j, but these would include fractions unless i and j are non-negative. Therefore, this construction doesn't work for the entire infinite chessboard.But maybe we can tile the chessboard with multiple quadrants, each using different prime factors. For example, divide the chessboard into four quadrants: northeast (i≥0,j≥0), northwest (i≤0,j≥0), southeast (i≥0,j≤0), southwest (i≤0,j≤0). In each quadrant, use a different pair of primes. For example:- NE: 2^i * 3^j- NW: 5^{-i} * 7^j- SE: 11^i * 13^{-j}- SW: 17^{-i} * 19^{-j}But this would result in numbers that are products of different primes depending on the quadrant, ensuring uniqueness across the entire board. However, moving from one quadrant to another would cross regions where numbers are not multiples of each other, potentially violating the k=2 condition. For example, a cell on the boundary between NE and NW would have neighbors in different quadrants with different prime factors, hence not multiples. Therefore, this approach may not maintain the k=2 condition across the entire board.Alternatively, use a single prime for each axis, extending to negative indices with negative exponents, but map to positive integers by using absolute values or different encoding. However, this is not straightforward since negative exponents lead to fractions, which are not positive integers.Another idea: instead of using exponents, use a different encoding for negative directions. For example, represent each cell (i,j) as follows:- If i is positive, multiply by 2^i; if i is negative, multiply by 3^{|i|}.- If j is positive, multiply by 5^j; if j is negative, multiply by 7^{|j|}.Thus, each cell (i,j) would be assigned the number 2^{i+} * 3^{i-} * 5^{j+} * 7^{j-}, where i+ = max(i,0), i- = max(-i,0), similarly for j. This way, each cell's number is a unique positive integer, and moving in the positive x-direction multiplies by 2, negative x-direction multiplies by 3, positive y-direction multiplies by 5, negative y-direction multiplies by 7. Therefore, each neighbor in the positive x or y direction is a multiple (by 2 or 5), and each neighbor in the negative x or y direction is a factor (divided by 3 or 7). However, this construction ensures that each cell has two multiples (right and up) and two factors (left and down), similar to the original quadrant idea but extended to all directions using different primes. Let's verify:Take a cell (i,j). Its right neighbor is (i+1,j), which is 2^{i+1} * 3^{i-} * 5^{j+} * 7^{j-} = 2 * (current number), so a multiple. Its left neighbor is (i-1,j), which is 2^{i-1} * 3^{i-} * 5^{j+} * 7^{j-} = (current number)/2 if i > 0, but if i ≤ 0, then i- increases, so it becomes 3^{|i-1|} * ... which is 3 * current number if i was negative. Wait, actually, this encoding might not work as intended.Wait, let's clarify. If i is positive, then i+ = i and i- =0. If i is negative, i+ =0 and i- = |i|. So for cell (i,j):- Right neighbor (i+1,j): - If i+1 ≥0, then i+ becomes i+1, and i- becomes 0. - Number becomes 2^{i+1} * 3^0 * 5^{j+} *7^{j-} = 2 * 2^i * 5^{j+} *7^{j-} = 2 * current number. - If i+1 <0, then i+1 is negative, so i+ =0, i- = |i+1| = |i| -1. For example, if i = -2, then i+1 = -1, so number becomes 2^0 *3^{1} *5^{j+}*7^{j-} = 3 * current number /3^{2} (since original number was 3^2 * ...). Wait, this doesn't multiply by 3 but divides by 3. Hmm, this is problematic.Similarly, moving left from (i,j):- Left neighbor (i-1,j): - If i-1 ≥0: 2^{i-1} *3^{0} *5^{j+} *7^{j-} = current number /2. - If i-1 <0: i becomes negative, so i+ =0, i- = |i-1|. Original number was 2^{i} *3^{0}... if i was positive. Wait, no. If i was positive and we move left into negative i, the left neighbor would be 3^{1} *5^{j+} *7^{j-} since i-1 = -1, so i- =1. But the original number was 2^i * ..., so this is a different prime factor, not a multiple or factor. Hence, this approach doesn't maintain the multiple/factor relationship across the axes.Therefore, this encoding doesn't work as intended. Moving in negative directions from positive i leads to numbers with different prime factors, which are neither multiples nor factors of the original number. Therefore, this doesn't satisfy the k=2 condition.Hmm, this is tricky. Maybe the initial idea only works in the first quadrant, but not the entire board. However, the problem requires the chessboard to be infinite in all directions. So we need a different approach.Wait, perhaps instead of using multiplicative factors, use additive properties. But since the problem involves multiples, additive approaches might not help.Another thought: consider using a hexagonal tiling instead of square, but the problem specifies a chessboard, which is square.Alternatively, abandon the strict grid structure and think combinatorially. Since the chessboard is infinite, maybe we can assign numbers in such a way that every number m is part of a chain where each element has two multiples adjacent. For example, m -> 2m -> 4m -> ... and each step branches out. However, on a 2D grid, we can have multiple branches.Alternatively, consider that each number m is placed such that it has two outgoing edges (multiples) and two incoming edges (factors). This forms a directed graph where each node has out-degree 2 and in-degree 2. If such a graph can be embedded in the grid, then k=2 is possible. However, embedding a directed graph with these properties on a 2D grid is non-trivial. But perhaps using a diagonal lattice where each node points to two multiples in one diagonal direction and receives two factors from the opposite diagonal. This might work, but I need to visualize it. For example, place m at (0,0). Then, 2m and 3m are placed at (1,1) and (-1,1), while factors m/2 and m/3 are placed at (-1,-1) and (1,-1). But m/2 and m/3 must be integers, so m must be divisible by 2 and 3. This requires m to be a multiple of 6. Then, m=6 at (0,0). 2m=12 at (1,1), 3m=18 at (-1,1). Factors 3 and 2 at (-1,-1) and (1,-1). Then, for 12 at (1,1), its factors would be 6 (already at (0,0)) and 4 at (2,2), and multiples 24 and 36. This seems to create a expanding grid where each node branches into multiples and factors. However, ensuring uniqueness and covering the entire infinite grid is challenging.But since the chessboard is infinite, maybe this fractal-like expansion works. Each number branches into two larger multiples and two smaller factors, ensuring that all numbers are unique and cover the entire grid. However, this requires that every number is part of such a quadruple, which might be possible. For example, starting with 6:- 6 has multiples 12 and 18, and factors 2 and 3.- 12 has multiples 24 and 36, and factors 6 and 4.- 18 has multiples 36 and 54, and factors 6 and 9.- 2 has multiples 4 and 6, and factors 1 and ... but 1 can't be used again. Wait, 1 can only be used once. So this is a problem.If we start with 6, its factors are 2 and 3. Then, 2 needs factors, but the only factor is 1, which can only be placed once. Therefore, this approach hits a snag because primes cannot have two factors (other than 1 and themselves), but 1 is unique. Hence, primes would need to have their factors as 1 and themselves, but since they can't repeat, they can't have two factors adjacent. Therefore, this construction fails for primes.This suggests that numbers with few factors (like primes) pose a problem because they can't have two factors adjacent (since 1 is unique and they can't have another factor besides themselves). Therefore, maybe k=2 is not possible because primes cannot satisfy the condition of having two multiples adjacent. Wait, no. The condition is that each number m must have at least k multiples adjacent. For primes, they need two multiples adjacent. Since primes are only divisible by 1 and themselves, their multiples are 2p, 3p, etc. If we can arrange two multiples of p adjacent to p, then p's condition is satisfied. The problem arises with the factors of p, but the factors don't need to be adjacent to p. Wait, no. The factors of p would be 1 and p, but since p is already placed, and 1 can only be placed once, p's other neighbors can be any numbers. However, p's other neighbors (the two not multiples) can be any numbers, but those numbers are not required to be factors of p. They just need to not be multiples of p (since only two of the four neighbors need to be multiples). Wait, no. The condition is that at least k neighbors are multiples. The other neighbors can be any numbers, whether they are factors or not. The key is that for p, two of its neighbors must be multiples of p (e.g., 2p and 3p), and the other two can be any numbers not divisible by p. However, those other two numbers (say q and r) have their own conditions: each needs two multiples adjacent to them. If q and r are not multiples of p, then their multiples must come from their other neighbors. This seems feasible as long as q and r are chosen such that their multiples are placed elsewhere.However, primes still pose a problem because placing two multiples for each prime requires an infinite number of multiples, but since the board is infinite, this could be manageable. For example, each prime p is placed with two multiples 2p and 3p adjacent. Then, 2p and 3p each need two multiples adjacent to them, which could be 4p, 5p, 6p, etc. However, this leads to an infinite chain of multiples for each prime, which would require an infinite number of unique numbers. But since the board is infinite, and numbers are positive integers, which are also infinite, this is possible. However, coordinating this across the entire board without overlap is challenging.Alternatively, use a different prime for each "branch" to avoid overlap. For example, assign different primes to different regions of the board. But ensuring that all numbers are unique globally would require a systematic assignment.This is getting quite abstract. Maybe it's time to look for existing mathematical literature or known results. However, since I can't access external resources, I need to reason it out.Another angle: think of the problem as a graph labeling problem where each node must be labeled with a unique positive integer such that each node has at least k neighbors labeled with multiples of its own label. The question is about the maximum k for such a labeling exists on the infinite 2D grid.From graph theory, certain labelings are possible depending on the graph's structure. For an infinite grid, it's a 4-regular graph. The problem resembles a constraint satisfaction problem where each node imposes constraints on its neighbors.For k=2, if we can find a labeling that satisfies the condition, then k=2 is possible. The earlier attempt with the quadrant using 2^i*3^j works for k=2 in the first quadrant but doesn't extend to the entire grid. However, if we can find a similar construction that covers the entire grid, then k=2 is achievable.Alternatively, use a more complex base. For example, assign to each cell (i,j) the number 2^i * 3^j * 5^{s(i,j)} * 7^{t(i,j)} where s and t are functions that alternate signs based on the quadrant. But this is getting too vague.Wait, here's another approach: use the fundamental theorem of arithmetic. Every positive integer can be uniquely represented as a product of primes. Assign to each cell a unique combination of primes such that moving in certain directions multiplies by a specific prime, ensuring that two neighbors are multiples.For example, choose two primes, say 2 and 3. Assign to each cell (i,j) the number 2^i * 3^j. This is similar to the earlier quadrant idea but extended to all integers i and j. However, for negative i or j, this would result in fractions. To avoid fractions, map the grid to positive integers by using absolute values and separate primes for negative directions. For example:- For the x-axis: moving right multiplies by 2, moving left multiplies by 3.- For the y-axis: moving up multiplies by 5, moving down multiplies by 7.Thus, each cell (i,j) is labeled with 2^{max(i,0)} * 3^{max(-i,0)} * 5^{max(j,0)} * 7^{max(-j,0)}. This ensures that all labels are positive integers. Now, let's check the neighbors:- From cell (i,j), moving right (i+1,j): - If i ≥0: multiply by 2. - If i <0: i+1 is closer to zero, so decrease the exponent of 3. However, this doesn't result in a multiple unless i ≥0. Similarly for other directions. This approach doesn't guarantee that neighbors are multiples, as moving left from a negative i would involve multiplying by 3, which isn't a multiple of the current number. Hence, this doesn't satisfy the k=2 condition.Perhaps another way: assign to each cell (i,j) a unique number in such a way that moving right or up always multiplies by a prime, and left or down divides by a prime. For example, use a different prime for each row and column. However, this would require an infinite number of primes and careful assignment to avoid duplicates.Alternatively, use a pairing function to map each cell (i,j) to a unique positive integer, and then assign numbers based on that mapping. For example, use the Cantor pairing function to assign n = f(i,j), and then assign the number 2^n * 3^m where m is another pairing. But this doesn't inherently ensure the multiple condition.At this point, it's clear that constructing an explicit example for k=2 is non-trivial, but the earlier quadrant-based approach shows that in a quadrant, k=2 is achievable. Extending this to the entire infinite chessboard requires a different labeling that covers all directions without overlap or contradiction. If such a labeling exists, then k=2 is possible. Otherwise, k=1 might be the answer.But wait, in the quadrant example, each number has two multiples (right and up) and two factors (left and down). If we can mirror this structure in all four quadrants, perhaps using different prime factors for each quadrant, we could cover the entire board. For example:- In the NE quadrant (i≥0, j≥0): numbers are 2^i * 3^j- In the NW quadrant (i≤0, j≥0): numbers are 5^{-i} * 7^j- In the SE quadrant (i≥0, j≤0): numbers are 11^i * 13^{-j}- In the SW quadrant (i≤0, j≤0): numbers are 17^{-i} * 19^{-j}This way, each quadrant uses a distinct set of primes, ensuring that numbers are unique across the entire board. Now, in each quadrant, moving right or up increases the exponent of the respective prime, resulting in multiples, while moving left or down decreases the exponent, resulting in factors. However, when crossing quadrant boundaries, the numbers switch to different primes, which are not multiples of each other. Therefore, cells on the quadrant boundaries would have neighbors from different quadrants, which might not be multiples.For example, consider the cell (0,0) with number 1 (assuming we start at 2^0*3^0=1). Its right neighbor is 2^1*3^0=2 (NE quadrant), and its up neighbor is 2^0*3^1=3 (NE quadrant). Its left neighbor is 5^1*7^0=5 (NW quadrant), and its down neighbor is 11^0*13^1=13 (SE quadrant). Therefore, cell (0,0)=1 has neighbors 2, 3, 5, 13. Among these, 2 and 3 are multiples of 1 (trivially), but 5 and 13 are not multiples of 1. Wait, but 5 and 13 are multiples of 1 since every number is a multiple of 1. Therefore, actually, all four neighbors are multiples of 1. But since we need unique numbers, they are different, which is allowed. However, 1's condition is satisfied because all four neighbors are multiples (k=4), but we saw earlier that k=4 is impossible. Wait, no. The problem allows k=4 for m=1 because all numbers are multiples of 1. However, other numbers would require k=4, which is impossible. But in this construction, only 1 has k=4; other numbers have k=2.Wait, let's check another cell. Take cell (1,0)=2 (NE quadrant). Its neighbors are:- Right: (2,0)=4 (NE) – multiple of 2- Left: (0,0)=1 (NE) – factor of 2- Up: (1,1)=6 (NE) – multiple of 2- Down: (1,-1)=11^1*13^1=143 (SE) – not a multiple of 2Therefore, cell 2 has two multiples (4 and 6) and two non-multiples (1 and 143). Thus, k=2 for cell 2. Similarly, cell (0,1)=3 (NE quadrant) has:- Right: (1,1)=6 (multiple)- Left: (-1,1)=5*7=35 (NW) – not a multiple of 3- Up: (0,2)=9 (multiple)- Down: (0,0)=1 (factor)Thus, two multiples (6 and 9) and two non-multiples (35 and 1). So k=2.Similarly, cell 5 in NW quadrant (i=-1,j=0)=5 has neighbors:- Right: (0,0)=1 (NE) – not multiple of 5- Left: (-2,0)=25 (NW) – multiple of 5- Up: (-1,1)=5*7=35 (NW) – multiple of 5- Down: (-1,-1)=17*19=323 (SW) – not multiple of 5Thus, two multiples (25 and 35) and two non-multiples (1 and 323). So k=2.Similarly, cell 13 in SE quadrant (i=0,j=-1)=13 has neighbors:- Right: (1,-1)=143 (SE) – multiple of 13 (143=11*13)- Left: (-1,-1)=323 (SW) – not multiple of 13- Up: (0,0)=1 (NE) – not multiple- Down: (0,-2)=169 (SE) – multiple of 13Thus, two multiples (143 and 169) and two non-multiples (323 and 1). Therefore, k=2.This seems to work! By dividing the board into four quadrants, each using different prime pairs, we can ensure that each number has exactly two multiples adjacent (in the same quadrant) and two non-multiples (from other quadrants). Since the other quadrants use different primes, their numbers won't be multiples of the current number, satisfying the k=2 condition. This construction works because:1. Each quadrant uses distinct primes, ensuring global uniqueness.2. Within each quadrant, moving right or up increases the exponent of the respective prime, generating multiples.3. Crossing quadrant boundaries introduces numbers from different prime factors, which are not multiples.Therefore, this construction satisfies the conditions for k=2. Moreover, since all numbers are unique and positive integers, this arrangement is valid. Therefore, k=2 is achievable.But what about k=3? Earlier attempts showed that k=3 leads to exponential growth of required multiples, and dependencies that are difficult to satisfy. Even if we tried a similar quadrant approach for k=3, each number would need three multiples adjacent, which would require more primes or a denser arrangement. However, each number has only four neighbors, so needing three to be multiples leaves only one neighbor that can be a non-multiple. This makes the construction much harder, as each multiple would itself need three multiples, leading to a very rapid expansion of numbers. Additionally, primes would again pose a problem because they would require three multiples, but their multiples are spaced out, leading to conflicts and overlaps. Therefore, it's highly likely that k=3 is impossible.Thus, the greatest value of k is 2.Final AnswerThe greatest value of ( k ) is (boxed{2}).

✅:To determine the greatest value of ( k ) for which Alex the T-rex can place positive integers on an infinite chessboard such that no two numbers are equal and each number ( m ) on a square ( C ) requires that at least ( k ) of the squares orthogonally adjacent to ( C ) have multiples of ( m ), we analyze the constraints and possible constructions.1. Impossibility of ( k = 4 ): - If ( k = 4 ), every neighbor of a square must be a multiple of the number on that square. This is impossible because each number is unique, and arranging four distinct multiples around each square leads to contradictions.2. Challenges with ( k = 3 ): - Each number would need three of its four neighbors to be multiples. This leads to exponential growth in required multiples and dependencies that are difficult to satisfy, especially for prime numbers. Constructing such an arrangement is highly complex and likely impossible.3. Feasibility of ( k = 2 ): - A construction using a quadrant-based approach with different prime factors for each quadrant ensures that each number has two multiples adjacent. For example: - Assign numbers in each quadrant using distinct primes (e.g., ( 2^i cdot 3^j ) for the first quadrant). - Moving right or up within a quadrant increases the prime exponent, generating multiples. - Crossing quadrant boundaries introduces numbers from different primes, which are not multiples, ensuring the ( k = 2 ) condition is met.4. Verification: - Each number in the constructed grid has exactly two multiples adjacent (within the same quadrant) and two non-multiples (from other quadrants). This satisfies the ( k = 2 ) condition while maintaining unique positive integers across the entire infinite chessboard.Thus, the greatest value of ( k ) for which this is possible is (boxed{2}).

🔑:1. Understanding the Problem: We need to place positive integers on an infinite chessboard such that: - No two numbers are equal. - For any number ( m ) placed on a square ( C ), at least ( k ) of the squares orthogonally adjacent to ( C ) have a multiple of ( m ) written on them.2. Analyzing the Constraints: - If ( k = 4 ), then all four orthogonally adjacent squares must have multiples of ( m ). This is impossible because two adjacent squares cannot both divide each other. Hence, ( k neq 4 ). - If ( k = 3 ), consider a square with a number ( a ). Suppose the number ( ka ) is written next to it. Without loss of generality, let the number not divisible by ( a ) be to its left. Below ( ka ), there must be a multiple of it, say ( akb ). The square to the left of this must be a multiple of both ( a ) and ( akb ), which is not possible since it cannot divide both.3. Constructing a Solution for ( k = 2 ): - We need to show that it is possible to place numbers such that each number ( m ) has at least 2 orthogonally adjacent squares with multiples of ( m ). - Start by placing the numbers ( 1, 2, 4, 8 ) in a ( 2 times 2 ) square: [ begin{array}{|c|c|} hline 1 & 2 hline 4 & 8 hline end{array} ] - We can extend this pattern by adding new rows or columns while maintaining the condition. For example, add a new column to the right: [ begin{array}{|c|c|c|} hline 1 & 2 & 16 hline 4 & 8 & 32 hline end{array} ] Here, ( 16 ) is a multiple of ( 2 ) and ( 8 ), and ( 32 ) is a multiple of ( 4 ) and ( 8 ).4. Generalizing the Construction: - For any new row or column, pick the top right square and add a multiple of it to its right. Below it, take a multiple of the product of numbers on the left and top (choose multiples large enough to ensure distinctness). - This ensures that each number has at least 2 orthogonally adjacent squares with multiples of it.The final answer is ( boxed{ k = 2 } ).

🧠:Okay, so I have this problem here where I need to find the area of a region R in the Cartesian plane. The region is defined by three conditions: x is greater than or equal to 0, y is greater than or equal to 0, and x + y + the floor of x plus the floor of y is less than or equal to 5. Hmm, floor functions can complicate things because they introduce piecewise behavior. Let me break this down step by step.First, let's recall that the floor function, denoted by ⌊x⌋, gives the greatest integer less than or equal to x. So for any real number x, ⌊x⌋ is the integer part of x, effectively stripping away the decimal part. For example, ⌊3.7⌋ = 3, ⌊5⌋ = 5, and ⌊2.999⌋ = 2. Similarly for y.Given that x and y are both non-negative, their floor functions will be non-negative integers as well. The inequality we have is x + y + ⌊x⌋ + ⌊y⌋ ≤ 5. Let's denote ⌊x⌋ as m and ⌊y⌋ as n, where m and n are non-negative integers (since x and y are non-negative). Then, the inequality becomes x + y + m + n ≤ 5.But since m = ⌊x⌋ and n = ⌊y⌋, we know that x is in the interval [m, m + 1) and y is in [n, n + 1). So, for each pair of integers m and n, we can define a region where x is between m and m + 1, y is between n and n + 1, and x + y + m + n ≤ 5.Therefore, the region R is a union of such regions for all valid m and n. To find the area of R, I need to consider all possible pairs (m, n) where m and n are non-negative integers, and for each pair, determine the area of the region where x ∈ [m, m + 1), y ∈ [n, n + 1), and x + y ≤ 5 - m - n. Then, sum up all these areas.Wait, let me verify that. If x is in [m, m + 1) and y is in [n, n + 1), then m and n are fixed, so the inequality x + y + m + n ≤ 5 can be rewritten as x + y ≤ 5 - m - n. But since x and y are in their respective intervals, x + y will range from m + n to (m + 1) + (n + 1) - ε (where ε is a small positive number). Therefore, the maximum possible x + y in each such square [m, m + 1) × [n, n + 1) is m + n + 2 - ε. Therefore, for the inequality x + y ≤ 5 - m - n to hold in that square, we need that 5 - m - n ≥ m + n. Wait, no. That might not be the correct approach. Let me think again.Actually, for each square [m, m + 1) × [n, n + 1), the inequality x + y + m + n ≤ 5 becomes x + y ≤ 5 - m - n. So within each such square, we have the linear inequality x + y ≤ C, where C = 5 - m - n. The area in that square satisfying x + y ≤ C can be found by geometry, depending on the value of C.But since x and y are in [m, m + 1) and [n, n + 1) respectively, shifting coordinates might help. Let me define u = x - m and v = y - n, where u and v are in [0, 1). Then the inequality becomes (u + m) + (v + n) + m + n ≤ 5, which simplifies to u + v + 2m + 2n ≤ 5. Therefore, u + v ≤ 5 - 2m - 2n. So the problem reduces to, for each m and n, finding the area within the unit square [0,1) × [0,1) where u + v ≤ S, with S = 5 - 2m - 2n. Then, summing these areas over all valid m and n.But we must have S ≥ 0, because u + v can't be negative. So 5 - 2m - 2n ≥ 0 => 2(m + n) ≤ 5. Since m and n are non-negative integers, 2(m + n) must be less than or equal to 5. Since m + n must be an integer or a half-integer? Wait, m and n are integers, so m + n is an integer. Then 2(m + n) is even, so 2(m + n) ≤ 5 implies that 2(m + n) ≤ 4, so m + n ≤ 2.5, but since m + n is an integer, m + n ≤ 2. Therefore, m + n can be 0, 1, or 2.Wait, let's check. If m and n are integers, then 2(m + n) is even. The inequality 2(m + n) ≤ 5. Since 5 is odd, 2(m + n) can be at most 4, which is even. Therefore, m + n ≤ 2. So possible values of m + n are 0, 1, 2.Therefore, the possible pairs (m, n) are those where m + n ≤ 2. So m and n are non-negative integers such that m + n ≤ 2. Let me list all possible (m, n):- When m + n = 0: (0, 0)- When m + n = 1: (0,1), (1,0)- When m + n = 2: (0,2), (1,1), (2,0)Therefore, in total, there are 1 + 2 + 3 = 6 squares to consider. Each of these corresponds to different regions in the first quadrant, each shifted by their m and n.For each of these (m, n), we need to find the area within the unit square [m, m +1) × [n, n +1) where x + y + m + n ≤ 5. But as we transformed earlier, substituting u = x - m and v = y - n, the inequality becomes u + v ≤ 5 - 2m - 2n. Let's denote S = 5 - 2m - 2n. Then, since u and v are in [0,1), the region u + v ≤ S can be a triangle or a trapezoid or the entire square or empty, depending on the value of S.Wait, S must be non-negative, so as we saw before, m + n ≤ 2. So S = 5 - 2(m + n). For each (m, n):1. (0, 0): S = 5 - 0 = 52. (0, 1): S = 5 - 2*1 = 33. (1, 0): S = 5 - 2*1 = 34. (0, 2): S = 5 - 2*2 = 15. (1, 1): S = 5 - 2*2 = 16. (2, 0): S = 5 - 2*2 = 1Wait, but when m + n = 2, S = 1, so the inequality becomes u + v ≤ 1. Similarly, for m + n =1, S=3, but u and v are each in [0,1), so u + v is in [0,2). So S=3 would mean that the inequality u + v ≤ 3 is always true in the unit square. Wait, but u + v in the unit square can be at most 2 (if u and v approach 1). So if S=3, the inequality u + v ≤3 is always satisfied in the unit square. Therefore, the area would be 1 (the entire unit square).Similarly, S=5 for (0,0). Since u and v are in [0,1), u + v can be at most 2, so 5 is way larger. Therefore, u + v ≤5 is always true. So for (0,0), the entire unit square is included.But wait, let's check each case:For each (m, n):1. (0, 0): S = 5. Since u + v ≤5. But u and v are each in [0,1), so u + v < 2. Therefore, 2 < 5, so the entire square is included. Area =1.2. (0,1): S =3. Similarly, u + v ≤3. Since u + v <2, again, the entire square is included. Area=1.3. (1,0): S=3. Same as above. Entire square included. Area=1.4. (0,2): S=1. Now, u and v are in [0,1). The inequality u + v ≤1. So within the unit square, this is a triangular region with vertices at (0,0), (0,1), (1,0). The area is 0.5.5. (1,1): S=1. Similarly, u + v ≤1. So again, a triangle with area 0.5.6. (2,0): S=1. Same as above, area 0.5.Wait, but hold on. The original problem was x ≥0 and y ≥0. So when m and n are 0,1,2, but in the original problem, we have to consider all points (x,y) where x and y are non-negative. So each (m, n) corresponds to the unit square starting at (m, n). For example, (0,0) is the square [0,1) × [0,1), (0,1) is [0,1) × [1,2), (0,2) is [0,1) × [2,3), (1,0) is [1,2) × [0,1), (1,1) is [1,2) × [1,2), (2,0) is [2,3) × [0,1). But does the original inequality x + y + floor(x) + floor(y) ≤5 allow for x and y beyond these squares? For example, x could be in [3,4), but then floor(x)=3, and floor(y) would contribute as well. Wait, but earlier we concluded that m + n ≤2, so floor(x) + floor(y) ≤2. Therefore, x and y can't be in regions where floor(x) + floor(y) ≥3. Let's verify this.Suppose m + n ≥3. Then 2(m + n) ≥6, so S=5 -2(m +n) ≤5 -6= -1. But S =5 -2(m +n) would be negative, so the inequality u + v ≤ S would be impossible because u + v ≥0. Therefore, such regions (m +n ≥3) contribute nothing. Therefore, the only regions that contribute are those with m +n ≤2, as we listed earlier. Therefore, the possible (m,n) pairs are indeed the six I listed.But wait, perhaps we need to check if higher m or n could still satisfy the original inequality. Let me test with an example. Let's take m=3, n=0. Then floor(x)=3, floor(y)=0. Then the original inequality would be x + y +3 +0 ≤5 =>x + y ≤2. But x is in [3,4), so x ≥3. Therefore, x + y ≥3 +0=3, but the inequality requires x + y ≤2, which is impossible. Therefore, m=3, n=0 gives no contribution. Similarly for m=0, n=3. So indeed, only m +n ≤2 contributes.Therefore, the region R is the union of the six unit squares (each of size 1x1, but some with only partial areas) corresponding to (m,n) = (0,0), (0,1), (1,0), (0,2), (1,1), (2,0). For each of these, we need to find the area where x + y + m +n ≤5, which translates to u + v ≤5 -2m -2n. But as we saw, for (0,0), (0,1), (1,0), the entire unit square is included. For (0,2), (1,1), (2,0), only a triangle with area 0.5 is included.Therefore, the total area would be:For (0,0), (0,1), (1,0): each contributes 1, so 3*1=3.For (0,2), (1,1), (2,0): each contributes 0.5, so 3*0.5=1.5.Total area = 3 + 1.5 =4.5=9/2.Wait, 4.5 is 9/2? Wait, 4.5 is 9/2? No, 4.5 is 9/2 divided by 2. Wait, 4.5 is equal to 9/2? Wait, 9 divided by 2 is 4.5, yes. So 4.5 is 9/2. So the area is 9/2. Therefore, the answer is boxed{dfrac{9}{2}}.But let me verify this again, step by step, to make sure I didn't miss anything.First, for each (m,n):1. (0,0): x in [0,1), y in [0,1). Inequality x + y +0 +0 ≤5 =>x + y ≤5. Since x + y <2 here, so entire square included. Area=1.2. (0,1): x in [0,1), y in [1,2). Inequality x + y +0 +1 ≤5 =>x + y ≤4. Here, x in [0,1), y in [1,2), so x + y in [1,3). Since 3 <4, so entire square included. Area=1.3. (1,0): x in [1,2), y in [0,1). Similar to (0,1), inequality x + y +1 +0 ≤5 =>x + y ≤4. x in [1,2), y in [0,1). x + y in [1,3). Again, 3 <4, so entire square included. Area=1.4. (0,2): x in [0,1), y in [2,3). Inequality x + y +0 +2 ≤5 =>x + y ≤3. Here, x in [0,1), y in [2,3). x + y in [2,4). So the inequality x + y ≤3 cuts through this square. The intersection is where x + y ≤3. Let me visualize this square: x from 0 to1, y from2 to3. The line x + y =3 would go from (0,3) to (3,0), but within this square, the line x + y=3 intersects the square at (0,3) and (1,2). However, y is in [2,3), so the line x + y =3 intersects the left side at (0,3) and the bottom side at (1,2). But wait, the square is [0,1) × [2,3). So in this square, the region x + y ≤3 is the area below the line from (0,3) to (1,2). But since the square is from y=2 to y=3, actually, the line x + y=3 crosses the square at (0,3) and (1,2). But within the square, the points (x,y) satisfy x ∈ [0,1) and y ∈ [2,3). So for each x in [0,1), y can be from 2 up to min(3, 3 -x). But since x ∈ [0,1), 3 -x ∈ (2,3]. Therefore, the upper bound for y is 3 -x. However, y is already bounded above by 3. So the upper limit for y is 3 -x, but since y <3, and 3 -x >2 (because x <1), we have that y must be from 2 to 3 -x.Therefore, the region is a trapezoid? Wait, no. Let's parametrize x from 0 to1. For each x, y ranges from 2 to 3 -x. So integrating over x from 0 to1, the length in y is (3 -x) -2 =1 -x. Therefore, the area is the integral from 0 to1 of (1 -x) dx = [x - (x^2)/2] from 0 to1 = (1 - 0.5) -0 =0.5. So yes, area 0.5.Similarly for (1,1): x in [1,2), y in [1,2). Inequality x + y +1 +1 ≤5 =>x + y ≤3. x in [1,2), y in [1,2). So x + y ranges from 2 to 4. The inequality x + y ≤3 would cut this square. The line x + y=3 goes from (1,2) to (2,1). But in the square [1,2) × [1,2), the line x + y=3 divides the square into two regions. The region where x + y ≤3 is a triangle with vertices at (1,1), (1,2), (2,1). Wait, but x and y are in [1,2), so the intersection points are at (1,2) and (2,1), but those are the corners. Wait, but in the square [1,2) × [1,2), the line x + y=3 is from (1,2) to (2,1). However, the square is open on the right and top, so the line is actually from (1,2) to (2,1), but within the square, the region x + y ≤3 is the area below the line. For x in [1,2), y is in [1,2), and x + y ≤3. So for each x in [1,2), y can go from 1 to 3 -x. Since x is in [1,2), 3 -x is in (1,2]. Therefore, y ranges from 1 to 3 -x. The area is the integral from x=1 to x=2 of (3 -x -1) dx = integral from1 to2 of (2 -x) dx = [2x - (x^2)/2] from1 to2 = (4 -2) - (2 -0.5) = 2 -1.5 =0.5. So again, area 0.5.Similarly for (2,0): x in [2,3), y in [0,1). Inequality x + y +2 +0 ≤5 =>x + y ≤3. x ∈ [2,3), y ∈ [0,1). So x + y ranges from2 to4. The inequality x + y ≤3 requires y ≤3 -x. For x in [2,3), 3 -x is in (0,1]. So y must be from0 to3 -x. The area is the integral from x=2 tox=3 of (3 -x -0) dx = integral from2 to3 of (3 -x) dx = [3x - (x^2)/2] from2 to3 = (9 -4.5) - (6 -2) =4.5 -4=0.5. So indeed, area 0.5.Therefore, adding up all these:(0,0):1(0,1):1(1,0):1(0,2):0.5(1,1):0.5(2,0):0.5Total area=1+1+1+0.5+0.5+0.5= 4.5=9/2.Therefore, the area is 9/2. But wait, let me check again if I considered all possible regions. Since m and n are non-negative integers with m +n ≤2, and we listed all 1 +2 +3=6 pairs, which seems correct. For m +n=0:1 pair, m +n=1:2 pairs, m +n=2:3 pairs. Yes, that's correct. Each of these contributes as calculated. So total area is indeed 9/2.But let me think if there's another way to approach this problem to confirm. Maybe by graphing or considering different regions based on the integer parts.Alternatively, consider that x and y are non-negative, and the floor functions break the plane into unit squares. In each unit square, floor(x)=m and floor(y)=n, so the inequality becomes x + y ≤5 -m -n. But since x =m +u, y =n +v, with u, v ∈[0,1), then substituting back:(m +u) + (n +v) +m +n ≤5 =>2m +2n +u +v ≤5.So, 2(m +n) +u +v ≤5. Since u +v <2, then 2(m +n) <5 - (u +v). But since u +v <2, then 5 - (u +v) >3. Therefore, 2(m +n) <5 - (u +v) <5. But since m and n are integers, 2(m +n) must be even, so 2(m +n) ≤4. Hence, m +n ≤2, as before.Therefore, only m +n=0,1,2 are possible. Therefore, the same six squares. Then, the area calculation proceeds as before. So this confirms the previous approach.Alternatively, maybe consider all possible values of floor(x)=m and floor(y)=n where m and n are non-negative integers such that 2(m +n) ≤5. As established, m +n ≤2. So m and n as before.Alternatively, think of the original inequality x + y + floor(x) + floor(y) ≤5. Let me define m = floor(x), n = floor(y). Then x =m +a, y =n +b, where a,b ∈[0,1). Then the inequality becomes (m +a) + (n +b) +m +n ≤5 =>2m +2n +a +b ≤5. Therefore, a +b ≤5 -2m -2n. As a and b are in [0,1), the maximum of a +b is less than 2. Therefore, for each m,n, if 5 -2m -2n ≥2, then the entire unit square [m, m +1) × [n, n +1) is included. If 0 ≤5 -2m -2n <2, then only the region a +b ≤5 -2m -2n is included. If 5 -2m -2n <0, then no area is included.So, for each m,n:- If 5 -2m -2n ≥2, include entire square (area1).- If 0 ≤5 -2m -2n <2, include the region a +b ≤5 -2m -2n (area to calculate).- Else, include nothing.So let's categorize the (m,n) pairs based on 5 -2m -2n:For m +n=0: 5 -0=5 ≥2, entire square included.For m +n=1:5 -2=3 ≥2, entire square included.For m +n=2:5 -4=1, which is between0 and2, so include region a +b ≤1.For m +n=3:5 -6= -1 <0, include nothing.Therefore, the same as before. So the total area is:For m +n=0:1 square, area1.For m +n=1:2 squares, each area1.For m +n=2:3 squares, each area0.5.Total area=1+2*1 +3*0.5=1+2+1.5=4.5=9/2.Thus, the answer is boxed{dfrac{9}{2}}.But wait, just to be thorough, let me check the calculation for the areas where a +b ≤ S, with S=1. For example, in the case where S=1, which occurs when m +n=2, the region a +b ≤1 within the unit square [0,1) ×[0,1) is a right-angled triangle with legs of length1, so area=0.5. Correct.Similarly, if S were between1 and2, the area would be a trapezoid or another shape, but in our problem S=1 for m +n=2, so we have the triangle. For S≥2, the entire square is included. For S<0, nothing is included. So our calculations hold.Therefore, after all these checks and multiple approaches, I'm confident that the area of region R is 9/2.Final AnswerThe area of ( R ) is (boxed{dfrac{9}{2}}).

✅:Given the region ( R ) in the Cartesian plane defined by ( x geq 0 ), ( y geq 0 ), and ( x + y + lfloor x rfloor + lfloor y rfloor leq 5 ), we need to determine its area.1. Understanding the Floor Functions: - Let ( m = lfloor x rfloor ) and ( n = lfloor y rfloor ). These are non-negative integers. - The inequality becomes ( x + y + m + n leq 5 ). - For each pair ((m, n)), ( x ) and ( y ) lie in the intervals ([m, m+1)) and ([n, n+1)) respectively.2. Transforming Coordinates: - Substitute ( x = m + u ) and ( y = n + v ) where ( u, v in [0, 1) ). - The inequality transforms to ( u + v leq 5 - 2m - 2n ).3. Valid Pairs of ((m, n)): - The inequality ( 2(m + n) leq 5 ) must hold. Since ( m ) and ( n ) are integers, ( m + n leq 2 ). - Valid pairs ((m, n)) are: ((0, 0)), ((0, 1)), ((1, 0)), ((0, 2)), ((1, 1)), ((2, 0)).4. Calculating Areas: - For pairs ((0, 0)), ((0, 1)), ((1, 0)): The inequality ( u + v leq 5 - 2m - 2n ) covers the entire unit square, area = 1 each. - For pairs ((0, 2)), ((1, 1)), ((2, 0)): The inequality ( u + v leq 1 ) forms a right-angled triangle with area = 0.5 each.5. Summing Areas: - Total area = ( 3 times 1 + 3 times 0.5 = 3 + 1.5 = 4.5 ).Thus, the area of the region ( R ) is (boxed{dfrac{9}{2}}).

🔑:1. Understanding the problem: We need to find the area of the region ( R ) in the Cartesian plane where ( (x, y) ) satisfies: [ x geq 0, quad y geq 0, quad text{and} quad x + y + lfloor x rfloor + lfloor y rfloor leq 5 ] Here, ( lfloor x rfloor ) and ( lfloor y rfloor ) denote the floor functions of ( x ) and ( y ), respectively.2. Simplifying the inequality: Since ( lfloor x rfloor ) and ( lfloor y rfloor ) are integers, let: [ lfloor x rfloor = a quad text{and} quad lfloor y rfloor = b ] where ( a ) and ( b ) are non-negative integers. Then the inequality becomes: [ x + y + a + b leq 5 ] Since ( a leq x < a+1 ) and ( b leq y < b+1 ), we can rewrite the inequality as: [ a + b + (x - a) + (y - b) leq 5 implies x + y leq 5 - (a + b) ]3. Case analysis based on ( a + b ): - Case 1: ( a + b = 0 ) [ x + y leq 5 ] Here, ( a = 0 ) and ( b = 0 ), so ( 0 leq x < 1 ) and ( 0 leq y < 1 ). The region is the unit square with vertices ((0,0)), ((1,0)), ((0,1)), and ((1,1)). The area is: [ text{Area} = 1 ] - Case 2: ( a + b = 1 ) [ x + y leq 4 ] Here, ( a ) and ( b ) can be either ( (0,1) ) or ( (1,0) ). The regions are: - For ( a = 0 ) and ( b = 1 ): ( 0 leq x < 1 ) and ( 1 leq y < 2 ) - For ( a = 1 ) and ( b = 0 ): ( 1 leq x < 2 ) and ( 0 leq y < 1 ) Each region is a unit square, and both satisfy ( x + y leq 4 ). The total area is: [ text{Area} = 2 ] - Case 3: ( a + b = 2 ) [ x + y leq 3 ] Here, ( a ) and ( b ) can be ( (0,2) ), ( (1,1) ), or ( (2,0) ). The regions are: - For ( a = 0 ) and ( b = 2 ): ( 0 leq x < 1 ) and ( 2 leq y < 3 ) - For ( a = 1 ) and ( b = 1 ): ( 1 leq x < 2 ) and ( 1 leq y < 2 ) - For ( a = 2 ) and ( b = 0 ): ( 2 leq x < 3 ) and ( 0 leq y < 1 ) Each region is a unit square, and the intersection with ( x + y leq 3 ) forms an isosceles right triangle with legs of length 1. The area of each triangle is: [ text{Area of each triangle} = frac{1}{2} ] Since there are three such triangles, the total area is: [ text{Area} = 3 times frac{1}{2} = frac{3}{2} ]4. Summing the areas: The total area of the region ( R ) is: [ 1 + 2 + frac{3}{2} = frac{9}{2} ]The final answer is (boxed{frac{9}{2}})