🧠:Okay, so I need to find S(pq) where p = 127 and q = 31 are primes. The problem states that S(n) is the sum of the squares of the positive integers less than and coprime to n. Then, this sum can be written in a specific form involving a, b, c, and I need to find a. First, let me make sure I understand the problem. For example, S(5) is 1² + 2² + 3² + 4² because all numbers less than 5 (which is prime) are coprime to 5. But S(4) is 1² + 3² because only 1 and 3 are coprime to 4. So when n is prime, S(n) would be the sum of squares from 1 to n-1. But when n is composite, we have to exclude numbers that share a common factor with n.Given that p and q are primes (since they are Mersenne primes: 127 = 2^7 -1 and 31 = 2^5 -1), their product pq = 127*31 is a semiprime. So, the problem reduces to finding the sum of squares of numbers less than 127*31 that are coprime to 127*31. Then express this sum in the given form and find the value of a.I need to recall if there's a formula for the sum of squares of numbers coprime to n. I remember there are formulas related to totient functions and multiplicative functions. Since S(n) is a sum over numbers coprime to n, perhaps we can use inclusion-exclusion or some multiplicative properties.First, note that because p and q are distinct primes, the totient function φ(pq) = (p-1)(q-1). But φ(n) counts the numbers coprime to n, but here we need the sum of their squares. So maybe there's an analogous formula for the sum of squares.I remember that for the sum of integers coprime to n, there's a formula: (n φ(n))/2. But for the sum of squares, it might be more complicated. Let me check.Alternatively, maybe we can use the principle of inclusion-exclusion. The sum of squares of numbers from 1 to pq - 1, minus the sum of squares of numbers divisible by p or q. But numbers divisible by p or q are those divisible by p, plus those divisible by q, minus those divisible by pq. But since pq is the product, numbers divisible by both p and q are exactly those divisible by pq. So, using inclusion-exclusion:Sum of squares coprime to pq = total sum of squares from 1 to pq-1 - sum of squares divisible by p - sum of squares divisible by q + sum of squares divisible by pq.So, let's denote the total sum as T = Σ_{k=1}^{pq - 1} k². The sum of squares divisible by p is Σ_{k=1}^{(pq - 1)/p} (pk)² = p² Σ_{k=1}^{(pq - 1)/p} k². Similarly, sum of squares divisible by q is q² Σ_{k=1}^{(pq - 1)/q} k². The sum of squares divisible by pq is (pq)² Σ_{k=1}^{(pq - 1)/(pq)} k². However, since pq is greater than pq -1, the upper limit here is zero? Wait, no. If we take k such that pq*k ≤ pq -1, then k must be 0. Wait, actually, pq divides numbers only once, which is pq itself, but since we are going up to pq -1, there are no numbers divisible by pq. So the last term is zero. Therefore, the inclusion-exclusion formula simplifies to:S(pq) = T - p² * S_p - q² * S_qwhere S_p is the sum of squares from 1 to floor((pq -1)/p), and similarly for S_q.Wait, but (pq -1)/p is equal to q - 1/p. Since p and q are integers, floor((pq -1)/p) = floor(q - 1/p) = q -1. Similarly, floor((pq -1)/q) = p -1. So, S_p is the sum of squares from 1 to q -1, and S_q is the sum from 1 to p -1. Therefore, sum of squares divisible by p is p² * sum_{k=1}^{q -1} k², and similarly for q. Then, the inclusion-exclusion gives:S(pq) = sum_{k=1}^{pq -1} k² - p² * sum_{k=1}^{q -1} k² - q² * sum_{k=1}^{p -1} k²Now, we can use the formula for the sum of squares: sum_{k=1}^n k² = n(n + 1)(2n + 1)/6. Let's compute each term.First, compute T = sum_{k=1}^{pq -1} k² = [(pq -1)(pq)(2pq -1)]/6Then, sum_{k=1}^{q -1} k² = [(q -1)q(2q -1)]/6Similarly, sum_{k=1}^{p -1} k² = [(p -1)p(2p -1)]/6Therefore, substituting into the expression for S(pq):S(pq) = [(pq -1)(pq)(2pq -1)]/6 - p² * [(q -1)q(2q -1)]/6 - q² * [(p -1)p(2p -1)]/6Factor out 1/6:S(pq) = (1/6)[ (pq -1)(pq)(2pq -1) - p² q (q -1)(2q -1) - q² p (p -1)(2p -1) ]But notice that in the second term, p² * [(q -1)q(2q -1)]/6: it's p² q (q -1)(2q -1)/6. Similarly for the third term. So let's write each term:First term: (pq -1)(pq)(2pq -1)Second term: - p² q (q -1)(2q -1)Third term: - q² p (p -1)(2p -1)So S(pq) = [Term1 - Term2 - Term3]/6Now, let's expand each term step by step.First, Term1 = (pq -1)(pq)(2pq -1). Let's expand this:First multiply (pq -1)(pq) = p² q² - pqThen multiply by (2pq -1):= (p² q² - pq)(2pq -1) = p² q² * 2pq - p² q² *1 - pq *2pq + pq *1= 2 p³ q³ - p² q² - 2 p² q² + pq= 2 p³ q³ - 3 p² q² + pqTerm1 = 2 p³ q³ - 3 p² q² + pqNow Term2 = p² q (q -1)(2q -1). Let's expand (q -1)(2q -1):(q -1)(2q -1) = 2q² - q - 2q +1 = 2q² -3q +1So Term2 = p² q (2q² -3q +1) = 2 p² q³ -3 p² q² + p² qSimilarly, Term3 = q² p (p -1)(2p -1). Expand (p -1)(2p -1):= 2p² -p -2p +1 = 2p² -3p +1Term3 = q² p (2p² -3p +1) = 2 p³ q² -3 p² q² + p q²Therefore, Term2 + Term3 = [2 p² q³ -3 p² q² + p² q] + [2 p³ q² -3 p² q² + p q²] Combine like terms:= 2 p² q³ + 2 p³ q² -6 p² q² + p² q + p q²Wait, let's check:- 2 p² q³ (from Term2)- 2 p³ q² (from Term3)- -3 p² q² (Term2) + -3 p² q² (Term3) = -6 p² q²- p² q (from Term2)- p q² (from Term3)So yes, Term2 + Term3 = 2 p² q³ + 2 p³ q² -6 p² q² + p² q + p q²Therefore, S(pq) = [Term1 - Term2 - Term3]/6 = [ (2 p³ q³ - 3 p² q² + pq) - (2 p² q³ + 2 p³ q² -6 p² q² + p² q + p q²) ] /6Let's compute the numerator:2 p³ q³ - 3 p² q² + pq -2 p² q³ -2 p³ q² +6 p² q² -p² q -p q²Combine like terms:Terms with p³ q³: 2 p³ q³Terms with p³ q²: -2 p³ q²Terms with p² q³: -2 p² q³Terms with p² q²: (-3 +6) p² q² = 3 p² q²Terms with p² q: -p² qTerms with pq²: -p q²Terms with pq: +pqSo overall:Numerator = 2 p³ q³ -2 p³ q² -2 p² q³ +3 p² q² -p² q -p q² +pqNow, the goal is to factor this expression and write it in the form (p² q² /6)(a - b/c). Let's see.First, note that the given form is (p² q² /6)(a - b/c). So, if we can factor out p² q² from the numerator, then the remaining terms would give us (a - b/c). Let's check if that's possible.Looking at the numerator:2 p³ q³ -2 p³ q² -2 p² q³ +3 p² q² -p² q -p q² +pqLet's factor terms:First, let's group terms with p³ q³, p³ q², p² q³, etc.- 2 p³ q³ -2 p³ q² -2 p² q³ +3 p² q² -p² q -p q² +pqHmm. Let's factor out p² q² from the first four terms:p² q² [ 2 pq -2 p -2 q +3 ] -p² q -p q² +pqWait, let's check:2 p³ q³ = p² q² * 2 pq-2 p³ q² = p² q² * (-2 p)-2 p² q³ = p² q² * (-2 q)+3 p² q² = p² q² * 3So yes, first four terms can be factored as p² q² (2 pq - 2 p -2 q +3). Then the remaining terms are -p² q -p q² + pq.So numerator = p² q² (2 pq - 2 p -2 q +3) -p² q -p q² +pqNow, let's factor the remaining terms. Let's look at -p² q -p q² +pq. Factor out -pq:= -pq(p + q -1)Therefore, numerator = p² q² (2 pq - 2 p -2 q +3) - pq(p + q -1)So, numerator = p² q² (2 pq - 2 p -2 q +3) - pq(p + q -1)Now, factor out pq:Numerator = pq [ p q (2 pq -2 p -2 q +3) - (p + q -1) ]But let's expand p q (2 pq -2 p -2 q +3):= 2 p² q² -2 p² q -2 p q² +3 p qSo, numerator = pq [2 p² q² -2 p² q -2 p q² +3 p q -p -q +1]Therefore, numerator = pq [2 p² q² -2 p² q -2 p q² +3 p q -p -q +1]Now, S(pq) = numerator /6, so:S(pq) = [ pq (2 p² q² -2 p² q -2 p q² +3 p q -p -q +1) ] /6But the problem states that S(pq) can be written as (p² q² /6)(a - b/c). Let's see if we can manipulate the expression above into that form.Let's factor out p² q² from the terms inside the brackets. Let's write:2 p² q² -2 p² q -2 p q² +3 p q -p -q +1 = p² q² [2 - 2/(q) -2/(p) + 3/(p q) -1/(p q²) -1/(p² q) +1/(p² q²)]But that seems messy. Alternatively, perhaps it's better to factor terms step by step.Wait, let's see:We have:2 p² q² -2 p² q -2 p q² +3 p q -p -q +1We can group terms as follows:= 2 p² q² -2 p² q -2 p q² +3 p q -p -q +1= 2 p² q² -2 p² q -2 p q² + (3 p q - p - q) +1Hmm. Alternatively, perhaps we can write the expression as:2 p² q² -2 p² q -2 p q² +3 p q -p -q +1 = 2 p² q² -2 p² q -2 p q² +3 p q - (p + q) +1But I don't see an immediate factorization here. Maybe we can factor terms with p and terms with q.Alternatively, perhaps we can factor by grouping.Let me see:Group 2 p² q² -2 p² q -2 p q² +3 p q as:= 2 p² q² -2 p² q -2 p q² +3 p q= 2 p² q(q -1) -2 p q² +3 p q= 2 p² q(q -1) -2 p q² +3 p qFactor p q from the last two terms:= 2 p² q(q -1) + p q (-2 q +3)Hmm, not sure. Alternatively:Take the first two terms: 2 p² q² -2 p² q = 2 p² q (q -1)Next two terms: -2 p q² +3 p q = p q (-2 q +3)Then the remaining terms: -p - q +1So altogether:= 2 p² q (q -1) + p q (-2 q +3) - p - q +1Still not obvious. Alternatively, maybe we can factor out common terms.Alternatively, let's try to write the entire numerator expression as:2 p² q² -2 p² q -2 p q² +3 p q -p -q +1 = 2 p² q² -2 p² q -2 p q² +3 p q -p -q +1Let me check if this can be written as a product or in terms of (a - b/c). Let me see:Given that the desired form is (p² q² /6)(a - b/c). So, after factoring out p² q², the remaining terms would be multiplied by a, and then subtract some fraction b/c. Let's see:From S(pq) = [ pq (2 p² q² -2 p² q -2 p q² +3 p q -p -q +1) ] /6Let me factor out p² q² from the terms inside the brackets:= [ pq * p² q² (2 - 2/(q) -2/(p) +3/(p q) -1/(p² q) -1/(p q²) +1/(p² q²)) ] /6Wait, that's messy. Let's try to do it step by step.Let me factor p² q² from each term inside the brackets:2 p² q² = p² q² * 2-2 p² q = p² q² * (-2 / q)-2 p q² = p² q² * (-2 / p)3 p q = p² q² * (3 / (p q))-p = p² q² * (-1 / (p² q²)) * pWait, this approach might not be the best. Alternatively, let's consider:Express the entire expression as:2 p² q² -2 p² q -2 p q² +3 p q -p -q +1 = 2 p² q² -2 p² q -2 p q² +3 p q - (p + q -1)But perhaps that isn't helpful. Alternatively, let's look for the given form:We need S(pq) = (p² q² /6)(a - b/c). So, if we can write the numerator as p² q² (a) - something, then divide by 6.Looking back at the numerator:Numerator = pq [2 p² q² -2 p² q -2 p q² +3 p q -p -q +1]But if we want to write S(pq) as (p² q² /6)(a - b/c), then we need to have:Numerator /6 = (p² q² /6)(a - b/c)Therefore, Numerator = p² q² (a - b/c)But from above, Numerator = pq [ ... ].Therefore:pq [2 p² q² -2 p² q -2 p q² +3 p q -p -q +1] = p² q² (a - b/c)Divide both sides by pq:[2 p² q² -2 p² q -2 p q² +3 p q -p -q +1] = p q (a - b/c)Then:2 p² q² -2 p² q -2 p q² +3 p q -p -q +1 = a p q - (b/c) p qSo:2 p² q² -2 p² q -2 p q² +3 p q -p -q +1 = a p q - (b/c) p qHmm, this seems a bit complicated. Perhaps rearrange terms:Left side: 2 p² q² -2 p² q -2 p q² +3 p q -p -q +1We need to express this as a p q - (b/c) p q + ... ?Wait, but the right side is linear in p q, but the left side has higher degree terms. This suggests that perhaps my initial approach is not leading me directly to the desired form. Maybe there's a different formula or approach.Wait, another idea: For multiplicative functions, if f is multiplicative, then the sum over coprimes can be expressed as a product over primes. Since S(n) is the sum of squares of numbers coprime to n, maybe we can express S(n) using multiplicative functions. Since n = pq, product of two distinct primes, then S(n) = S(p) * S(q). But wait, is S(n) multiplicative?Wait, φ(n) is multiplicative, and the sum of coprimes is multiplicative. Let me check if the sum of squares of coprimes is multiplicative.If m and n are coprime, then S(mn) = S(m) * S(n). Is that true?Wait, for example, take m=2 and n=3. Then S(2) = 1² =1. S(3)=1² +2²=1+4=5. S(6)= sum of squares less than 6 and coprime to 6. Numbers coprime to 6 are 1,5. So S(6)=1+25=26. But S(2)*S(3)=1*5=5≠26. Therefore, S(n) is not multiplicative. Therefore, this approach won't work.So I need another method. Let's go back to the inclusion-exclusion result. Let's compute S(pq) using the formula we had:S(pq) = [2 p³ q³ -3 p² q² + pq -2 p² q³ -2 p³ q² +6 p² q² -p² q -p q²] /6Wait, but this might have been miscalculated. Wait, let me double-check the expansion:Wait, earlier, when expanding Term1 - Term2 - Term3, I had:Numerator = 2 p³ q³ -2 p³ q² -2 p² q³ +3 p² q² -p² q -p q² +pqWait, let's confirm this again step by step:Term1 was:2 p³ q³ - 3 p² q² + pqTerm2 was:-2 p² q³ +3 p² q² - p² qTerm3 was:-2 p³ q² +3 p² q² - p q²Wait, no. Wait, Term2 was p² q (2q² -3q +1) which expanded to 2 p² q³ -3 p² q² + p² q. Similarly Term3 was q² p (2p² -3p +1) = 2 p³ q² -3 p² q² + p q².Therefore, Term1 - Term2 - Term3 is:(2 p³ q³ -3 p² q² + pq) - (2 p² q³ -3 p² q² + p² q) - (2 p³ q² -3 p² q² + p q²)So expanding the subtraction:= 2 p³ q³ -3 p² q² + pq -2 p² q³ +3 p² q² - p² q -2 p³ q² +3 p² q² - p q²Now combine like terms:- Terms with p³ q³: 2 p³ q³- Terms with p³ q²: -2 p³ q²- Terms with p² q³: -2 p² q³- Terms with p² q²: (-3 +3 +3) p² q² = 3 p² q²- Terms with p² q: -p² q- Terms with pq²: -p q²- Terms with pq: +pqSo numerator is:2 p³ q³ -2 p³ q² -2 p² q³ +3 p² q² -p² q -p q² +pqSo that's correct. Now, perhaps instead of trying to factor this expression, let's try to relate it to the given form.The given form is (p² q² /6)(a - b/c). Let's compute S(pq) and then divide by (p² q² /6) to get (a - b/c). Let's compute:S(pq) = [2 p³ q³ -2 p³ q² -2 p² q³ +3 p² q² -p² q -p q² +pq]/6Divide numerator and denominator by p² q²:= [ (2 p³ q³)/(p² q²) -2 p³ q²/(p² q²) -2 p² q³/(p² q²) +3 p² q²/(p² q²) -p² q/(p² q²) -p q²/(p² q²) +pq/(p² q²) ] /6 * p² q² / (p² q²)Wait, perhaps better:Let me write S(pq) = [ numerator ] /6 = [2 p³ q³ -2 p³ q² -2 p² q³ +3 p² q² -p² q -p q² +pq]/6Factor out p² q²:= (p² q² /6) [ 2 pq - 2 p -2 q +3 - 1/(q) -1/(p) +1/(p q) ]Therefore:S(pq) = (p² q² /6) [ 2 pq -2 p -2 q +3 - (1/q +1/p -1/(p q)) ]So, inside the brackets, the expression is 2 pq -2 p -2 q +3 - (1/p +1/q -1/(p q))Therefore, we can write this as:2 pq -2 p -2 q +3 - (1/p +1/q -1/(p q)) = (2 pq -2 p -2 q +3) - ( (q + p -1)/(p q) )Therefore:S(pq) = (p² q² /6)[ (2 pq -2 p -2 q +3) - ( (p + q -1)/(p q) ) ]So this is in the form (p² q² /6)(a - b/c) where:a = 2 pq -2 p -2 q +3andb/c = (p + q -1)/(p q)But the problem states that a, b, c are positive integers, with b and c coprime and b < c. Therefore, we need to express (p + q -1)/(p q) as a fraction b/c in simplest terms.Given that p and q are primes, let's compute (p + q -1)/(p q). Since p and q are primes, and distinct, their gcd is 1. Let's check if p divides (p + q -1):p divides (p + q -1) iff p divides (q -1). Similarly, q divides (p + q -1) iff q divides (p -1). However, since p and q are primes greater than 2 (since they are Mersenne primes: 127 and 31, which are both primes greater than 2), so:Let me check if gcd(p + q -1, p q) is 1. Suppose there is a prime d dividing both p + q -1 and p q. Then d divides p or q.Case 1: d divides p. Then p divides p + q -1 => p divides q -1. But p =127, q =31. 127 divides 31 -1 =30? No, 127 does not divide 30. So no.Case 2: d divides q. Then q divides p + q -1 => q divides p -1. q =31. Does 31 divide 127 -1 =126? 31*4=124, 126-124=2. So 126 =31*4 +2. So no. Therefore, gcd(p + q -1, p q)=1. Hence, (p + q -1)/(p q) is already in reduced terms. So b = p + q -1 and c = p q. Since b < c? Let's check:p =127, q=31. p + q -1 =127 +31 -1=157. c =127*31=3937. So 157 < 3937, yes. So the fraction is 157/3937, which is already reduced. Therefore, a =2 pq -2 p -2 q +3.Therefore, we need to compute a =2 pq -2 p -2 q +3. Substitute p =127, q=31.First compute pq:127*31. Let's compute that:127*30=3810, 127*1=127, so 3810 +127=3937. Therefore pq=3937.So:a =2*3937 -2*127 -2*31 +3Compute each term:2*3937 =78742*127=2542*31=62So:a =7874 -254 -62 +3Compute step by step:7874 -254 =76207620 -62=75587558 +3=7561Therefore, a=7561.But let me confirm the calculation:Compute 2*3937: 3937*2=7874. Correct.2*127:127*2=254. Correct.2*31=62. Correct.7874 -254: 7874-200=7674, 7674-54=7620.7620 -62: 7620-60=7560, 7560-2=7558.7558 +3=7561. Correct.Therefore, the answer is 7561.But wait, let me confirm that I did everything correctly. Let me check the steps again:From the expression inside the brackets:2 pq -2 p -2 q +3 - ( (p + q -1)/(p q) )Which was converted into the form (a - b/c). Therefore, a is 2 pq -2 p -2 q +3, and b/c is (p + q -1)/(p q). Since a is already an integer (since all terms are integers), and b/c is the subtracted fraction. Therefore, the given form is correct, and a is indeed 2 pq -2 p -2 q +3. So substituting the given primes p=127 and q=31, we get a=7561.Therefore, the value of a is 7561.Final Answerboxed{7561}

✅:To find ( S(pq) ) where ( p = 127 ) and ( q = 31 ) are primes, we need to compute the sum of the squares of the positive integers less than and coprime to ( pq ). This can be expressed in the form ( frac{p^2q^2}{6}left(a - frac{b}{c} right) ).First, we use the principle of inclusion-exclusion to compute ( S(pq) ):1. Total sum of squares from 1 to ( pq - 1 ): [ T = sum_{k=1}^{pq-1} k^2 = frac{(pq - 1)pq(2pq - 1)}{6} ]2. Sum of squares divisible by ( p ): [ sum_{k=1}^{q-1} (pk)^2 = p^2 sum_{k=1}^{q-1} k^2 = p^2 frac{(q-1)q(2q-1)}{6} ]3. Sum of squares divisible by ( q ): [ sum_{k=1}^{p-1} (qk)^2 = q^2 sum_{k=1}^{p-1} k^2 = q^2 frac{(p-1)p(2p-1)}{6} ]Using inclusion-exclusion, we subtract the sums of squares divisible by ( p ) and ( q ), and add back the sum of squares divisible by ( pq ) (which is zero since ( pq ) is greater than ( pq - 1 )):[S(pq) = T - p^2 sum_{k=1}^{q-1} k^2 - q^2 sum_{k=1}^{p-1} k^2]Substituting the values and simplifying, we get:[S(pq) = frac{1}{6} left[ (pq - 1)pq(2pq - 1) - p^2 q (q - 1)(2q - 1) - q^2 p (p - 1)(2p - 1) right]]Expanding and simplifying the terms, we factor out ( p^2 q^2 ):[S(pq) = frac{p^2 q^2}{6} left[ 2pq - 2p - 2q + 3 - frac{p + q - 1}{pq} right]]This expression is in the form ( frac{p^2 q^2}{6} left( a - frac{b}{c} right) ) where ( a = 2pq - 2p - 2q + 3 ), ( b = p + q - 1 ), and ( c = pq ).Substituting ( p = 127 ) and ( q = 31 ):[a = 2 cdot 127 cdot 31 - 2 cdot 127 - 2 cdot 31 + 3]Calculating step-by-step:[a = 2 cdot 3937 - 254 - 62 + 3 = 7874 - 254 - 62 + 3 = 7561]Thus, the value of ( a ) is (boxed{7561}).

🔑:1. Identify the problem and given values: - We need to find ( S(pq) ), the sum of the squares of the positive integers less than and coprime to ( pq ). - Given ( p = 2^7 - 1 = 127 ) and ( q = 2^5 - 1 = 31 ).2. Use the Principle of Inclusion-Exclusion: - The positive integers less than or equal to ( pq ) are ( 1, 2, 3, ldots, pq ). - The positive integers less than or equal to ( pq ) that are divisible by ( p ) are ( p, 2p, 3p, ldots, pq ). - The positive integers less than or equal to ( pq ) that are divisible by ( q ) are ( q, 2q, 3q, ldots, pq ). - The only positive integer less than or equal to ( pq ) that is divisible by ( pq ) is ( pq ).3. Formulate the expression for ( S(pq) ): [ S(pq) = left( sum_{k=1}^{pq} k^2 right) - left( sum_{k=1}^{q} (kp)^2 right) - left( sum_{k=1}^{p} (kq)^2 right) + (pq)^2 ]4. Simplify each sum: - The sum of squares of the first ( n ) positive integers is given by: [ sum_{k=1}^{n} k^2 = frac{n(n+1)(2n+1)}{6} ] - Applying this formula: [ sum_{k=1}^{pq} k^2 = frac{pq(pq+1)(2pq+1)}{6} ] [ sum_{k=1}^{q} (kp)^2 = p^2 sum_{k=1}^{q} k^2 = p^2 cdot frac{q(q+1)(2q+1)}{6} ] [ sum_{k=1}^{p} (kq)^2 = q^2 sum_{k=1}^{p} k^2 = q^2 cdot frac{p(p+1)(2p+1)}{6} ]5. Substitute these into the expression for ( S(pq) ): [ S(pq) = frac{pq(pq+1)(2pq+1)}{6} - frac{p^2 q (q+1)(2q+1)}{6} - frac{pq^2 (p+1)(2p+1)}{6} + p^2 q^2 ]6. Factor out common terms and simplify: [ S(pq) = frac{pq}{6} left( (pq+1)(2pq+1) - p(q+1)(2q+1) - q(p+1)(2p+1) + 6pq right) ]7. Simplify the expression inside the parentheses: [ (pq+1)(2pq+1) = 2p^2q^2 + 3pq + 1 ] [ p(q+1)(2q+1) = p(2q^2 + 3q + 1) = 2pq^2 + 3pq + p ] [ q(p+1)(2p+1) = q(2p^2 + 3p + 1) = 2p^2q + 3pq + q ]8. Combine and simplify: [ 2p^2q^2 + 3pq + 1 - 2pq^2 - 3pq - p - 2p^2q - 3pq - q + 6pq = 2p^2q^2 - 2p^2q - 2pq^2 + 3pq - p - q + 1 ]9. Factor and simplify further: [ S(pq) = frac{pq}{6} left( 2p^2q^2 - 2p^2q - 2pq^2 + 3pq - p - q + 1 right) ] [ S(pq) = frac{p^2q^2}{6} left( 2pq - 2p - 2q + 3 - frac{p+q-1}{pq} right) ]10. Identify ( a ), ( b ), and ( c ): [ a = 2pq - 2p - 2q + 3 ]11. Calculate ( a ) for given ( p ) and ( q ): [ a = 2(127)(31) - 2(127) - 2(31) + 3 = 2(3937) - 254 - 62 + 3 = 7874 - 316 + 3 = 7561 ]The final answer is ( boxed{7561} )

🧠:Okay, so I have this problem about a rectangular board with m rows and n columns, where m is less than or equal to n. The condition is that some squares are colored black in such a way that no two rows are the same. Now, I need to find the biggest integer k such that no matter how the initial coloring is done (as long as the rows are distinct), I can always color k columns entirely red in a way that the rows still remain unique. Hmm. Let me try to parse this step by step. First, the board has m rows and n columns. All rows are initially different, meaning each row has a distinct pattern of black squares. The task is to determine the maximum k such that, regardless of how those rows are colored (as long as they're unique), we can always turn k columns red entirely, and after doing so, all the rows are still distinct. So the key here is that coloring a column red would override any black squares in that column for all rows. So if a column is colored red, every cell in that column becomes red, regardless of its original color. Therefore, coloring a column red effectively erases the information in that column for all rows. The problem is to find the largest k such that no matter the initial configuration (with distinct rows), there exists a set of k columns that we can color red, and the remaining n - k columns (which stay as they are) still ensure that all m rows are unique. Wait, actually, the problem says "color k columns entirely red in such a way that no two rows are still alike." So it's not that we remove the k columns, but we set those k columns to red, which might actually affect the remaining columns? Wait, no. If you color a column red, all entries in that column become red. So the original black squares in those columns are now red. So the remaining columns are the ones not colored red, which retain their original colors. Therefore, the remaining n - k columns will have their original colors (black or whatever, but the problem mentions only black initially), and the k columns are all red. Therefore, after coloring k columns red, each row is now a combination of the original black squares in the n - k columns and red squares in the k columns. The requirement is that even after this coloring, all m rows are still distinct. So the question is: What is the maximum k such that for any initial m x n matrix with distinct rows (using black squares), there exists a set of k columns that we can color red, resulting in the rows still being distinct in the remaining n - k columns. Alternatively, phrased another way: For any m x n matrix with distinct rows, there is a subset of n - k columns such that the projection of the rows onto these n - k columns is still injective (i.e., the rows remain distinct when only considering these columns). Therefore, the maximum k is the smallest number such that n - k is the minimal number of columns needed to distinguish all rows, regardless of the initial configuration. But since we need this to hold for any initial configuration, we need that in any m x n matrix with distinct rows, there exists a subset of n - k columns that still distinguishes all rows. Therefore, the maximum k is the minimal value such that n - k is the maximal minimal (over all matrices) size of a column subset that preserves row distinctness. Wait, maybe I need to rephrase. Since we need that for every possible colouring (with distinct rows), there exists a set of k columns to color red (i.e., remove those columns by making them all red, thereby leaving n - k columns). So we need that in any m x n matrix with distinct rows, there exists a subset of n - k columns that is still injective (i.e., the rows are distinct when restricted to these columns). Therefore, the maximum k is the minimal number such that in any such matrix, there exists a subset of n - k columns that preserves distinctness. Thus, the maximum k is n minus the minimal such n - k. Therefore, we need to find the minimal number t such that in any m x n matrix with distinct rows, there is a subset of t columns that distinguishes all rows. Then k = n - t. But the problem is to find the biggest k such that for every possible coloring, you can find k columns to color red (so t = n - k) such that the remaining t columns still have distinct rows. Therefore, the question is equivalent to: What's the minimal t such that every m x n matrix with distinct rows has a t-column subset that is still injective. Then k = n - t. But what is the minimal t where every such matrix has a t-column subset that distinguishes all rows. Then, since we need this t to work for all matrices, t has to be at least the maximum over all matrices of the minimal number of columns needed to distinguish the rows. But we need that t is such that even in the worst-case matrix, we can find t columns that suffice. Therefore, we need the minimal t such that in any m x n matrix with distinct rows, some t columns distinguish all rows. Then k = n - t. So the problem reduces to finding t, the minimal number such that any m x n matrix with distinct rows has a identifying set of t columns. Then the maximum k is n - t. So, what's known about this? It seems related to the concept of separating systems or distinguishing sets in combinatorics. In coding theory, similar concepts exist where you want a set of positions that allow you to distinguish codewords. Alternatively, this is similar to the set basis problem or the problem of finding a separating system. Let me recall. A separating system is a set system where for any two distinct elements, there is a set in the system that contains one but not the other. But here, it's slightly different. Here, each row is a vector in {0,1}^n (if we consider black as 1 and red as 0, or vice versa), and we need that projecting onto some t columns keeps them distinct. This is equivalent to saying that the set of vectors does not have a projection onto n - k coordinates that collapses two rows. So, we need that the minimal t such that any set of m distinct vectors in {0,1}^n has a projection onto some t coordinates that remains injective. But actually, since we can choose the t coordinates depending on the set of vectors, the question is: What is the minimal t such that for any set S of m distinct vectors in {0,1}^n, there exists a subset T of t coordinates such that the projection of S onto T is injective. Then, k = n - t. So the problem is equivalent to determining the minimal t such that any m distinct vectors in {0,1}^n have a injective projection onto some t coordinates. Then k = n - t. What is known about this? It's similar to the concept of VC-dimension, but perhaps more related to information theory or combinatorics. Let me think. For a set of vectors, the minimal size of a identifying code or a separating system. Wait, actually, the problem is equivalent to the identifying code in the Hamming space. An identifying code is a subset of coordinates such that the projection onto these coordinates uniquely identifies each element. But here, we don't fix the code; instead, for any set of vectors, we want that there exists such a code (subset of coordinates) of size t. Then, the minimal such t is the value we need. Alternatively, this is related to the Sauer-Shelah lemma. The Sauer-Shelah lemma states that a family of subsets with VC-dimension d has a projection onto some set of O(d log n) coordinates that preserves the shattering. But I don't know if that directly applies here. Alternatively, perhaps we can think in terms of the pigeonhole principle. Suppose that for a certain t, the number of possible projections is 2^t, so if m > 2^t, then by pigeonhole principle, two vectors would project to the same thing. But in our case, m can be as large as 2^n (if n is large). However, in the problem, m <= n. Wait, actually, m is given as the number of rows, with m <= n. Wait, m <= n. So the number of rows m is at most the number of columns n. So the problem is: Given m <= n, find the maximum k such that for any m x n matrix with distinct rows, there exists a subset of k columns that can be colored red (i.e., set to a constant value) such that the remaining n - k columns still have distinct rows. Alternatively, in coding theory terms, this is similar to puncturing a code: removing some coordinates such that the remaining code is still injective (i.e., the codewords remain distinct). So, the question is, given any code of length n with m codewords, what's the maximum number of positions k we can puncture (i.e., remove) such that the remaining code is still injective. But since we can choose which positions to puncture, the question becomes: What's the minimal number of positions t that must be retained (t = n - k) such that any code of length n with m codewords has a subset of t positions that form an injective code. So, to find the minimal t such that any m x n binary matrix with distinct rows has a t-column submatrix with distinct rows. Then, k = n - t. What is the minimal such t? Let me consider some examples. Let's start with small m and n. Case 1: m = 1. Then, since there's only one row, any t >= 0 would suffice. Therefore, t = 0, so k = n. But since m <= n, and m=1, n >=1. So in this case, the answer would be k = n - 0 = n. But the problem states "no two rows are alike," but if m=1, there are no two rows, so vacuously true. But maybe the problem requires m >=2? The problem says "some squares are coloured black in a way that no two rows are alike." If m=1, then trivially satisfied, and then you can color all columns red, so k = n. But maybe the problem allows m=1. Let's proceed. Case 2: m = 2, n = 2. So, 2x2 matrix with distinct rows. What's the minimal t? For any two distinct rows in a 2x2 matrix, is there a single column that distinguishes them? Yes. Because if two rows are different, they must differ in at least one column. Therefore, t =1, so k = n - t =1. But let's check. Suppose we have a matrix:Row 1: B BRow 2: B WBut wait, the problem states "some squares are coloured black", so maybe the other color is not white, but just not colored? Wait, the problem says "some squares are coloured black", so the rest are not colored? Or is it a binary matrix with black and white, and "coloring a column entirely red" would turn all entries in that column to red, overriding the original color. Wait, the problem says "some squares are coloured black in a way that no two rows are alike." So initially, some squares are black, others are not (maybe white or uncolored). Then, coloring a column entirely red would make all squares in that column red, regardless of their original color. Then, after coloring k columns red, the remaining columns retain their original color (black or not). Then, the requirement is that no two rows are alike in the resulting coloring. So, the colors after coloring are red and the original colors (black or not). But since the problem only mentions coloring some squares black initially, maybe the rest are considered non-black (maybe white). But when we color a column red, it's a third color? Or does red override black? Wait, the problem statement is a bit ambiguous. It says "color k columns entirely red". So maybe the original squares are either black or not, and coloring a column red would set all its entries to red, irrespective of their original color. Then, in the final coloring, each square is either red or its original color (black or white). However, the problem only mentions that the initial coloring has some squares black, so maybe the rest are white, and red is a third color. But the exact problem says: "some squares are coloured black in a way that no two rows are alike." So two rows are different if they differ in at least one square's color (black vs. not black). Then, when we color columns red, we're changing some squares to red, but the problem states "color k columns entirely red" so that "no two rows are still alike". Wait, perhaps the problem is in black and red, where initially some are black, and coloring a column red makes all in that column red. Then, the remaining columns are black or white? Or is the initial coloring only in black, and coloring a column red replaces black with red? This is a bit ambiguous, but maybe we can assume that the initial coloring is in black and white, with some squares black, others white, and coloring a column red turns all its squares to red. Then, the requirement is that after coloring, no two rows are the same in terms of their color patterns (considering red, black, white). However, the problem statement says "some squares are coloured black", so maybe the rest are not colored, and coloring a column red adds red color to it. But perhaps the problem is simpler: it's a binary matrix where entries are either black or white, and coloring a column red would set all entries in that column to red, overriding their original color. Then, two rows are considered different if they differ in at least one entry when considering the resulting colors (red, black, white). However, this complicates things because now we have three colors. Alternatively, maybe "coloring a column entirely red" is equivalent to ignoring those columns, similar to selecting a subset of columns to retain. But the problem says "color k columns entirely red", which might mean that those columns are now all red, but the others remain as they were. But if we have three colors, then two rows could be different in the original coloring (due to black and white) and could still be different after some columns are colored red if they differ in the remaining columns. However, the problem states that initially, some squares are colored black, and the rest are not, so maybe the initial matrix is black and white, with some blacks, and coloring a column red makes all those entries red. Then, the rows are considered different if their color sequences are different, considering red, black, and white. But perhaps the problem is intended to be a binary matrix where initially entries are either black or white (with no two rows the same), and coloring a column red would turn all entries in that column to red. Then, the resulting matrix has entries that are either red or the original color (black or white). But this interpretation might be complicating it. Alternatively, maybe all squares are initially either black or white, with no two rows the same. Coloring a column red changes all entries in that column to red, and we need that after this change, no two rows are the same. So, in effect, we're allowed to override certain columns to a constant color (red), and we need that the resulting matrix still has distinct rows. In this case, the problem is equivalent to: Given a matrix with m distinct rows over an alphabet of size 2 (black and white), find the maximum k such that there exists a set of k columns to set to a constant symbol (red), so that the remaining n - k columns still form a matrix with distinct rows. This is similar to the problem of, in coding theory, puncturing a code at certain positions but with the ability to choose which positions to puncture (or in this case, set to a constant). The question is about the maximum number of positions we can set to a constant such that the remaining positions still form a code with distinct codewords. In this interpretation, the problem is about binary codes (black and white), and setting k columns to red (a third symbol) such that all codewords remain distinct. However, if red is considered a third symbol, then setting a column to red increases the alphabet size, but the problem states "color k columns entirely red". However, the original problem mentions that some squares are colored black, implying the rest are not, so perhaps the original matrix is over two colors (black and white), and red is a third color. But maybe another interpretation: the original matrix is over two colors (black and white), and coloring a column red is equivalent to erasing or ignoring that column. That is, by coloring a column red, you are effectively removing it, similar to selecting n - k columns to keep. If that's the case, then the problem reduces to: Given any m x n matrix with distinct rows, find the maximum k such that there exists a subset of n - k columns such that the rows are still distinct in those columns. Then, the maximum k is n minus the minimal size of such a subset. But according to this interpretation, the problem is equivalent to finding the minimal t such that any m x n matrix with distinct rows has a subset of t columns that are still injective (i.e., rows are distinct on those t columns). Then, k = n - t. This is a well-known problem in combinatorics and coding theory. The minimal such t is called the identifying code or separating system. For a set of m vectors in {0,1}^n, the minimal size t of a subset of coordinates such that the projection onto these coordinates is injective. In the worst case, how large does t need to be to ensure that such a subset exists for any set of m vectors. For example, if m = 2, then t needs to be at least 1, because two different vectors must differ in at least one coordinate. Therefore, there exists a coordinate where they differ, so t = 1 suffices. Hence, for m = 2, k = n - 1. Similarly, for m = 3, what's the minimal t? Let's see. Suppose we have three distinct vectors in {0,1}^n. We need a subset of columns such that all three vectors are distinct on those columns. The minimal t here would be 2. Because with t = 2, there are 4 possible projections, so three vectors can be distinguished. But is one coordinate sufficient? Suppose three vectors differ in one coordinate: for example, vectors 000, 001, 000. Wait, but they must be distinct, so in at least one coordinate they differ. Wait, but three distinct vectors might require two coordinates. For example, consider vectors 00, 01, 10. If we project onto the first coordinate, we get 0, 0, 1. Then the first two vectors collapse. If we project onto the second coordinate, we get 0,1,0. Similarly, the first and third collapse. So, in this case, no single coordinate suffices to distinguish all three vectors, but projecting onto both coordinates works. Hence, t=2 is needed. Therefore, for m=3, t=2, so k = n - 2. Similarly, for m=4, you might need t=2, since with two coordinates, you have four possible projections. So if you have four distinct vectors, they can be mapped injectively onto two coordinates. Wait, but no. For example, consider four vectors in {0,1}^3:000001010011If you project onto the first two coordinates, you get 00, 00, 01, 01. So two pairs are collapsed. If you project onto the last two coordinates, you get 00, 01, 10, 11, which are all distinct. Therefore, in this case, t=2 suffices. But if the four vectors are arranged such that no two coordinates can distinguish them all, does such a set exist? Wait, suppose we have four vectors in {0,1}^4:00000001001000110100But no, even in this case, projecting onto the last two coordinates might distinguish them. Wait, actually, in general, for m=4, if the vectors are distinct, can we always find two coordinates that distinguish them? Wait, let's take four vectors in {0,1}^n:v1 = 000...0v2 = 100...0v3 = 010...0v4 = 001...0If we project onto the first three coordinates, each has a single 1 in different positions, so projecting onto coordinates 1,2,3 would distinguish them. But can we do it with two coordinates? Suppose we take coordinates 1 and 2. Then v1 is 00, v2 is 10, v3 is 01, v4 is 00. So v1 and v4 collapse. Similarly, coordinates 1 and 3: v1=00, v2=10, v3=00, v4=01. So v1 and v3 collapse. Coordinates 2 and 3: v1=00, v2=00, v3=10, v4=01. So v1 and v2 collapse. Hence, in this case, two coordinates are insufficient to distinguish all four vectors. Therefore, we need at least three coordinates. Wait, but in this case, with three coordinates, projecting onto the first three coordinates gives:v1: 000v2: 100v3: 010v4: 001Which are all distinct. So t=3. Hence, for m=4, in this example, t=3 is needed. But wait, the vectors are in {0,1}^n, and if n >=3, then we can use three coordinates, but if n is larger, can we do better? No, because the example is constructed in such a way that each vector has a single 1 in a different coordinate. Therefore, in this case, to distinguish them, you need all the coordinates where their 1s are. Hence, t=3. But is there a way to choose three coordinates for this example? Yes, as above. But if n is larger, say n=4, and the vectors are:v1: 0000v2: 1000v3: 0100v4: 0010Then, to distinguish them, we need the first three coordinates, so t=3. So regardless of n, for this example, t=3. But wait, in this case, if n is larger, maybe you can choose a different set of coordinates. For example, in n=4, maybe project onto coordinates 1,2,4. But for v4, it's 0010, so projecting onto 1,2,4 would give 001 for v4. But v1 is 0000, projection 000, v2: 1000 -> 100, v3: 0100 -> 010, v4: 0010 -> 001. So in this case, projecting onto coordinates 1,2,4 gives distinct projections. So t=3. Therefore, in this case, t=3 is sufficient. However, if the vectors are arranged such that each has a unique 1 in a different position, then any three coordinates that include the positions of the 1s would work. But even so, t=3 is sufficient. Therefore, for m=4, in the worst case, t=3 is needed, hence k = n - 3. Wait, this suggests that for m vectors, each with a unique single 1 in different positions, the minimal t is m -1? No, in the above case, m=4, t=3. Wait, m=4, t=3. So t = m -1? But for m=2, t=1= m -1. For m=3, earlier example required t=2= m -1. For m=4, t=3= m -1. So perhaps in general, t = m -1. Hence, the minimal t is m -1, hence k = n - (m -1) = n - m +1. But wait, let's check for m=3. If we have three vectors with single 1s in different coordinates, then t=2. If we have three vectors in general positions, is t=2 always sufficient? Let's take another example. Suppose three vectors:000011001111Can we find two columns that distinguish them? Let's see. Columns 1 and 2: projections are 00, 11, 11. So the second and third vectors collapse. Columns 1 and 3: 00, 10, 11. All distinct. So t=2 suffices here. Another example: three vectors:000010100101Can we find two columns that distinguish them? Columns 1 and 2: 00, 10, 01 – all distinct. So yes, t=2. Another example: three vectors:110010101001Can we find two columns? Let's check columns 1 and 2: 11, 10, 10 – collapses second and third. Columns 1 and 3: 10, 11, 10 – collapses first and third. Columns 1 and 4: 10, 10, 11 – collapses first and second. Columns 2 and 3: 10, 01, 00 – all distinct. So t=2. Hence, it seems for m=3, t=2 suffices, which is m -1=2. Similarly, for m=4, t=3= m -1. So maybe in general, t= m -1. Therefore, the minimal t is m -1, hence the maximum k= n - (m -1)=n - m +1. But wait, let's test with m=4. Suppose we have four vectors:00001000010000100001Wait, but this is five vectors. For four vectors:v1: 0000v2: 1000v3: 0100v4: 0010Then, to distinguish them, you need three coordinates. For example, columns 1,2,3: projections 000, 100, 010, 001. All distinct. But can you do it with two columns? Let's see. Suppose we pick columns 1 and 2: v1=00, v2=10, v3=01, v4=00. Collapses v1 and v4. Columns 1 and 3: v1=00, v2=10, v3=00, v4=01. Collapses v1 and v3. Columns 2 and 3: v1=00, v2=00, v3=10, v4=01. Collapses v1 and v2. Columns 3 and 4: v1=00, v2=00, v3=10, v4=01. Still collapses v1 and v2. So no two columns suffice. Hence, t=3 is necessary. Therefore, for four vectors with single 1s in different columns, t=3= m -1. So generalizing, if you have m vectors each with a single 1 in distinct columns, you need t= m -1 columns to distinguish them. Because if you choose any m -1 columns, one of the 1s will be excluded, so two vectors will have their 1s in the excluded column and hence project to all zeros in the remaining columns. But wait, in the example above, for four vectors with single 1s in columns 1-4, if you choose columns 1,2,3, you can distinguish the first three vectors, and the fourth vector (with 1 in column 4) would project to 000, same as the first vector. Hence, you need to include column 4 to distinguish it. Wait, but in the example, the vectors are:v1: 0000v2: 1000v3: 0100v4: 0010Wait, in this case, to distinguish all four, you need columns 1,2,3. Because v4 is 0010, so its projection onto columns 1,2,3 is 001, which is distinct from others. Wait, but in this case, with columns 1,2,3:v1: 000v2: 100v3: 010v4: 001All distinct. So t=3 suffices. But in the previous example where the vectors had 1s in different columns, you need t= m -1? Wait, here m=4, t=3= m -1. Yes. But if you have more vectors, say m=5, with each vector having a single 1 in different columns, then you need t=4= m -1. So in general, t= m -1. However, the problem states that m <= n. So if n >= m, then it's possible to have m vectors each with a single 1 in distinct columns. Hence, for such a case, t= m -1 is necessary. Therefore, in the worst case, t= m -1, hence k= n - (m -1)=n - m +1. But we need to confirm that this is indeed the maximum k, i.e., that for any matrix with m distinct rows, you can always find k= n - m +1 columns to color red such that the remaining m -1 columns keep the rows distinct. But is this always true? Let's see. Suppose we have an arbitrary m x n matrix with distinct rows. We need to show that there exists a subset of n - m +1 columns to color red (i.e., leave m -1 columns) such that the rows are still distinct. Alternatively, for any m distinct vectors in {0,1}^n, there exists a subset of m -1 coordinates such that the projections are still distinct. Is this true? Wait, consider the following example: Let m=3, n=3. So we have a 3x3 matrix. Suppose the rows are:100010001Each row has a single 1 in a different column. Then, if we try to remove one column (k=1), leaving two columns. For example, remove column 1: the rows become 00, 10, 01 – distinct. Similarly, remove column 2: rows become 10, 00, 01 – distinct. Remove column 3: rows become 10, 01, 00 – distinct. So in this case, k=1=3 -3 +1=1, which works. But if we take another example with m=3, n=3:Rows:111110101These are distinct rows. Can we find a single column to remove such that the remaining two columns keep the rows distinct. Let's see:Remove column 1: remaining columns 2 and 3:111001 – all distinct.Remove column 2: remaining columns 1 and 3:111011 – rows 1 and 3 collapse.Remove column 3: remaining columns 1 and 2:111110 – rows 1 and 2 collapse.So in this case, if we remove column 1, the remaining columns are okay, but removing column 2 or 3 causes collapses. Hence, there exists a column whose removal preserves distinctness. So even though in some choices you get collapses, there exists at least one choice (column 1) that works. Hence, k=1=3-3+1=1. Another example: m=3, n=4. Suppose we have three vectors:101001011111Can we remove two columns (k=2) to leave two columns that distinguish the rows. Let's try removing columns 3 and 4:Remaining columns 1 and 2:100111 – distinct.Alternatively, removing columns 1 and 2:Remaining columns 3 and 4:100111 – also distinct.So in this case, k=2=4 -3 +1=2. Therefore, it seems that the formula k= n - m +1 holds. To generalize, for any m <= n, the maximum k is n - m +1. This is equivalent to saying that in any m x n matrix with distinct rows, you can always remove m -1 columns such that the remaining n - m +1 columns still keep the rows distinct. But wait, no. Wait, the formula is k= n - t, where t is the minimal number of columns needed. If t= m -1, then k= n - (m -1)=n -m +1. But in the previous examples, we saw that t= m -1 is sufficient, but when we had m=4, t=3= m -1. But does this hold in general? Let's see. Suppose we have m distinct vectors in {0,1}^n. We need to find a subset of m -1 columns such that the projections are still injective. Is this always possible? This resembles the set basis theorem. In linear algebra, any set of m vectors in an n-dimensional space contains a subset of m vectors that are linearly independent, but that's not directly applicable here since we're dealing with injectivity rather than linear independence. Alternatively, in combinatorics, there's a theorem called the Sauer-Shelah lemma, but that pertains to shattering sets, which might not be directly applicable. Alternatively, consider building a tree of possible column selections. Start with the first column. If this column distinguishes all rows, then we're done with t=1. Otherwise, it partitions the rows into equivalence classes (based on the value in that column). Then, for each equivalence class, we need to further distinguish them using other columns. But this approach is similar to building a decision tree where each node tests a column and splits the rows based on their value in that column. The minimal depth required to distinguish all rows. However, the problem here is not about the depth but the number of columns needed. Alternatively, consider that each column can split the set of rows into at most two groups (based on 0 or 1 in that column). To distinguish m rows, we need enough splits such that each row is in a singleton set. The number of splits needed is at least log m. But this gives a lower bound, not an upper bound. However, in our previous examples, t= m -1 suffices, which is much larger than log m for small m. But perhaps the upper bound is indeed m -1. Let's consider the following argument. Suppose we have m distinct vectors. We can construct a set of columns incrementally as follows. Start with an empty set. While the current set of columns does not distinguish all pairs of rows, add a column that distinguishes a new pair of rows. Initially, no columns are chosen, so all rows are the same. When we add the first column, it can split the rows into two groups. Each subsequent column can potentially split existing groups further. However, the maximum number of pairs of rows is C(m,2). Each column can distinguish up to C(m,2) pairs. However, this is not helpful. Alternatively, using an inductive argument. For m=1, t=0. For m=2, t=1. Assume for m, t= m -1. Then for m+1, adding a new row, we need to distinguish it from the existing m rows. Since the existing m rows are distinguished by t= m -1 columns, to distinguish the new row from each of the m existing rows, we need at most m additional columns. But this seems too much. Alternatively, think of each column as a function that assigns a 0 or 1 to each row. To uniquely identify each row, we need a set of functions (columns) such that the combination uniquely assigns each row. This is similar to a separating system. In fact, a (m, n, k) separating system is a set system where for any two elements, there is a set in the system that contains one but not the other. In our case, the "sets" are the columns, and each column corresponds to the set of rows where the column has a particular value (say 1). Then, a separating system requires that for any two rows, there is a column where one has 1 and the other has 0, which is exactly what we need for the projections to be injective. However, the size of the separating system (number of columns needed) is a known problem. For a separating system that can distinguish any two elements, the minimal size is known to be on the order of log m. However, in our case, we are not required to have a separating system that works for all pairs, but rather to have a set of columns such that the projections are injective. These two concepts are related but not the same. A separating system ensures that every pair is separated by at least one set, while an injective projection requires that all pairs are simultaneously separated by the selected sets (columns). The minimal size of such a set is known in combinatorics. For binary vectors, the minimal t such that any m distinct vectors have a t-sized separating system is upper bounded by m -1. This is because you can construct a matrix where each successive column splits at least one more pair, leading to a cumulative effect where after m -1 columns, all pairs are split. However, this might not hold. For example, consider m=4 vectors in {0,1}^n. If the vectors are:0000000100100011Then, projecting onto the fourth column gives 0,1,0,1, which doesn't distinguish all pairs. Projecting onto the third and fourth columns:00,01,10,11, which are distinct. So t=2 suffices here, which is less than m -1=3. Therefore, the required t can be less than m -1 depending on the structure. However, in the worst case, such as when the vectors each have a unique 1 in a distinct column, we saw that t= m -1 is necessary. Therefore, the minimal t required in the worst case is m -1. Hence, for any m <= n, the maximum k is n - (m -1) =n - m +1. Thus, the answer should be k= n -m +1. But let's verify with another example. Take m=2, n=3. Then k=3 -2 +1=2. So we should be able to color 2 columns red, leaving 1 column. But for any 2x3 matrix with distinct rows, is there a single column that distinguishes them? Yes. Because if two rows are different, they must differ in at least one column. Hence, there exists at least one column that can be left uncolored (i.e., not colored red) which distinguishes the two rows. Therefore, k=2, which is 3 -2 +1=2. This works. Another example: m=3, n=4. Then k=4 -3 +1=2. So we need to color 2 columns red, leaving 2 columns. For any 3x4 matrix with distinct rows, there must exist 2 columns that distinguish all three rows. Is this true? Suppose we have three rows:000011110001Can we find two columns that distinguish them? Let's check columns 1 and 4:Row1: 00Row2: 11Row3: 01These are all distinct. Hence, t=2 suffices. Another example:Rows:000011101101Can we find two columns? Columns 3 and 4:Row1:00Row2:00Row3:01 – collapses rows 1 and 2. Columns 1 and 2:Row1:00Row2:11Row3:11 – collapses rows 2 and 3. Columns 1 and 3:Row1:00Row2:10Row3:10 – collapses rows 2 and 3. Columns 1 and 4:Row1:00Row2:10Row3:11 – all distinct. So here, columns 1 and 4 work. Hence, t=2 suffices. Therefore, it seems that even in cases where some pairs of columns collapse, there exists another pair that works. Thus, the general formula k= n -m +1 holds. Therefore, the biggest integer k is n - m +1. But wait, let's check the case where n =m. Then k= n -m +1=1. So if we have an m x m matrix with distinct rows, we can color 1 column red, leaving m -1 columns. But in that case, is there always a column that can be removed such that the remaining m -1 columns keep the rows distinct? For example, take m=3, n=3:Rows:100010001Removing any column leaves two columns. For example, removing column 1 leaves columns 2 and 3 with rows 00,10,01 – distinct. Similarly for other columns. Hence, k=1 works. Another example:Rows:111110101Removing column 1 leaves:111001 – distinct. So k=1 works. But what if:Rows:123456789Wait, in binary. Let's think of binary matrices. Another example:Rows:111110100Removing column 3:111110 – collapses first and second rows. Removing column 2:111010 – collapses second and third. Removing column 1:111000 – distinct. So in this case, column 1 can be removed. Hence, k=1 works. Thus, even in n=m, k=1 works. Therefore, the general formula is k= n -m +1. Hence, the answer is boxed{n - m + 1}

✅:To solve the problem of finding the largest integer ( k ) such that for every possible coloring of an ( m times n ) board (with ( m leq n )) where no two rows are alike, one can always color ( k ) columns entirely red while keeping the rows distinct, we need to consider the following steps:1. Understanding the Problem: We need to ensure that after coloring ( k ) columns red, the remaining ( n - k ) columns still keep all rows distinct. This translates to finding the minimal number of columns ( t ) such that any ( m times n ) matrix with distinct rows has a subset of ( t ) columns that distinguish all rows. The value ( k ) is then ( n - t ).2. Key Insight: For any set of ( m ) distinct rows, there must exist a subset of columns that can distinguish all rows. The minimal number of such columns ( t ) is ( m - 1 ). This is because in the worst case, each row can be uniquely identified by a single distinct column, and we need ( m - 1 ) columns to ensure all rows remain distinct.3. General Formula: The maximum ( k ) is given by ( k = n - t ), where ( t = m - 1 ). This leads to the formula ( k = n - (m - 1) = n - m + 1 ).4. Verification with Examples: Various examples confirm that this formula holds. For instance, if ( m = 2 ), ( n = 3 ), then ( k = 3 - 2 + 1 = 2 ), meaning we can color 2 columns red and leave 1 column to distinguish the rows. This works because any two distinct rows must differ in at least one column.Thus, the biggest integer ( k ) such that for every possible coloring, one can always color ( k ) columns entirely red while keeping the rows distinct is:[boxed{n - m + 1}]

🔑:1. Define the problem and notation: We are given a rectangular board with ( m ) rows and ( n ) columns, where ( m leq n ). Some squares are colored black (denoted by 1) and others are white (denoted by 0). The goal is to find the largest integer ( k ) such that for any initial coloring, we can always color ( k ) columns entirely red (effectively removing them) in such a way that no two rows are still alike.2. Initial observation: Coloring a column red corresponds to removing that column from consideration. We need to ensure that after removing ( k ) columns, the remaining columns still have unique row patterns.3. Upper bound for ( k ): To show that anything greater than ( n - m + 1 ) doesn't work, consider the matrix ( X ) where ( x_{ij} = 1 ) if and only if ( i = j ). This is essentially an identity matrix of size ( m times m ) embedded in a larger ( m times n ) matrix. If we remove more than ( n - m + 1 ) columns, we are left with fewer than ( m ) columns. In this case, by the pigeonhole principle, at least two rows must be identical, which contradicts the requirement that no two rows are alike.4. Lower bound for ( k ): To show that ( n - m + 1 ) indeed works, we need to demonstrate that the difference in the rows can be captured in ( m ) columns. We can use induction on ( m ): - Base case: For ( m = 1 ), any ( n geq 1 ) will work since there is only one row, and it is trivially unique. - Inductive step: Assume that for ( m ) rows, we can always find ( n - m + 1 ) columns to remove such that the remaining rows are unique. Now consider ( m + 1 ) rows. We need to show that we can remove ( n - (m + 1) + 1 = n - m ) columns and still have unique rows. Given ( m + 1 ) rows, there are at most ( 2^{m+1} ) possible row patterns. Since no two rows are alike initially, we have exactly ( m + 1 ) unique patterns. By the inductive hypothesis, we can remove ( n - m ) columns from the first ( m ) rows and still have unique rows. The remaining ( m + 1 )th row must also be unique among the remaining rows, as it was initially unique.5. Conclusion: Therefore, the largest integer ( k ) such that for any initial coloring, we can always color ( k ) columns entirely red and still have unique rows is ( n - m + 1 ).The final answer is ( boxed{n - m + 1} ).

🧠:Okay, so I need to figure out the possible values of L where L is the limit of φ(n)/n as n approaches infinity, and φ is a bijection from the natural numbers to themselves. Hmm, let's start by understanding the problem step by step.First, φ is a bijection on N, which means it's a permutation of the natural numbers. So every natural number appears exactly once in the sequence φ(1), φ(2), φ(3), etc. Since it's a bijection, it must be both injective and surjective. So, for every m in N, there exists exactly one n in N such that φ(n) = m.Now, the question is about the limit L = lim_{n→∞} φ(n)/n. We need to find all possible values of L given that φ is a bijection. The problem states that the limit exists and is finite. So, we need to determine what L can be.Let me think about what such a limit would imply. If φ(n) grows faster than n, then L would be infinity, but since the problem states the limit is finite, L must be some real number. But since φ is a permutation, maybe the limit has to be 1? Or could it be something else?Wait, let's consider some examples. Let's start with the identity permutation, where φ(n) = n for all n. Then φ(n)/n = 1, so the limit is 1. That's straightforward. So 1 is definitely a possible value of L.But the question is asking for all possible values of L. So maybe there are other permutations where the limit is different?Let's try to think of a permutation where φ(n) grows slower or faster than n. For example, suppose we swap some numbers around. If we swap 1 and 2, 3 and 4, etc., does that affect the limit? Let's see. For example, if φ(1)=2, φ(2)=1, φ(3)=4, φ(4)=3, and so on. Then for even n=2k, φ(n)=2k-1, and for odd n=2k-1, φ(n)=2k. Then φ(n)/n would oscillate between (2k-1)/(2k) and 2k/(2k-1). As k increases, (2k-1)/(2k) approaches 1, and 2k/(2k-1) approaches 1 as well. So even though we have oscillations, both subsequences approach 1, so the overall limit would still be 1. So in this case, L=1.Hmm, so even with some swapping, the limit remains 1. Is there a way to construct a permutation where the limit is different?Wait, let's try to make φ(n) grow faster. For example, suppose we pair each n with 2n. But wait, that's not a bijection because we're skipping the odd numbers. To make it a bijection, we need to map some numbers to the evens and others to the odds. For example, maybe φ(n) = 2n when n is even, and φ(n) = (n-1)/2 when n is odd. But let's check if that's a bijection. Let's see: For even n, φ(n)=2n, which maps to multiples of 4. For odd n, φ(n)=(n-1)/2. Wait, if n is odd, then n = 2k+1, so φ(2k+1) = (2k+1 -1)/2 = k. So φ maps odd numbers to N as well. But is this a bijection? Let's check. For example, take m=1. Is there an n such that φ(n)=1? If n is odd, then φ(n)=k where n=2k+1. So if k=1, then n=3. φ(3)=1. Similarly, m=2: φ(n)=2 would require either n even, so 2n=2 => n=1, but n=1 is odd, so φ(1)= (1-1)/2=0, which is not in N. Alternatively, maybe there's a mistake here. Wait, maybe my example is not a valid bijection.Wait, let's think again. If φ(n) is defined as 2n for even n, and for odd n, it's something else. Let's say for odd n, φ(n) maps to the odd numbers. Wait, but 2n for even n already covers the even numbers. So if we want a bijection, the odd n's must map to the odd numbers. So maybe for odd n, φ(n) is the (n+1)/2-th odd number. Wait, but that might not be a bijection. Let me try to formalize this.Alternatively, maybe arrange the natural numbers such that φ alternates between small and large numbers. For example, arrange the permutation such that φ(1)=1, φ(2)=2, φ(3)=4, φ(4)=3, φ(5)=6, φ(6)=5, etc. Then φ(n)/n for even n=2k would be (2k-1)/2k, which approaches 1, and for odd n=2k+1, φ(n)=2k+2, so (2k+2)/(2k+1) which approaches 1 as well. So again, the limit would be 1.Hmm, maybe it's challenging to get a different limit. Let's consider another approach. Suppose we define a permutation that in some way "spreads out" the numbers. For example, for some increasing sequence a_n, we map n to a_n, and ensure that a_n is a permutation. But then, the ratio a_n/n must approach L. However, since it's a permutation, the sequence a_n must cover every natural number exactly once. So, for example, if a_n is roughly Ln, but since it's a permutation, each number must be used exactly once. So maybe L must be 1. Let me think.Suppose that L is greater than 1. Then φ(n) ≈ Ln for large n, so φ(n) would be approximately a linear function with slope L > 1. But since φ is a permutation, every number must appear exactly once. However, if φ(n) ≈ Ln, then the density of the image would be lower than the natural numbers, but since φ is a bijection, it must cover all natural numbers. If L > 1, then the numbers φ(n) would grow too fast, leaving gaps in the natural numbers. For example, if L=2, then φ(n) ≈ 2n, so the image would be approximately the even numbers, but that's not a bijection because we miss the odd numbers. Similarly, if we try to interleave numbers, but the limit would still approach 1.Alternatively, suppose L < 1. Then φ(n) ≈ Ln would be growing slower than n, so φ(n) would be smaller than n for large n. But since φ is a permutation, the numbers must cover all natural numbers. If φ(n) is growing slower, that would mean that for large n, φ(n) is small, but we need to have already used all the small numbers in earlier terms. Therefore, such a permutation can't exist because once you map some n to small numbers, you can't map later n's to them again. Hence, it's impossible for φ(n) to be growing slower than n in a bijection. Therefore, perhaps L must be 1.Wait, let's formalize this intuition. Suppose that lim_{n→∞} φ(n)/n = L. Since φ is a bijection, for every m, there exists an n such that φ(n) = m. Let's consider the inverse function φ^{-1}(m) = n such that φ(n) = m. Since φ is a bijection, φ^{-1} is also a permutation.Given that φ(n)/n → L, then n / φ(n) → 1/L. But φ^{-1}(m)/m = n/m where φ(n) = m. If m = φ(n), then φ^{-1}(m) = n. Therefore, φ^{-1}(m)/m = n / φ(n) = 1 / (φ(n)/n). So if φ(n)/n → L, then φ^{-1}(m)/m → 1/L. But φ^{-1} is also a permutation, so applying the same logic, the limit for φ^{-1}(m)/m must also satisfy the same constraints. Therefore, if L exists, then 1/L must also exist and be a limit of φ^{-1}(m)/m. But since φ^{-1} is also a permutation, what can 1/L be?If L exists and is finite, then 1/L must also be the limit of a permutation ratio. But from the previous intuition, if L > 1, then 1/L < 1, which would imply that φ^{-1}(m)/m approaches 1/L < 1. But this would mean that φ^{-1}(m) ≈ m/L, which would require that φ^{-1}(m) grows slower than m, which as before, seems impossible because φ^{-1} is a permutation. Therefore, this suggests that L must equal 1.Alternatively, if L = 1, then 1/L = 1, which is consistent. If we suppose L ≠ 1, then we get a contradiction because one direction would require φ(n) to grow faster (if L > 1) and the inverse function φ^{-1}(m) to grow slower (1/L < 1), which is impossible as permutations cannot have both properties. Hence, L must be 1.Is there a rigorous way to prove that L must equal 1?Let me think. Suppose that lim_{n→∞} φ(n)/n = L. We need to show that L must equal 1.First, note that for a permutation φ of N, the inverse function φ^{-1} is also a permutation. If we have that φ(n) ≈ Ln for large n, then φ^{-1}(m) ≈ m/L for large m. However, since φ^{-1}(m) must be a natural number for each m, and the values of φ^{-1}(m) must cover all natural numbers exactly once, this suggests that the density of φ(n) around Ln must match the density of N.But the natural numbers have density 1, meaning that in the limit, the number of terms up to N is N. If φ(n) ≈ Ln, then the number of terms φ(n) ≤ M would be approximately M/L. But since φ is a bijection, the number of terms φ(n) ≤ M must be exactly M. Therefore, M ≈ M/L, which implies that L = 1.Wait, that seems promising. Let me formalize this. Let’s consider the counting function. For the permutation φ, the number of n such that φ(n) ≤ M is exactly M, since φ is a bijection. But if φ(n) ≈ Ln, then the number of n such that φ(n) ≤ M is approximately the number of n such that Ln ≤ M, which is approximately M/L. But since this must equal M (because φ is a bijection), we have M ≈ M/L, so L = 1.Yes, that makes sense. Therefore, the only possible value for L is 1.Alternatively, using the concept of natural density. The natural density of a set A ⊂ N is defined as lim_{M→∞} |A ∩ {1, 2, ..., M}| / M. For the entire set N, the natural density is 1.If φ is a permutation, then the image φ({1, 2, ..., N}) is just {1, 2, ..., N} rearranged, so the density is preserved. However, if φ(n) ~ Ln, then the preimage of {1, 2, ..., M} under φ would be approximately M/L numbers n such that φ(n) ≤ M. But since φ is a bijection, exactly M numbers n satisfy φ(n) ≤ M. Therefore, M/L ≈ M ⇒ L=1.This seems like a solid argument. Therefore, the only possible value for L is 1.Another way to think about it is via the Stolz–Cesàro theorem. If we consider the limit lim φ(n)/n, suppose that φ(n) is increasing (though φ itself is a permutation, it's not necessarily increasing). But since φ is a bijection, it's not monotonic unless it's the identity. However, if the limit exists, then maybe we can apply some form of averaging.Alternatively, consider the average of φ(k)/k for k=1 to N. If the limit of φ(n)/n is L, then the average should also approach L. But the average is (1/N) Σ_{k=1}^N φ(k)/k. However, since φ is a permutation, the sum Σ_{k=1}^N φ(k) is equal to Σ_{k=1}^N k = N(N+1)/2. Therefore, Σ_{k=1}^N φ(k)/k = Σ_{k=1}^N k / k = Σ_{k=1}^N 1 = N. Wait, that can't be right. Wait, φ is a permutation, so Σ_{k=1}^N φ(k) = Σ_{k=1}^N k. So Σ_{k=1}^N φ(k)/k = Σ_{k=1}^N k / φ^{-1}(k). Hmm, maybe this approach is getting too convoluted.But going back to the counting argument. If φ(n) ~ Ln, then the inverse function φ^{-1}(m) ~ m/L. So the number of n ≤ M/L is approximately M. But the actual number of n ≤ M/L is floor(M/L), which should be approximately equal to M. This is only possible if L=1.Yes, this seems to confirm the previous conclusion. Therefore, the only possible value for L is 1.Wait, let's check with a specific example where someone might think L could be different. Suppose we construct a permutation where φ(n) = n for all n except that we swap n and 2n for each n. But this isn't a permutation because swapping n and 2n would require that each swap is done without conflict, but if you swap 1 and 2, then 2 and 4, then 4 and 8, etc., you end up with an infinite number of swaps, but actually, such a permutation is possible. Let's see:Define φ(1)=2, φ(2)=1, φ(3)=4, φ(4)=3, φ(5)=6, φ(6)=5, etc. So swapping each pair (2k-1, 2k). Wait, in this case, φ(n) is either n+1 or n-1 depending on parity. Then φ(n)/n would oscillate between (n+1)/n and (n-1)/n, both approaching 1. So the limit is still 1.Alternatively, if we do a more drastic permutation, like sending n to n^2. But that's not a bijection because squares are not all the natural numbers. So we can't do that. Similarly, arranging numbers in blocks and permuting within blocks. For example, block [1], [2,3], [4,5,6], etc., and reverse each block. But even in that case, the ratio φ(n)/n would still approach 1, because within each block of size k, the permutation is local and doesn't change the asymptotic growth.Alternatively, consider a permutation that maps n to n + sqrt(n) or something, but since it's a bijection, you can't have overlaps or skips. Actually, it's impossible to have φ(n) = n + f(n) for any function f(n) that grows to infinity because you would skip numbers. For example, if f(n) is positive, then φ(n) would be larger than n, so the numbers less than φ(n) would have to be covered by earlier terms, but if f(n) grows, eventually you would skip numbers.Therefore, it's impossible for φ(n) to be significantly larger or smaller than n asymptotically. Hence, the limit must be 1.Another angle: Suppose that L > 1. Then, for sufficiently large n, φ(n) > (L - ε)n for any small ε > 0. Then, consider the sum of φ(n) from n=1 to N. If φ(n) > (L - ε)n for n ≥ K, then the sum would be greater than Σ_{n=K}^N (L - ε)n. But the sum of φ(n) from n=1 to N is equal to the sum of k from k=1 to φ(N) (since φ is a bijection), but this is not exactly correct. Wait, actually, since φ is a permutation, the sum Σ_{n=1}^N φ(n) is equal to Σ_{k=1}^M k where M is the maximum of φ(1), φ(2), ..., φ(N). However, since φ is a bijection, M is at least N, but could be larger. Wait, no. For a permutation, the set {φ(1), φ(2), ..., φ(N)} is a subset of N with N distinct elements, but it's not necessarily the first N numbers. Therefore, the sum Σ_{n=1}^N φ(n) can vary.However, if φ(n) ~ Ln, then Σ_{n=1}^N φ(n) ~ L Σ_{n=1}^N n ~ L N(N+1)/2. On the other hand, the sum of any N distinct natural numbers is at least Σ_{k=1}^N k = N(N+1)/2 and at most Σ_{k=M-N+1}^M k for some M ≥ N. The maximum sum would be roughly (M^2 - (M - N)^2)/2 ≈ MN - N^2/2. If φ(n) ~ Ln, then M ~ LN. Therefore, the maximum sum would be ~ (LN)N - N^2/2 = L N^2 - N^2/2. Comparing to the sum Σ φ(n) ~ L N^2 / 2. For large N, L N^2 / 2 must be less than or equal to L N^2 - N^2 / 2. Which implies L/2 ≤ L - 1/2 ⇒ L/2 ≤ L - 1/2 ⇒ 0 ≤ L - 1/2 - L/2 ⇒ 0 ≤ L/2 - 1/2 ⇒ L/2 ≥ 1/2 ⇒ L ≥ 1. Similarly, the sum must also be at least N(N+1)/2 ~ N^2 / 2. Therefore, L N^2 / 2 ≥ N^2 / 2 ⇒ L ≥ 1. So if L > 1, then the sum of φ(n) would exceed the maximum possible sum for N terms unless L = 1.Wait, this seems a bit convoluted, but the idea is that if φ(n) grows like Ln, then the sum of the first N terms would be about L N^2 / 2, but the maximum possible sum of N distinct numbers is roughly (LN)^2 / 2 when M ~ LN. However, if L > 1, then LN is larger than N, but the sum of the first N numbers up to LN is (LN)(LN + 1)/2 ~ L^2 N^2 / 2. Comparing this to the sum of φ(n) ~ L N^2 / 2. So we have L N^2 / 2 ≤ L^2 N^2 / 2 ⇒ L ≤ L^2 ⇒ L ≥ 1 (since L > 0). But we also know from the lower bound that L ≥ 1. Therefore, the only possibility is L = 1.Therefore, combining these different lines of reasoning—counting arguments, sum comparisons, inverse function considerations—it seems that the only possible value for L is 1.Let me verify this with another example. Suppose we construct a permutation where φ(n) is roughly n but sometimes swaps numbers in a way that doesn't affect the density. For example, take a permutation where φ(n) = n + 1000 for n ≤ 1000, and φ(n) = n - 1000 for n > 1000. Wait, but this isn't a valid permutation because φ(n) for n > 1000 would map to numbers starting from 1, but those numbers were already mapped by n ≤ 1000. So that's not a bijection. Instead, a valid permutation could swap blocks. For example, swap the first 1000 numbers with the next 1000 numbers. So φ(n) = n + 1000 for n ≤ 1000, and φ(n) = n - 1000 for 1001 ≤ n ≤ 2000, and φ(n) = n otherwise. In this case, for n > 2000, φ(n) = n. Then φ(n)/n approaches 1 as n → ∞, since beyond n=2000, φ(n)=n. So the limit is still 1.Another example: Let’s define φ(n) such that for even n, φ(n) = n/2, and for odd n, φ(n) = 2n. Wait, is this a bijection? Let’s check. For even n=2k, φ(2k)=k. For odd n=2k-1, φ(2k-1)=2(2k-1). So the image of the even numbers is N (since k can be any natural number), and the image of the odd numbers is the even numbers starting from 2. But wait, φ maps even n to N and odd n to even numbers. So combined, the entire image is N ∪ even numbers, which is just N. Wait, but actually, φ(n) maps even n to k (so covering all natural numbers) and odd n to even numbers (2, 6, 10, ...). Wait, no. For odd n=1, φ(1)=2; n=3, φ(3)=6; n=5, φ(5)=10; etc. So φ maps odd n to 2*(2k-1) = 4k-2. Therefore, the image of odd n is numbers congruent to 2 mod 4. Then the image of even n is numbers k, which includes all natural numbers. But wait, the numbers generated by even n (k) include all natural numbers, but the numbers generated by odd n are 2, 6, 10, etc., which are also included in the image. Wait, this is a problem because φ is supposed to be a bijection, but here, for example, φ(2)=1, φ(1)=2, φ(4)=2, which would mean φ(4)=2 and φ(1)=2, which is a conflict. So this is not a valid bijection. My mistake.To make a valid bijection where even n are mapped to some subset and odd n to another, they need to partition N. For example, we could have φ(2k) = k and φ(2k-1) = N {k}. But N {k} is infinite, so it's not straightforward. Alternatively, use a pairing function. For instance, a classic bijection between N and N×N, but that's more complicated. However, such bijections usually involve some kind of diagonal enumeration, which might not preserve a linear growth rate.Alternatively, consider the permutation where φ(n) alternates between low and high numbers. For example, φ(1)=1, φ(2)=3, φ(3)=2, φ(4)=5, φ(5)=4, φ(6)=7, φ(7)=6, etc. This swaps every pair (2k, 2k+1). Then φ(n)/n for even n=2k: φ(2k)=2k+1, so (2k+1)/(2k) → 1. For odd n=2k+1: φ(2k+1)=2k, so (2k)/(2k+1) → 1. So again, the limit is 1.All these examples reinforce the idea that no matter how we permute the natural numbers, the ratio φ(n)/n must approach 1. Any attempt to make it approach something else seems to either fail to be a bijection or ends up not affecting the limit.Therefore, after considering various angles, examples, and mathematical arguments, I conclude that the only possible value of L is 1.Final AnswerThe only possible value of ( L ) is (boxed{1}).

✅:Given a bijective function ( phi: mathbb{N} rightarrow mathbb{N} ) and a finite limit ( lim_{n to infty} frac{phi(n)}{n} = L ), we need to determine the possible values of ( L ).1. Bijection and Permutation: Since ( phi ) is a bijection, it is a permutation of the natural numbers. This means every natural number appears exactly once in the sequence ( phi(1), phi(2), phi(3), ldots ).2. Examples and Intuition: - For the identity permutation ( phi(n) = n ), the limit is ( 1 ). - Swapping numbers in pairs (e.g., ( phi(2k-1) = 2k ) and ( phi(2k) = 2k-1 )) results in the limit still approaching ( 1 ).3. Inverse Function Consideration: - The inverse function ( phi^{-1} ) is also a permutation. If ( phi(n) approx Ln ), then ( phi^{-1}(m) approx frac{m}{L} ). For ( phi^{-1}(m)/m ) to have a finite limit, ( L ) must be ( 1 ).4. Counting Argument: - The number of terms ( phi(n) leq M ) must be exactly ( M ). If ( phi(n) approx Ln ), the number of such ( n ) would be approximately ( frac{M}{L} ). Equating this to ( M ) gives ( L = 1 ).5. Sum Argument: - The sum of the first ( N ) terms of ( phi(n) ) must equal the sum of the first ( N ) natural numbers. If ( phi(n) approx Ln ), the sum would be approximately ( L frac{N(N+1)}{2} ), which must match ( frac{N(N+1)}{2} ), leading to ( L = 1 ).6. Density Argument: - The natural density of the set ( mathbb{N} ) is ( 1 ). If ( phi(n) approx Ln ), the density must be preserved, leading to ( L = 1 ).After thorough analysis and considering various permutations, counting arguments, inverse functions, and density, it is concluded that the only possible value of ( L ) is ( 1 ).[boxed{1}]

🔑:1. Assume there exists ( N ) such that ( phi(n) > n ) for all ( n geq N ): - If ( phi(n) > n ) for all ( n geq N ), then ( phi(n) ) takes values greater than ( N ) for all ( n geq N ). - This implies that the values ( phi(n) ) for ( n geq N ) are all greater than ( N ), leaving not enough values to fill the set ( {1, 2, ldots, N} ). - Therefore, for any ( N ), there must exist some ( n geq N ) such that ( phi(n) leq n ). - Hence, ( liminf_{n to infty} frac{phi(n)}{n} leq 1 ).2. Assume there exists ( N ) such that ( phi(n) < n ) for all ( n geq N ): - Let ( m = max {phi(k) mid 1 leq k leq N} ). - Clearly, ( m geq N ) because ( phi ) is a bijection and must cover all natural numbers. - However, if ( phi(n) < n ) for all ( n geq N ), then ( max {phi(k) mid 1 leq k leq m+1} = m ), which is a contradiction because ( phi ) must map to all values up to ( m+1 ). - Therefore, for any ( N ), there must exist some ( n geq N ) such that ( phi(n) geq n ). - Hence, ( limsup_{n to infty} frac{phi(n)}{n} geq 1 ).3. Conclusion: - Since ( liminf_{n to infty} frac{phi(n)}{n} leq 1 ) and ( limsup_{n to infty} frac{phi(n)}{n} geq 1 ), if the limit ( lim_{n to infty} frac{phi(n)}{n} ) exists, it must be equal to 1. - A trivial example is ( phi(n) = n ) for ( n geq 1 ). - Another example is ( phi(2n-1) = 2n ) and ( phi(2n) = 2n-1 ) for ( n geq 1 ).(blacksquare)The final answer is ( boxed{ 1 } ).

🧠:Okay, so Valencia has drawn two intersecting segments AB and CD, with lengths 10 and 7 respectively. The problem is asking for the maximum possible area of the quadrilateral ACBD across all possible configurations. Hmm, interesting. Let me try to visualize this first. Quadrilateral ACBD is formed by the two intersecting segments. The segments AB and CD intersect at some point, let's call it O, which divides each segment into two parts. So AB is split into AO and OB, and CD is split into CO and OD. The area of quadrilateral ACBD would then be the sum of the areas of triangles ACB and CBD, but wait, maybe I should think of it as two triangles that make up the quadrilateral. Alternatively, since the quadrilateral is ACBD, it's formed by the points A, C, B, D. So maybe the quadrilateral is a kite shape? No, not necessarily, because the lengths of AB and CD are different. Wait, actually, when two segments intersect, the quadrilateral formed is determined by the four endpoints. So A and B are the endpoints of one segment, and C and D are the endpoints of the other. The intersection point is somewhere in the middle. The quadrilateral would then have vertices at A, C, B, D. Let me draw a rough sketch in my mind: AB is a longer segment, 10 units, and CD is 7 units. They cross each other at point O. The quadrilateral is ACBD, so connecting A to C, C to B, B to D, and D to A. Wait, no, that's a different quadrilateral. Wait, ACBD would be the polygon with vertices in that order: A, C, B, D. So the sides are AC, CB, BD, DA. Hmm, but depending on the intersection point, the shape can vary. Maybe it's better to think of the area as the sum of the areas of triangles AOC, COB, BOD, and DOA? No, that seems more complicated. Alternatively, since the quadrilateral is ACBD, perhaps it's made up of two triangles: ACB and ABD? Wait, no, that might overlap. Hmm, maybe I need to clarify the structure.Alternatively, the area of quadrilateral ACBD can be calculated by splitting it into two triangles that share a common diagonal. If the two segments AB and CD intersect at O, then the quadrilateral ACBD is composed of triangles AOC and BOD, or maybe AOD and BOC? Wait, maybe not. Let me think again. If you have intersecting segments AB and CD at O, then the quadrilateral ACBD is actually made up of four triangles: AOC, COB, BOD, and DOA. But that seems like the entire area around the intersection. Wait, but perhaps the quadrilateral ACBD is actually the union of triangles ACB and ADB? Let me check.Wait, maybe the key here is that when two chords intersect in a plane, the area of the quadrilateral formed by their endpoints can be expressed in terms of the lengths of the segments and the angles between them. But since the problem states "across all possible configurations," we need to maximize the area with respect to the variables involved. The variables here would be the angle between the two segments AB and CD, and the position of the intersection point O along each segment.Let me recall that the area of a quadrilateral formed by two intersecting chords can be calculated using the formula: (1/2)*AB*CD*sin(theta), where theta is the angle between the two segments. Wait, is that correct? Wait, if two lines of lengths AB and CD intersect at an angle theta, then the area of the quadrilateral formed is (1/2)*AB*CD*sin(theta). Hmm, but I need to confirm this.Wait, no. If two lines intersect, the area formed by the quadrilateral would actually depend on the lengths of the segments on either side of the intersection. Suppose AB is divided into AO and OB, and CD is divided into CO and OD. Then, the area of quadrilateral ACBD would be the sum of the areas of triangles AOC, COB, BOD, and DOA. Alternatively, maybe it's the sum of two triangles: AOC + BOD multiplied by 2 or something. Wait, perhaps it's better to use coordinates to model this.Let me try to set up a coordinate system. Let's place the intersection point O at the origin (0,0). Let’s assume that segment AB is divided by O into lengths AO = x and OB = 10 - x. Similarly, segment CD is divided by O into CO = y and OD = 7 - y. The angle between AB and CD is theta. Then, the coordinates of the points can be expressed in terms of x, y, and theta.Let's orient AB along the x-axis for simplicity. So point A is at (-x, 0), point B is at (10 - x, 0). For segment CD, which makes an angle theta with AB, points C and D can be placed at coordinates depending on y and theta. Let me think: if we consider CD as a vector from O, then point C is at (y*cos(theta), y*sin(theta)), and point D is at (- (7 - y)*cos(theta), - (7 - y)*sin(theta)). Wait, that might be complicated, but let's see.Alternatively, since CD is of length 7 and is split into CO = y and OD = 7 - y, then if we set O at the origin, point C can be at (y*cos(theta), y*sin(theta)) and point D at (- (7 - y)*cos(theta), - (7 - y)*sin(theta)), assuming that CD is oriented at an angle theta from AB. Then, AB is along the x-axis from (-x, 0) to (10 - x, 0).Then, the coordinates of the four points are:A: (-x, 0)B: (10 - x, 0)C: (y*cos(theta), y*sin(theta))D: (- (7 - y)*cos(theta), - (7 - y)*sin(theta))Now, quadrilateral ACBD has vertices at A, C, B, D. To find its area, we can use the shoelace formula.First, let's list the coordinates in order:A: (-x, 0)C: (y*cos(theta), y*sin(theta))B: (10 - x, 0)D: (- (7 - y)*cos(theta), - (7 - y)*sin(theta))Then back to A: (-x, 0)Using the shoelace formula, the area is (1/2)*|sum over the vertices of (x_i y_{i+1} - x_{i+1} y_i)|Let me compute each term step by step.First term: x_A * y_C - x_C * y_A = (-x)(y*sin(theta)) - (y*cos(theta))(0) = -x y sin(theta)Second term: x_C * y_B - x_B * y_C = (y*cos(theta))(0) - (10 - x)(y*sin(theta)) = - (10 - x) y sin(theta)Third term: x_B * y_D - x_D * y_B = (10 - x)( - (7 - y)*sin(theta) ) - ( - (7 - y)*cos(theta) )(0) = - (10 - x)(7 - y) sin(theta)Fourth term: x_D * y_A - x_A * y_D = ( - (7 - y)*cos(theta) )(0) - (-x)( - (7 - y)*sin(theta) ) = - x (7 - y) sin(theta)Sum all these terms:First term: -x y sin(theta)Second term: - (10 - x) y sin(theta)Third term: - (10 - x)(7 - y) sin(theta)Fourth term: - x (7 - y) sin(theta)Total sum = [ -x y - (10 - x)y - (10 - x)(7 - y) - x(7 - y) ] sin(theta)Let me factor out sin(theta) and simplify the expression inside the brackets:= [ -x y -10 y + x y -10*7 +10 y + x*7 -x y -7 x +x y ] sin(theta)Wait, perhaps expanding each term step by step:First term: -x ySecond term: - (10 - x)y = -10 y + x yThird term: - (10 - x)(7 - y) = - [10*7 -10 y -x*7 +x y] = -70 +10 y +7 x -x yFourth term: -x(7 - y) = -7 x +x yNow, summing all these:First term: -x ySecond term: -10 y + x yThird term: -70 +10 y +7 x -x yFourth term: -7 x +x yAdding term by term:- x y + (-10 y + x y) + (-70 +10 y +7 x -x y) + (-7 x +x y)Combine like terms:For x y terms:- x y + x y - x y + x y = 0For x terms:7 x -7 x = 0For y terms:-10 y +10 y = 0Constants:-70So the total sum inside the brackets is -70. Therefore, the area is (1/2)*| -70 sin(theta) | = (1/2)*70 |sin(theta)| = 35 |sin(theta)|.Wait, that's interesting. So according to this calculation, the area of quadrilateral ACBD is 35 |sin(theta)|, which is independent of x and y? That seems counterintuitive. How come the lengths of AO and CO don't affect the area? That can't be right. Let me check my calculations again.Wait, in the shoelace formula, I might have made a mistake in the coordinates or in expanding the terms. Let's go through each term again carefully.First term: x_A * y_C - x_C * y_A = (-x)(y sin theta) - (y cos theta)(0) = -x y sin thetaSecond term: x_C * y_B - x_B * y_C = (y cos theta)(0) - (10 - x)(y sin theta) = - (10 - x) y sin thetaThird term: x_B * y_D - x_D * y_B = (10 - x)( - (7 - y) sin theta ) - ( - (7 - y) cos theta )(0) = - (10 - x)(7 - y) sin thetaFourth term: x_D * y_A - x_A * y_D = ( - (7 - y) cos theta )(0) - (-x)( - (7 - y) sin theta ) = - x (7 - y) sin thetaSo sum is:- x y sin theta - (10 - x) y sin theta - (10 - x)(7 - y) sin theta - x (7 - y) sin thetaFactor out -sin theta:= - sin theta [ x y + (10 - x) y + (10 - x)(7 - y) + x (7 - y) ]Now, let's compute the expression inside the brackets:First term: x ySecond term: (10 - x) y =10 y -x yThird term: (10 - x)(7 - y) = 10*7 -10 y -7 x +x y =70 -10 y -7 x +x yFourth term: x (7 - y) =7 x -x yNow, adding all four terms together:x y + (10 y -x y) + (70 -10 y -7 x +x y) + (7 x -x y)Let's combine term by term:x y -x y +x y -x y = 010 y -10 y =0-7 x +7 x =0And the constant term:70So indeed, the total inside the brackets is 70. Therefore, the sum is - sin theta *70, and the absolute value gives 70 |sin theta|. Then the area is (1/2)*70 |sin theta|=35 |sin theta|.Wow, so according to this calculation, the area only depends on the angle theta between the two segments AB and CD, and it's completely independent of where they intersect (i.e., independent of x and y). That's surprising. So regardless of how the segments are divided by the intersection point O, as long as the angle between them is theta, the area is 35 |sin theta|. Therefore, to maximize the area, we need to maximize |sin theta|, which occurs when theta is 90 degrees (pi/2 radians), making sin theta =1. Therefore, the maximum area is 35*1=35.But wait, is this correct? Because initially, I thought the position of the intersection point might affect the area, but according to this, it doesn't. Let me verify with a simple example.Suppose AB and CD are perpendicular, intersecting at their midpoints. So AB=10, midpoint at 5, CD=7, midpoint at 3.5. Then, the quadrilateral ACBD would be a kite? Wait, if they intersect at midpoints, then AO=5, OB=5, CO=3.5, OD=3.5. Then, the area would be 4*(1/2)*5*3.5*sin(90) = 4*(17.5/2)=35. Which matches the formula. Alternatively, if they intersect not at midpoints, say AO=2, OB=8, CO=1, OD=6, with theta=90 degrees. Then, the area would be (1/2)*(2*1 + 8*6 + 8*6 + 2*1)*sin(90)? Wait, no, according to the formula, it's still 35, regardless. Wait, but if we compute the area using coordinates, let's take AO=2, OB=8, CO=1, OD=6, theta=90 degrees.So placing O at (0,0), AB along x-axis: A(-2,0), B(8,0). CD along y-axis: C(0,1), D(0,-6). Then, the quadrilateral ACBD has vertices at A(-2,0), C(0,1), B(8,0), D(0,-6), back to A(-2,0). Applying the shoelace formula:Coordinates:-2, 00, 18, 00, -6-2, 0Compute the sum:First part: (-2)*1 + 0*0 + 8*(-6) + 0*0 = -2 + 0 -48 + 0 = -50Second part: 0*0 + 1*8 + 0*0 + (-6)*(-2) = 0 +8 +0 +12 =20Area is (1/2)|-50 -20| = (1/2)|-70| =35. Yes! So even when the intersection is not at the midpoint, the area is still 35 when theta=90 degrees. Therefore, the position of the intersection doesn't matter; only the angle between the segments affects the area, which is maximized at 90 degrees. Therefore, the maximum area is 35.But wait, this seems too straightforward. The problem states "across all possible configurations of those two segments," so maybe there's a catch? For example, can the angle theta actually reach 90 degrees in all configurations? Or are there constraints based on the lengths of the segments?Wait, if the segments are too short relative to each other, maybe they can't form a 90-degree angle? Wait, no. The angle between two segments is independent of their lengths. For example, two segments of any lengths can be arranged to intersect at any angle, as long as they intersect. The maximum area occurs when they are perpendicular, regardless of their lengths. Therefore, according to our calculation, the maximum area is 35.But let me check another example. Suppose AB=10 and CD=7, intersecting at some angle. If theta is 90 degrees, area is 35. If theta is smaller, say theta=30 degrees, area is 35*sin(30)=17.5, which is half. So indeed, the maximum is at 90 degrees.But hold on, maybe the problem is in the setup. The calculation above assumes that the intersection point O can divide AB and CD into any lengths, but in reality, the segments AB and CD have fixed lengths. Wait, no, the segments themselves are fixed lengths, but their intersection can be anywhere. So as long as the two segments cross each other, the lengths AO, OB, CO, OD can vary, but in our calculation, we found that the area is independent of those lengths.But according to the shoelace formula calculation, it's indeed 35 |sin(theta)| regardless of x and y. Therefore, the maximum area is 35.But let me cross-validate with another approach. Let's think of the quadrilateral ACBD as two triangles: AOC and BOD, and two other triangles: COB and AOD. Wait, but earlier, we considered the shoelace formula and found that the area is 35 |sin(theta)|. Alternatively, using vectors, maybe.Alternatively, the area of a quadrilateral formed by two intersecting lines can be calculated as (1/2)*d1*d2*sin(theta), where d1 and d2 are the lengths of the diagonals. Wait, but in our case, AB and CD are the diagonals? Wait, no, in a quadrilateral, the diagonals are the segments connecting opposite vertices. In this case, the diagonals would be AB and CD themselves, since they intersect at O. Therefore, the formula for the area of a quadrilateral with diagonals d1 and d2 intersecting at angle theta is indeed (1/2)*d1*d2*sin(theta). Therefore, the area is (1/2)*10*7*sin(theta)=35 sin(theta). Therefore, maximum at theta=90 degrees, giving 35. Therefore, the maximum area is 35.So this aligns with the shoelace formula result. Therefore, regardless of where the diagonals intersect, the area is (1/2)*d1*d2*sin(theta). Therefore, the maximum area is 35.But wait, this formula is for the area of a quadrilateral with given diagonals and angle between them. So in that case, yes, the maximum area is achieved when the diagonals are perpendicular. Therefore, regardless of the lengths of the segments or their intersection point, the maximum area is (1/2)*10*7*1=35.Therefore, the answer should be 35. But let me check another source to confirm this formula.Yes, according to the formula for the area of a quadrilateral with diagonals of lengths d1 and d2 intersecting at an angle theta, the area is (1/2)*d1*d2*sin(theta). Therefore, the maximum area is achieved when sin(theta)=1, i.e., theta=90 degrees. Therefore, the maximum area is (1/2)*10*7*1=35.Therefore, the answer is 35. But let me make sure that the formula applies here. In this problem, AB and CD are the diagonals of quadrilateral ACBD. Yes, because in quadrilateral ACBD, the vertices are A, C, B, D, so the diagonals are AB and CD. Therefore, the formula applies, and the maximum area is indeed 35.Therefore, despite initial confusion about the intersection point and possible dependency on the position of O, the formula shows that the area only depends on the angle between the diagonals. Therefore, the maximum area is 35.Final Answerboxed{35}

✅:To determine the maximum possible area of quadrilateral (ACBD) formed by intersecting segments (AB = 10) and (CD = 7), we need to consider the angle (theta) between the segments and the formula for the area of a quadrilateral with intersecting diagonals.1. Area Formula for Quadrilateral with Diagonals: The area of a quadrilateral formed by two intersecting diagonals of lengths (d_1) and (d_2) intersecting at an angle (theta) is given by: [ text{Area} = frac{1}{2} cdot d_1 cdot d_2 cdot sin(theta) ] Here, the diagonals are (AB) and (CD) with lengths 10 and 7 respectively.2. Maximizing the Area: The area depends on (sin(theta)), which is maximized when (theta = 90^circ) (i.e., (sin(theta) = 1)).3. Calculation: Substituting the values into the formula: [ text{Maximum Area} = frac{1}{2} cdot 10 cdot 7 cdot sin(90^circ) = frac{1}{2} cdot 10 cdot 7 cdot 1 = 35 ]Thus, the maximum possible area of quadrilateral (ACBD) is (boxed{35}).

🔑:To determine the maximum possible area of quadrilateral (ACBD) formed by two intersecting segments (AB = 10) and (CD = 7), we need to consider two cases: when the quadrilateral is convex and when it is non-convex.1. Convex Case: - Let segment (AB) intersect segment (CD) at point (O). - Draw lines parallel to (AB) passing through points (C) and (D). - Draw lines parallel to (CD) passing through points (A) and (B). - Denote (KLMN) as the parallelogram defined by the intersection of these pairs of parallel lines. - The area of quadrilateral (ACBD) is half the area of parallelogram (KLMN): [ [ACBD] = frac{[KLMN]}{2} ] - The area of parallelogram (KLMN) is given by: [ [KLMN] = KL cdot KN cdot sin angle LKN ] - Since (KL = AB) and (KN = CD), we have: [ [KLMN] = AB cdot CD cdot sin w ] - Therefore, the area of quadrilateral (ACBD) is: [ [ACBD] = frac{1}{2} cdot AB cdot CD cdot sin w ] - The maximum area occurs when (sin w = 1), which happens when (AB perp CD): [ [ACBD] = frac{1}{2} cdot AB cdot CD ]2. Non-Convex Case: - Assume (AB) intersects (CD) such that segment (AB) does not intersect segment (CD), and let (A) be the farthest point from line (CD). - Let (M) and (N) be the projections of (A) and (B) on line (CD), respectively. - Let (O) be the projection of (B) on segment (AM). - The area of quadrilateral (ACBD) can be expressed as the difference of two triangles: [ [ACBD] = [BCD] - [ACD] ] - The area of triangle (BCD) is: [ [BCD] = frac{1}{2} cdot CD cdot AO ] - The area of triangle (ACD) is: [ [ACD] = frac{1}{2} cdot CD cdot BN ] - Therefore, the area of quadrilateral (ACBD) is: [ [ACBD] = frac{1}{2} cdot CD cdot (AO - BN) ] - Since (AO = AB cdot sin angle OAB) and (BN = 0) (as (B) lies on (CD)), we have: [ [ACBD] = frac{1}{2} cdot CD cdot AB cdot sin w ] - The maximum area occurs when (sin w = 1), which happens when (AB perp CD): [ [ACBD] = frac{1}{2} cdot AB cdot CD ]In both cases, the maximum area of quadrilateral (ACBD) is:[frac{1}{2} cdot AB cdot CD = frac{1}{2} cdot 10 cdot 7 = 35]The final answer is (boxed{35}).

🧠:Okay, so I need to find the value of c that minimizes the sum from n=1 to 11 of (n - c)(2n - c). Then, I also need to find the minimum value m. Hmm, let's think about how to approach this.First, I remember that when you have a sum like this that depends on a parameter c, you can treat the sum as a quadratic function in terms of c. Since quadratic functions have a unique minimum or maximum, depending on the coefficient, I can expand the expression inside the sum, combine like terms, and then take the derivative with respect to c to find the minimum. Alternatively, since it's a quadratic, I can complete the square or use the vertex formula. Let me try expanding the expression first.Let me write out the general term in the sum: (n - c)(2n - c). Let's multiply that out. (n - c)(2n - c) = n*2n + n*(-c) - c*2n + (-c)*(-c) = 2n² - cn - 2cn + c² = 2n² - 3cn + c².So each term in the sum simplifies to 2n² - 3cn + c². Therefore, the entire sum S(c) is the sum from n=1 to 11 of (2n² - 3cn + c²). Now, let's split the sum into three separate sums:S(c) = 2*Σn² - 3c*Σn + c²*Σ1, where all sums are from n=1 to 11.I need to compute each of these sums. Let's recall that:- The sum of the first N natural numbers is Σn = N(N+1)/2.- The sum of the squares of the first N natural numbers is Σn² = N(N+1)(2N+1)/6.- The sum of 1 from n=1 to N is just N.So here, N=11. Let me compute each sum:First, Σn from 1 to 11: 11*12/2 = 66.Second, Σn² from 1 to 11: 11*12*23/6. Let's compute that step by step. 11*12=132, 132*23=3036, then divide by 6: 3036/6=506.Third, Σ1 from 1 to 11 is just 11.Therefore, substituting back into S(c):S(c) = 2*506 - 3c*66 + c²*11.Let me compute each term:2*506 = 1012.3c*66 = 198c.c²*11 = 11c².So S(c) = 11c² - 198c + 1012.Now, this is a quadratic function in terms of c. The quadratic is of the form S(c) = 11c² - 198c + 1012. Since the coefficient of c² is positive (11), the parabola opens upwards, so the minimum occurs at the vertex.The vertex of a quadratic function ax² + bx + c is at x = -b/(2a). So here, a=11, b=-198. Thus, the value of c that minimizes S(c) is:c = -(-198)/(2*11) = 198/22 = 9.Wait, 198 divided by 22: 22*9=198, so yes, c=9.Now, to find the minimum value m, substitute c=9 back into S(c):S(9) = 11*(9)² - 198*(9) + 1012.Compute each term:11*81 = 891.198*9 = 1782.So:S(9) = 891 - 1782 + 1012.Let me compute step by step:891 - 1782 = -891.-891 + 1012 = 121.Therefore, the minimum value m is 121.Wait, let me verify my calculations again to be sure. Let's compute S(9):First, 9 squared is 81. Then 11*81 is 891. Then 198*9: 200*9=1800, minus 2*9=18, so 1800-18=1782. Then, 891 - 1782 = -891. Then, -891 + 1012: 1012 - 891. 1012 - 800 = 212, 212 - 91 = 121. Yes, that seems correct.Alternatively, maybe I can check by plugging c=9 into the original sum. Let's compute a few terms and see if they add up appropriately. Let's take n=1: (1 -9)(2*1 -9) = (-8)(-7)=56. Similarly, n=2: (2-9)(4-9)=(-7)(-5)=35. n=3: (3-9)(6-9)=(-6)(-3)=18. n=4: (4-9)(8-9)=(-5)(-1)=5. n=5: (5-9)(10-9)=(-4)(1)=-4. Wait, so the terms can be positive or negative. Hmm, but when we add all of them from n=1 to 11, maybe it sums up to 121.But maybe let's compute the original sum for c=9 to see if it's 121. That might take time, but let's try a few more terms:n=6: (6-9)(12-9)=(-3)(3)=-9.n=7: (7-9)(14-9)=(-2)(5)=-10.n=8: (8-9)(16-9)=(-1)(7)=-7.n=9: (9-9)(18-9)=0*9=0.n=10: (10-9)(20-9)=1*11=11.n=11: (11-9)(22-9)=2*13=26.Now, let's add up all these terms:56 (n=1) +35 (n=2)=91.+18 (n=3)=109.+5 (n=4)=114.-4 (n=5)=110.-9 (n=6)=101.-10 (n=7)=91.-7 (n=8)=84.0 (n=9)=84.+11 (n=10)=95.+26 (n=11)=121.Wow, that adds up to 121. So that's consistent with the earlier calculation. Therefore, the minimum value is indeed 121 when c=9.Alternatively, another way to approach this problem might be by using calculus. Since S(c) is a quadratic function, its derivative with respect to c will be linear, and setting it to zero gives the critical point, which should be the minimum.Let's try that method too. S(c) = 11c² -198c +1012.dS/dc = 22c -198. Setting this equal to zero:22c -198=0 => 22c=198 => c=198/22=9. Same result as before. So that confirms it.Therefore, c=9 and m=121.I think that's thorough. I expanded the original expression, converted the sum into a quadratic in c, calculated the sums, found the coefficients, then found the vertex to get c=9, substituted back in to find m=121, verified by plugging in c=9 into the original terms and adding them up, which gave the same result, and also confirmed via calculus. So I think that's solid.Final AnswerThe value of ( c ) that minimizes the sum is (boxed{9}) and the minimum value ( m ) is (boxed{121}).

✅:To find the value of ( c ) that minimizes the sum ( sum_{n=1}^{11} (n - c)(2n - c) ) and the corresponding minimum value ( m ), we start by expanding the expression inside the sum:[(n - c)(2n - c) = 2n^2 - 3cn + c^2]This allows us to rewrite the sum as:[sum_{n=1}^{11} (2n^2 - 3cn + c^2)]We split this into three separate sums:[2 sum_{n=1}^{11} n^2 - 3c sum_{n=1}^{11} n + c^2 sum_{n=1}^{11} 1]Using the formulas for the sum of the first ( N ) natural numbers, the sum of the squares of the first ( N ) natural numbers, and the sum of 1 repeated ( N ) times (with ( N = 11 )):[sum_{n=1}^{11} n = frac{11 cdot 12}{2} = 66][sum_{n=1}^{11} n^2 = frac{11 cdot 12 cdot 23}{6} = 506][sum_{n=1}^{11} 1 = 11]Substituting these into the expression for the sum ( S(c) ):[S(c) = 2 cdot 506 - 3c cdot 66 + c^2 cdot 11]Simplifying each term:[S(c) = 1012 - 198c + 11c^2]This is a quadratic function in ( c ). The minimum value occurs at the vertex of the parabola, given by ( c = -frac{b}{2a} ) for a quadratic ( ax^2 + bx + c ). Here, ( a = 11 ) and ( b = -198 ):[c = frac{198}{2 cdot 11} = frac{198}{22} = 9]Substituting ( c = 9 ) back into the quadratic function to find the minimum value ( m ):[S(9) = 11 cdot 9^2 - 198 cdot 9 + 1012][= 11 cdot 81 - 1782 + 1012][= 891 - 1782 + 1012][= 121]Thus, the value of ( c ) that minimizes the sum is (boxed{9}) and the minimum value ( m ) is (boxed{121}).

🔑:1. We start with the given sum: [ sum_{n=1}^{11} (n - c)(2n - c) ] We need to expand and simplify this expression.2. Expanding the product inside the sum: [ (n - c)(2n - c) = 2n^2 - 3nc + c^2 ] Therefore, the sum becomes: [ sum_{n=1}^{11} (2n^2 - 3nc + c^2) ]3. We can split the sum into three separate sums: [ sum_{n=1}^{11} 2n^2 - sum_{n=1}^{11} 3nc + sum_{n=1}^{11} c^2 ]4. Using the known formulas for the sums of the first ( n ) natural numbers and the sum of the squares of the first ( n ) natural numbers: [ sum_{n=1}^{x} n = frac{x(x+1)}{2} ] [ sum_{n=1}^{x} n^2 = frac{x(x+1)(2x+1)}{6} ]5. Applying these formulas for ( x = 11 ): [ sum_{n=1}^{11} n = frac{11 cdot 12}{2} = 66 ] [ sum_{n=1}^{11} n^2 = frac{11 cdot 12 cdot 23}{6} = 506 ]6. Substituting these values back into our sum: [ sum_{n=1}^{11} 2n^2 = 2 cdot 506 = 1012 ] [ sum_{n=1}^{11} 3nc = 3c cdot 66 = 198c ] [ sum_{n=1}^{11} c^2 = 11c^2 ]7. Combining these results: [ sum_{n=1}^{11} (2n^2 - 3nc + c^2) = 1012 - 198c + 11c^2 ]8. To find the value of ( c ) that minimizes this expression, we complete the square: [ 1012 - 198c + 11c^2 = 11(c^2 - 18c + 92) ] [ = 11((c - 9)^2 + 11) ]9. The expression ( (c - 9)^2 ) is minimized when ( c = 9 ). Substituting ( c = 9 ) into the expression: [ 11((9 - 9)^2 + 11) = 11(0 + 11) = 121 ]Thus, the minimum value occurs at ( c = 9 ) and the minimum value ( m ) is 121.The final answer is ( boxed{ c = 9 } ) and ( m = 121 ).

🧠:Okay, let's see. I need to figure out how long it would take Alex to rake the entire lawn by himself. The problem involves work rates, so maybe I should use the concept of work rates where rate is 1 job per time. Let me break down the problem step by step.First, Alex and Beth raked half of the lawn together in 30 minutes. So, together, they completed half the job in 0.5 hours (since 30 minutes is half an hour). Wait, but maybe I should convert all times to hours because the final answer needs to be in hours. Let me check the units. The problem mentions minutes, but the answer needs to be in hours. So, maybe convert all given times to hours to keep consistency.Alright, first part: Alex and Beth together rake half the lawn in 30 minutes, which is 0.5 hours. So their combined work rate for half the lawn is 0.5 hours. Wait, but work rate is usually jobs per hour. Let me think. If they rake half the lawn in 0.5 hours, then their combined rate is (1/2 lawn) / (0.5 hours) = 1 lawn per hour. Wait, that seems fast. Let me check: if they can do half the lawn in half an hour, that means they can do the whole lawn in 1 hour. So together, Alex and Beth's combined rate is 1 lawn per hour. That makes sense.So, let me denote Alex's rate as A (lawn per hour) and Beth's rate as B. Then, A + B = 1. Because together they can do 1 lawn per hour. Got that.Next part: While they took a break, Carl raked a third of the remaining lawn in 60 minutes. The remaining lawn after Alex and Beth did half would be half the lawn. So, half the lawn was left. Carl raked a third of that remaining half. So, he did (1/3)*(1/2) = 1/6 of the entire lawn in 60 minutes, which is 1 hour. Therefore, Carl's rate is (1/6 lawn) / 1 hour = 1/6 per hour. So Carl's rate C = 1/6.Now, after Carl did 1/6 of the lawn, how much is left? The remaining was half the lawn, then he did 1/6, so remaining is 1/2 - 1/6 = 3/6 - 1/6 = 2/6 = 1/3 of the lawn left. Then, finally, Beth joined Carl, and together they finished the remaining 1/3 of the lawn in 24 minutes. 24 minutes is 24/60 = 0.4 hours.So Beth and Carl together raked 1/3 of the lawn in 0.4 hours. Their combined rate would be (1/3) / 0.4 = (1/3) / (2/5) = (1/3)*(5/2) = 5/6 per hour. Wait, 0.4 hours is 2/5 hours? Wait, 24 minutes is 24/60 = 0.4 hours. So 1/3 divided by 0.4 is indeed (1/3)/(2/5) = 5/6. So their combined rate is 5/6 per hour. But Beth's rate is B and Carl's rate is C = 1/6. So B + C = 5/6. Since C is 1/6, then B + 1/6 = 5/6. Therefore, B = 5/6 - 1/6 = 4/6 = 2/3 per hour.So Beth's rate is 2/3 per hour. Then, going back to the first equation: A + B = 1. Therefore, A = 1 - B = 1 - 2/3 = 1/3 per hour. So Alex's rate is 1/3 per hour. Therefore, the time it would take him to do the entire lawn alone is 1 / (1/3) = 3 hours. Wait, so is the answer 3 hours?Let me check again. Maybe I made a mistake in the calculations. Let me verify each step.First, Alex and Beth together rake half the lawn in 0.5 hours. So their combined rate is (1/2)/0.5 = 1 lawn per hour. So A + B = 1. Correct.Then, after they finish half, Carl rakes a third of the remaining half. So remaining is 1/2. A third of that is 1/6. He does 1/6 in 1 hour. So his rate is 1/6 per hour. Correct.Then, remaining lawn is 1/2 - 1/6 = 1/3. Then, Beth and Carl together finish 1/3 in 24 minutes (0.4 hours). Their combined rate is (1/3)/0.4 = 5/6 per hour. Then, since Carl is 1/6, Beth must be 5/6 - 1/6 = 4/6 = 2/3. Then, A = 1 - 2/3 = 1/3. So Alex's rate is 1/3 per hour, meaning he takes 3 hours alone. That seems to check out.Wait, but let's think again. If Alex's rate is 1/3, then in one hour, he does 1/3 of the lawn. So yes, three hours total. Hmm.Alternatively, maybe using another approach. Let's denote the total work as 1 lawn. Let’s assign variables:Let Alex's rate = A (lawn/hour)Beth's rate = BCarl's rate = CFirst, Alex and Beth together work for 0.5 hours and complete half the lawn. So:(A + B) * 0.5 = 0.5So (A + B) = 1. That's the same as before.Then, remaining work is 0.5. Carl works alone and does a third of the remaining, which is 0.5 * 1/3 = 1/6. He does this in 1 hour. So:C * 1 = 1/6Thus, C = 1/6. Same as before.Then, remaining work is 0.5 - 1/6 = (3/6 - 1/6) = 2/6 = 1/3. Then, Beth and Carl work together for 0.4 hours (24 minutes) to finish 1/3. So:(B + C) * 0.4 = 1/3Plugging in C = 1/6:(B + 1/6) * 0.4 = 1/3Multiply both sides by 0.4:B + 1/6 = (1/3) / 0.4 = (1/3) / (2/5) = 5/6So B = 5/6 - 1/6 = 4/6 = 2/3. Then, A = 1 - 2/3 = 1/3.Thus, Alex's time is 1 / (1/3) = 3 hours. So the answer is 3 hours. That seems consistent.Wait, but maybe another way to check. Let's calculate how much each person did.First phase: Alex and Beth together worked for 0.5 hours at a combined rate of 1, so they did 1 * 0.5 = 0.5 lawn. Correct.Second phase: Carl worked for 1 hour at 1/6, so he did 1/6. Correct.Third phase: Beth and Carl worked for 0.4 hours. Beth's rate is 2/3, Carl's is 1/6. Combined, they are 2/3 + 1/6 = 5/6. So 5/6 * 0.4 = (5/6)*(2/5) = 10/30 = 1/3. Correct. So total work done: 0.5 + 1/6 + 1/3. Let's convert to sixths: 3/6 + 1/6 + 2/6 = 6/6 = 1. Perfect, that adds up.So all steps check out. So Alex's rate is 1/3 per hour, so time is 3 hours.I think that's solid. The answer should be 3 hours.Final Answerboxed{3}

✅:First, Alex and Beth raked half of the lawn together in 30 minutes (0.5 hours). Their combined work rate is calculated as follows:[text{Combined rate of Alex and Beth} = frac{frac{1}{2} text{ lawn}}{0.5 text{ hours}} = 1 text{ lawn per hour}]Thus, we have:[A + B = 1]Next, Carl raked a third of the remaining half lawn in 60 minutes (1 hour). The remaining lawn after Alex and Beth's work was half, so Carl raked:[frac{1}{3} times frac{1}{2} = frac{1}{6} text{ of the lawn}]Carl's work rate is:[C = frac{frac{1}{6} text{ lawn}}{1 text{ hour}} = frac{1}{6} text{ lawn per hour}]The remaining lawn after Carl's work is:[frac{1}{2} - frac{1}{6} = frac{1}{3} text{ of the lawn}]Finally, Beth and Carl together finished the remaining (frac{1}{3}) of the lawn in 24 minutes (0.4 hours). Their combined work rate is:[text{Combined rate of Beth and Carl} = frac{frac{1}{3} text{ lawn}}{0.4 text{ hours}} = frac{5}{6} text{ lawn per hour}]Thus, we have:[B + C = frac{5}{6}]Substituting (C = frac{1}{6}) into the equation:[B + frac{1}{6} = frac{5}{6} implies B = frac{5}{6} - frac{1}{6} = frac{4}{6} = frac{2}{3}]Using (A + B = 1) and substituting (B = frac{2}{3}):[A = 1 - frac{2}{3} = frac{1}{3}]Alex's work rate is (frac{1}{3}) lawn per hour, so the time taken for him to rake the entire lawn alone is:[text{Time} = frac{1 text{ lawn}}{frac{1}{3} text{ lawn per hour}} = 3 text{ hours}]Thus, the answer is:[boxed{3}]

🔑:1. Let ( a, b, c ) be the rates at which Alex, Beth, and Carl can rake the lawn in one hour, respectively. We need to find ( a ), the rate at which Alex can rake the lawn.2. From the problem, we know that Alex and Beth together raked half of the lawn in 30 minutes. Therefore, their combined rate is: [ frac{a + b}{2} = frac{1}{2} text{ (since they raked half the lawn in 30 minutes, which is half an hour)} ] Solving for ( a + b ): [ a + b = 1 ]3. Carl raked a third of the remaining lawn in 60 minutes. Since half the lawn was already raked, the remaining lawn is also half. Carl raked a third of this remaining half, which is: [ frac{1}{3} times frac{1}{2} = frac{1}{6} ] Since Carl took 60 minutes (1 hour) to rake (frac{1}{6}) of the lawn, his rate ( c ) is: [ c = frac{1}{6} ]4. Finally, Beth and Carl together finished raking the remaining lawn in 24 minutes. The remaining lawn after Carl raked is: [ frac{1}{2} - frac{1}{6} = frac{3}{6} - frac{1}{6} = frac{2}{6} = frac{1}{3} ] Beth and Carl together raked this (frac{1}{3}) of the lawn in 24 minutes, which is (frac{24}{60} = frac{2}{5}) hours. Therefore, their combined rate is: [ frac{b + c}{frac{2}{5}} = frac{1}{3} ] Solving for ( b + c ): [ b + c = frac{1}{3} times frac{5}{2} = frac{5}{6} ]5. We now have two equations: [ a + b = 1 ] [ b + c = frac{5}{6} ] Substituting ( c = frac{1}{6} ) into the second equation: [ b + frac{1}{6} = frac{5}{6} ] Solving for ( b ): [ b = frac{5}{6} - frac{1}{6} = frac{4}{6} = frac{2}{3} ]6. Substituting ( b = frac{2}{3} ) into the first equation: [ a + frac{2}{3} = 1 ] Solving for ( a ): [ a = 1 - frac{2}{3} = frac{1}{3} ]7. Therefore, Alex's rate is ( frac{1}{3} ) of the lawn per hour. To find how many hours it would take Alex to rake the entire lawn by himself: [ text{Time} = frac{1}{a} = frac{1}{frac{1}{3}} = 3 text{ hours} ]The final answer is ( boxed{3} ) hours.