🧠:Okay, let's try to tackle this problem. The question is asking for all positive integers n between 1 and 100 such that the expression (n! × (n+1)!)/2 is a perfect square. Then we need to find the sum of all such n. Hmm, that sounds a bit tricky, but let me break it down step by step.First, let's recall that a perfect square is a number that can be expressed as some integer squared. So, for the given expression to be a perfect square, all the exponents in its prime factorization must be even. That means we need to analyze the prime factors of n! and (n+1)! and see how they combine when multiplied together and divided by 2.Let me start by simplifying the expression. The given expression is (n! × (n+1)!)/2. Notice that (n+1)! is equal to (n+1) × n!. So substituting that in, the expression becomes:(n! × (n+1) × n!)/2 = (n!² × (n+1))/2So now we have (n!² × (n+1)) / 2. Since n!² is already a perfect square, the problem reduces to determining when (n+1)/2 is a perfect square. Wait, is that correct? Let me check.Actually, n!² is a perfect square, so if (n+1)/2 is also a perfect square, then their product would be a perfect square. However, (n+1)/2 needs to be an integer as well, right? Because n!² is an integer, so (n!² × (n+1))/2 must be an integer. Therefore, 2 must divide (n+1). So n+1 must be even, meaning n is odd. So n must be odd. So first condition: n is odd. Then (n+1)/2 must be an integer, and also a perfect square.Wait, but let's verify this. Let me take a small n to test. Let's say n=1. Then (1! × 2!)/2 = (1×2)/2 = 1, which is 1², a perfect square. So n=1 works. Here, n is odd, (n+1)/2=1, which is 1². That fits.Another example: n=3. Then (3! ×4!)/2 = (6×24)/2=72. Is 72 a perfect square? 8²=64, 9²=81, so no. Wait, 72 is not a perfect square. But according to the earlier reasoning, (n+1)/2=(3+1)/2=2, which is not a perfect square, so that's why 72 isn't a square. So that seems to check out.Wait, but let's take n=7. Then (7! ×8!)/2. Let's compute (7!² ×8)/2. 8/2=4, so 7!² ×4. That is (7! ×2)², which is a perfect square. Therefore, n=7 should work. Let's verify:7! = 5040, 8! = 40320. So (5040×40320)/2 = (5040×20160) = let's compute that. 5040×20160. But 5040×20160 = 5040×20160. Let's note that 5040×20160 = 5040×20160. Let me compute 5040×20000 = 100,800,000. 5040×160=806,400. So total is 100,800,000 + 806,400 = 101,606,400. The square root of that is sqrt(101,606,400). Let's see, 10,000²=100,000,000. 10,080²=101,606,400. Yes, exactly. So 10,080² is 101,606,400. Therefore, yes, n=7 works. So that works because (n+1)/2=8/2=4=2², which is a perfect square.So in that case, n=7 works. So that's correct. So the logic is, the expression is (n!² × (n+1))/2. For this to be a perfect square, (n+1)/2 must be a perfect square, because n!² is already a perfect square, and multiplying two perfect squares gives another perfect square. So if (n+1)/2 is a perfect square, then the entire expression is a perfect square. However, we need to ensure that (n+1)/2 is an integer. Therefore, n+1 must be even, so n must be odd. Then, (n+1)/2 = k², where k is a positive integer. Therefore, n+1 = 2k², so n = 2k² -1. Then since n must be between 1 and 100, inclusive, we need to find all integers k such that 2k² -1 ≤ 100. Let's solve for k.So 2k² -1 ≤ 100 ⇒ 2k² ≤ 101 ⇒ k² ≤ 50.5 ⇒ k ≤ sqrt(50.5). sqrt(50) is approximately 7.07, sqrt(50.5) is slightly more, around 7.1. So k can be 1,2,...,7. Therefore, the possible values of k are 1 through 7. Then n = 2k² -1.Calculating for k from 1 to 7:k=1: n=2(1)-1=1k=2: n=8-1=7k=3: n=18-1=17k=4: n=32-1=31k=5: n=50-1=49k=6: n=72-1=71k=7: n=98-1=97Wait, k=7 gives n=2(49)-1=97. Then k=8 would give n=2(64)-1=127, which is over 100, so stop at k=7.Therefore, the possible n are 1,7,17,31,49,71,97.Now, we need to verify if these all satisfy the original condition. Wait, we need to check if for these n, the expression is indeed a perfect square. Let's check another one. Let's take n=17. Then (17! ×18!)/2. According to the earlier logic, this should be (17!² ×18)/2=17!² ×9= (17! ×3)², which is a perfect square. Let's compute 17! ×3 squared. Yes, that would be (3×17!)². So indeed, the expression is a perfect square. Similarly, for n=31, (31! ×32!)/2 = (31!² ×32)/2 =31!² ×16= (31! ×4)², which is a perfect square. So this seems to hold.But wait, let's take n=1. Then (1! ×2!)/2=(1×2)/2=1=1². Correct. n=7 as checked earlier works. So all these n satisfy the condition. Therefore, the possible n are 1,7,17,31,49,71,97. Then the sum would be 1+7=8, 8+17=25, 25+31=56, 56+49=105, 105+71=176, 176+97=273.But wait, before finalizing, let me check if there could be other n where even though (n+1)/2 is not a perfect square, the entire expression might still be a perfect square. Because maybe the division by 2 affects the exponents in the prime factors in such a way that even if (n+1)/2 is not a square, combining with the n!²'s exponents could result in all exponents being even. Is this possible?Let me think. The expression is (n!² × (n+1)) / 2. For this to be a perfect square, all primes in its factorization must have even exponents. Since n!² already contributes even exponents for all primes up to n. Then, when we multiply by (n+1) and divide by 2, we need to adjust the exponents. Let's consider the prime factors of (n+1). Let's denote n+1 as m. So m = n+1. Then the expression becomes (n!² × m)/2.Since n! includes all primes up to n. If m is greater than n, which it is (since m = n+1), then m is a prime number if n+1 is prime, otherwise it's composite. So if m is prime, then m is a new prime not present in n! So when we multiply n!² by m, the exponent of m in the product becomes 2 (from n!²) times 0 (since m isn't in n!) plus 1 from multiplying by m, so total exponent 1. Then we divide by 2, which is a prime factor 2. So let's think in terms of prime exponents.Let me take an example where (n+1)/2 is not a perfect square. Let's take n=5. Then the expression is (5!×6!)/2 = (120×720)/2 = (120×360)=43200. Is 43200 a perfect square? sqrt(43200)=sqrt(432×100)=sqrt(432)×10≈20.78×10≈207.8, which is not an integer. So 43200 is not a perfect square. But n=5 is odd, (n+1)/2=3, which is not a perfect square, so the earlier logic holds. So in this case, even though (n+1)/2 is 3, which is not a square, the entire expression is not a square. So that seems to confirm that the initial logic is correct.Another example: n=9. Then (9!×10!)/2. Let's compute 9!² ×10 /2. 10/2=5. So 9!² ×5. 5 is a prime not squared here, so the exponent of 5 in the product would be 2*(exponent of 5 in 9!) +1. The exponent of 5 in 9! is floor(9/5)=1. So 2*1 +1=3. So exponent 3 is odd, so the entire expression is not a perfect square. So indeed, unless 5 is squared, which would require the exponent to be even. But here, we have exponent 3. So unless we can compensate by dividing by 2, but in this case, dividing by 2 only affects the exponent of 2. Let's check the exponent of 2 in the expression. In 9!², exponent of 2 is 2*(floor(9/2)+floor(9/4)+floor(9/8))=2*(4+2+1)=14. Then multiply by 10 (which has one 2), so exponent becomes 14 +1=15. Then divide by 2: exponent of 2 becomes 15 -1=14, which is even. The exponent of 5 is 2*(floor(9/5)) +1=2*1 +1=3, which is odd. Hence, the entire number cannot be a perfect square. Therefore, even if (n+1)/2 is not a square, the other exponents from (n+1) can introduce odd exponents. Therefore, unless (n+1)/2 is a perfect square, the entire expression cannot be a perfect square.Wait, but let's think again. Suppose (n+1) itself is a square multiplied by 2. For example, suppose n+1 = 2k², then (n+1)/2 =k². Wait, that's exactly the case we considered before. So in that case, when we have n+1=2k², then (n+1)/2=k², which is a square. Then when we multiply n!² by (n+1) and divide by 2, we get n!² ×k², which is (n! ×k)², a perfect square. So that works.But what if n+1 has other factors that are squares? For example, suppose n+1 is 8=2³. Then (n+1)/2=4=2², which is a square, so that would work. Wait, n+1=8, so n=7, which is in our list. If n+1=18=2×9=2×3². Then (n+1)/2=9=3². So n=17, which is in our list. So in this case, even if n+1 has square factors, as long as when divided by 2, the result is a square, then it works. So our initial logic is correct.But let me check a case where n+1 is a square times 2, but maybe with another square factor. For example, n+1=2×k²×m². Wait, if n+1=2×(k×m)², then (n+1)/2=(k×m)², so that would still be a square. So as long as (n+1)/2 is a square, regardless of the factors, it works. So our initial conclusion is that n+1 must be twice a perfect square, hence n=2k²-1. Therefore, the possible n are those where n+1 is twice a square. So our list of n is correct.But to be thorough, let's consider another example where (n+1)/2 is a square. Let's take k=3, so n+1=2×9=18, so n=17. Then the expression is (17!×18!)/2=17!²×18/2=17!²×9=(17!×3)². Which is a perfect square. Yes. Similarly, n=31: n+1=32=2×16=2×4², so (n+1)/2=16=4². Therefore, the expression is (31!×32!)/2=31!²×32/2=31!²×16=(31!×4)². Perfect square.Therefore, the logic holds. So all n where n+1 is twice a perfect square, i.e., n=2k²-1, and n ≤100, gives us the values 1,7,17,31,49,71,97. Let me verify the last one: k=7, n=2×49 -1=97. Then n+1=98, which is 2×49=2×7², so (n+1)/2=49=7². Hence, the expression is (97!×98!)/2=97!²×98/2=97!²×49=(97!×7)². Perfect square. Correct.Therefore, all these n satisfy the condition, and there are no others because for other n, (n+1)/2 is not a perfect square, which would introduce an odd exponent in the prime factorization when considering the multiplication by (n+1) and division by 2. Hence, only the n of the form 2k²-1 work.Therefore, the sum is 1 + 7 +17 +31 +49 +71 +97. Let's compute that step by step:1 +7 =88 +17=2525 +31=5656 +49=105105 +71=176176 +97=273So the sum is 273. But just to ensure there are no mistakes in the calculation, let me add them again:1,7,17,31,49,71,97.Adding 1 and 7: 88 +17: 2525 +31: 5656 +49: 105105 +71: 176176 +97: 273. Yes, correct.Therefore, the answer should be 273. However, let me check once more if all these n are within 1 to 100. The largest n is 97, which is within 100. The next k=8 would give n=2×64 -1=127, which is over 100, so excluded. Therefore, all seven values are valid.Final check: Is there any other possibility where even if (n+1)/2 is not a square, but when combined with the n!², the division by 2 could result in all exponents even?Suppose, for example, that n+1 has a factor that is a square and the division by 2 removes an odd exponent. Let's take an example. Suppose n+1=2×a²×b, where b is square-free. Then (n+1)/2=a²×b. If b is 1, then it's a square. But if b is not 1, then (n+1)/2 is not a square. However, when multiplying by n!², which already contains factors of a² and b (depending on n). Wait, but n! contains primes up to n, and n+1 is greater than n, so if n+1 has a prime factor larger than n, then that prime is not present in n! and hence when multiplying by (n+1), that prime's exponent becomes 1 (since it's not in n!²), then divided by 2 (which doesn't affect it if it's another prime). Therefore, unless that prime factor is squared in n+1, but since n+1 is being divided by 2, which is a separate prime.This seems too vague. Let's think with a concrete example. Suppose n+1=50=2×25=2×5². Then (n+1)/2=25=5², so n=49. Which is in our list. So this works.Another example: n+1= 72=2×36=2×6². Then n=71. Which is in our list.But suppose n+1= 12=2×6=2×2×3. Then (n+1)/2=6, which is not a square. So n=11. Let's check if the expression is a perfect square. (11!×12!)/2=11!²×12/2=11!²×6. 6 is 2×3. So the exponents of 2 and 3 in 11!² are even. Then multiplying by 6 adds 1 to the exponent of 2 and 1 to the exponent of 3. Hence, exponents of 2 and 3 become odd. Therefore, the entire number is not a perfect square. Hence, n=11 does not work.Alternatively, if n+1= 2×k²×m², but this is covered by our previous analysis. So unless m is 1, then we get back to the case where n+1=2k².Therefore, I think the initial reasoning is correct: only when (n+1)/2 is a perfect square, hence n=2k²-1, are the conditions satisfied. Therefore, the sum is indeed 273.

✅:The given expression (frac{n! times (n+1)!}{2}) can be rewritten as (frac{n!^2 times (n+1)}{2}). For this to be a perfect square, all prime exponents in its factorization must be even. Since (n!^2) is already a perfect square, ((n+1)/2) must also be a perfect square. This requires (n+1) to be even (so (n) is odd), and ((n+1)/2 = k^2) for some integer (k). Thus, (n = 2k^2 - 1).To find valid (n) within (1 leq n leq 100), solve (2k^2 - 1 leq 100):[2k^2 leq 101 implies k^2 leq 50.5 implies k leq 7]Thus, (k = 1, 2, ldots, 7), yielding (n = 2k^2 - 1):[n = 1, 7, 17, 31, 49, 71, 97]Summing these values:[1 + 7 + 17 + 31 + 49 + 71 + 97 = 273]Answer: (boxed{273})

🔑:1. Given the problem, we need to find all positive integers ( n ) such that the expression [ frac{n! times (n+1)!}{2} ] is a perfect square. 2. Start with simplifying the given expression: [ frac{n! times (n+1)!}{2} ] Rewrite ((n+1)!) as ((n+1) times n!): [ frac{n! times (n+1) times n!}{2} = frac{(n!)^2 times (n+1)}{2} ] Therefore, the expression becomes: [ frac{(n!)^2 times (n+1)}{2} = k^2 ] for some integer ( k ).3. For (frac{(n!)^2 times (n+1)}{2}) to be a perfect square, (frac{n+1}{2}) must also be a perfect square: [ frac{n+1}{2} = m^2 ] where ( m ) is some integer.4. Rearrange the above equation to find ( n ): [ n + 1 = 2m^2 ] [ n = 2m^2 - 1 ]5. We must find all suitable values of ( n ) within the given range, ( 1 leq n leq 100 ): [ n = 2m^2 - 1 ] Check the values one by one for ( m ): [ begin{aligned} &text{If } m = 1, && n = 2(1)^2 - 1 = 1, &text{If } m = 2, && n = 2(2)^2 - 1 = 7, &text{If } m = 3, && n = 2(3)^2 - 1 = 17, &text{If } m = 4, && n = 2(4)^2 - 1 = 31, &text{If } m = 5, && n = 2(5)^2 - 1 = 49, &text{If } m = 6, && n = 2(6)^2 - 1 = 71, &text{If } m = 7, && n = 2(7)^2 - 1 = 97. end{aligned} ] For ( m geq 8 ), ( n = 2m^2 - 1 ) would exceed 100, so we stop here.6. Add all these valid ( n ) values together: [ 1 + 7 + 17 + 31 + 49 + 71 + 97 ] [ = 273 ] Conclusion:[boxed{273}]

🧠:Okay, so the problem is: It's currently 1:00 a.m., and I need to figure out what time it will be 2023 minutes from now. Hmm, let me think. Alright, first, I remember that time calculations usually involve converting minutes into hours and minutes because clocks cycle every 12 or 24 hours. Since it's 1:00 a.m., I suppose we're using a 12-hour clock, but maybe I should consider the 24-hour format to avoid confusion between AM and PM.Let me start by breaking down 2023 minutes into hours and minutes. There are 60 minutes in an hour, so I can divide 2023 by 60 to get the number of hours and the remaining minutes. Let me do that division. 2023 divided by 60. Let's see, 60 times 33 is 1980. Because 60*30=1800, 60*3=180, so 1800+180=1980. Then 2023 minus 1980 is 43. So that would be 33 hours and 43 minutes. Wait, so 33 hours and 43 minutes added to 1:00 a.m. But adding 33 hours to 1:00 a.m. How does that work? Let me convert 33 hours into days and hours. There are 24 hours in a day. So 33 divided by 24 is 1 day and 9 hours. So 24 hours is one full day, which brings us back to the same time the next day. Then adding the remaining 9 hours. Starting at 1:00 a.m., adding 24 hours would bring us to 1:00 a.m. the next day. Then adding 9 hours to that: 1:00 a.m. + 9 hours = 10:00 a.m. Then we have the remaining 43 minutes. So 10:00 a.m. plus 43 minutes is 10:43 a.m. Wait, but hold on. Let me verify that step again. Wait, 33 hours is 1 day and 9 hours, right? So adding 1 day to 1:00 a.m. brings it to 1:00 a.m. the next day. Then adding 9 hours: 1:00 a.m. + 9 hours = 10:00 a.m. Then add 43 minutes: 10:43 a.m. So the total time after 2023 minutes would be 10:43 a.m. But let me check another way. Alternatively, maybe breaking it down step by step. Let's convert 2023 minutes into hours and minutes again. 2023 divided by 60 is 33 with a remainder of 43, so 33 hours and 43 minutes. Now, if I add 24 hours to 1:00 a.m., it becomes 1:00 a.m. the next day. So 33 hours is 24 + 9. So 24 hours later is same time next day, then 9 hours more. So 1:00 a.m. + 24 hours = 1:00 a.m. next day. Then adding 9 hours: 1:00 a.m. + 9 hours = 10:00 a.m. Then adding 43 minutes: 10:43 a.m. Alternatively, maybe using a different approach. Let's convert the current time to minutes since midnight, add the 2023 minutes, then convert back to time. Current time is 1:00 a.m., which is 1 hour after midnight. So in minutes, that's 1*60 = 60 minutes. Total minutes after adding 2023: 60 + 2023 = 2083 minutes. Now, convert 2083 minutes back into hours and minutes. Divide by 60. 2083 divided by 60: 60*34 = 2040, so 2083 - 2040 = 43. So 34 hours and 43 minutes. But 34 hours is 24 + 10 hours, which is 1 day and 10 hours. So starting from midnight, 34 hours later is midnight + 24 hours = next midnight, then 10 hours and 43 minutes. So midnight + 10 hours is 10:00 a.m., plus 43 minutes is 10:43 a.m. Hmm, same result. So that seems to confirm it. Wait, but let me check again. Maybe I made a mistake in the first step. Let me verify the division. 2023 divided by 60: 60*30 = 1800. 2023 - 1800 = 223. Then 60*3 = 180. 223 - 180 = 43. So total hours are 30 + 3 = 33, and 43 minutes. So yes, 33 hours and 43 minutes. Alternatively, maybe adding the hours step by step. Starting at 1:00 a.m., adding 24 hours gets to 1:00 a.m. next day. Then adding the remaining 9 hours (since 33 -24=9). So 1:00 a.m. +9 hours is 10:00 a.m., then add 43 minutes: 10:43 a.m. But let me think again. Another way: Suppose I add 24 hours to 1:00 a.m., which is the same time next day. Then 2023 minutes is 33 hours and 43 minutes. So 24 hours is one day, leaving 9 hours and 43 minutes. So adding 9 hours to 1:00 a.m. is 10:00 a.m., then 43 minutes gives 10:43 a.m. Alternatively, maybe using modulo 24 for the hours. So 33 hours is how many days? 33 divided by 24 is 1.375 days. So 1 day (24 hours) and 9 hours. So adding 1 day brings us to 1:00 a.m. next day, then 9 hours to 10:00 a.m., then 43 minutes. Another check: Let's count days. 2023 minutes divided by 60 is 33.716... hours. 33.716 divided by 24 is approximately 1.4048 days. So 1 day and about 0.4048 days. 0.4048 days times 24 hours is about 9.716 hours, which is 9 hours and 43 minutes (since 0.716*60 ≈ 43). So same result. Alternatively, perhaps using a 24-hour clock to make it easier. 1:00 a.m. is 01:00. Adding 33 hours: 01:00 + 33 hours. 33 hours is 24 +9. So 01:00 +24 hours is 01:00 next day. Then +9 hours is 10:00 (10:00 a.m.), then +43 minutes: 10:43. Alternatively, adding hours step by step. Let's say starting at 1:00 a.m. Adding 24 hours: 1:00 a.m. next day. Then 33-24=9 hours. So 1:00 a.m. +9 hours: 1+9=10, so 10:00 a.m. Then +43 minutes: 10:43 a.m. All methods lead to 10:43 a.m. So that seems consistent. Wait, but let me check if I converted the minutes correctly. 2023 divided by 60: 60*30=1800, 2023-1800=223. Then 60*3=180, 223-180=43. So 33 hours 43 minutes. Correct. Alternatively, 60*33=1980, 2023-1980=43. Yes. So 33 hours is 1 day and 9 hours. So adding 1 day to 1:00 a.m. gives 1:00 a.m. next day. Then 9 hours added to that is 10:00 a.m. Then 43 minutes added gives 10:43 a.m. Another way: Let's count the hours step by step. Starting at 1:00 a.m. If I add 12 hours, that would be 1:00 p.m. Adding another 12 hours would be 1:00 a.m. next day. That's 24 hours. Then we have 33-24=9 hours left. So from 1:00 a.m. next day, adding 9 hours: 1:00 a.m. + 9 hours = 10:00 a.m. Then add 43 minutes: 10:43 a.m. Yes. Alternatively, let's do it in smaller increments. Let's add 24 hours (1 day) first. Then add 9 hours and 43 minutes. 1:00 a.m. +24 hours = 1:00 a.m. next day. Then adding 9 hours: 1:00 a.m. +9 hours = 10:00 a.m. Then adding 43 minutes: 10:00 a.m. +43 minutes = 10:43 a.m. All right. So regardless of the method, the result is 10:43 a.m. Wait, but let me check if there's any daylight saving time or other factors. The problem doesn't mention anything like that, so I can ignore that. Also, the problem is phrased as "What time will it be 2023 minutes from now?" So the answer should be in the same time zone, no mention of any changes. So 10:43 a.m. is the answer. Wait, but let me make sure about the AM/PM. Since we started at 1:00 a.m., adding 33 hours and 43 minutes. Let's think in 24-hour format to avoid confusion. 1:00 a.m. is 01:00. Adding 33 hours: 01:00 +33:43. Wait, 33 hours is 33:00, plus 43 minutes is 33:43. So adding 33:43 to 01:00. In 24-hour time: 01:00 + 33 hours. 33 hours is 1 day and 9 hours. So 01:00 +1 day = 01:00 next day. Then +9 hours = 10:00. Then +43 minutes = 10:43. So 10:43 in 24-hour time is 10:43 a.m. Yes. Another sanity check: Let's add 2023 minutes step by step. First, convert 2023 minutes to days: 2023 / (60*24) = 2023 / 1440 ≈ 1.4048 days. So 1 day and 0.4048 days. 0.4048 days *24 hours ≈ 9.716 hours, which is 9 hours and 43 minutes, as before. So adding 1 day, 9 hours, 43 minutes. Starting at 1:00 a.m., add 1 day: 1:00 a.m. next day. Add 9 hours: 10:00 a.m. Add 43 minutes: 10:43 a.m. Yes. Therefore, the final answer is 10:43 a.m.

✅:To determine the time 2023 minutes after 1:00 a.m., follow these steps:1. Convert 2023 minutes to hours and minutes: - Divide 2023 by 60: ( 2023 div 60 = 33 text{ hours (quotient)}, quad 2023 - (60 times 33) = 43 text{ minutes (remainder)} ). Thus, 2023 minutes = 33 hours and 43 minutes.2. Break down 33 hours into days and hours: - ( 33 div 24 = 1 text{ day (quotient)}, quad 33 - (24 times 1) = 9 text{ hours (remainder)} ). So, 33 hours = 1 day and 9 hours.3. Add the time incrementally: - Start at 1:00 a.m. - Add 1 day (24 hours): ( 1:00 text{a.m.} + 24 text{hours} = 1:00 text{a.m.} text{(next day)} ). - Add 9 hours: ( 1:00 text{a.m.} + 9 text{hours} = 10:00 text{a.m.} ). - Add 43 minutes: ( 10:00 text{a.m.} + 43 text{minutes} = 10:43 text{a.m.} ).Final Answer: 10:43 a.m.

🔑:1. We start by determining how many hours and minutes are contained in 2023 minutes. To do this, we divide 2023 minutes by 60 (since there are 60 minutes in an hour): [ 2023 div 60 = 33 text{ hours with a remainder of } 43 text{ minutes} ] This can be expressed as: [ 2023 = 33 times 60 + 43 ] Thus, 2023 minutes is equal to 33 hours and 43 minutes.2. Next, we consider how this converts to the time of day starting from 1:00 a.m. Adding 33 hours will cycle past one full day, as 24 hours constitutes a day. So, we break down the 33 hours as: [ 33 text{ hours} = 24 text{ hours} + 9 text{ hours} ]3. After 24 hours from 1:00 a.m., it will be 1:00 a.m. again, because 24 hours is exactly one day.4. We then add the remaining 9 hours to this time: [ 1:00 text{ a.m.} + 9 text{ hours} = 10:00 text{ a.m.} ]5. Finally, we add the 43 minutes to 10:00 a.m.: [ 10:00 text{ a.m.} + 43 text{ minutes} = 10:43 text{ a.m.} ] Conclusion:[boxed{10:43 text{ a.m.}}]

🧠:Okay, so I need to prove that when you rotate a line a around the axis of rotation, which is perpendicular to a, the image of a (which is a') lies in the same plane as a, and the angle between a and a' is equal to the angle of rotation. Hmm, let me start by visualizing this.First, let's recall what a rotation in 3D space does. An axis of rotation is a line about which all points rotate by a certain angle. If the line a is perpendicular to this axis, then every point on a should move in a circular path around the axis. Since the axis is perpendicular to a, the line a must lie in a plane that's perpendicular to the axis. Wait, so if I imagine the axis of rotation as, say, the z-axis in a coordinate system, then a line perpendicular to the z-axis would be in the xy-plane. Rotating this line around the z-axis by an angle θ should keep it in the xy-plane because the rotation around the z-axis doesn't affect the z-coordinate. So the image a' of the line a after rotation should still lie in the xy-plane, which is the same plane as a. That makes sense for the first part.But I need to formalize this. Let me set up a coordinate system. Let’s assume the axis of rotation is the z-axis. Then a line perpendicular to the z-axis would lie in a plane parallel to the xy-plane. Wait, but if the line is in the xy-plane, then rotating it around the z-axis would keep it in the xy-plane. Therefore, both a and a' are in the xy-plane, so they lie in the same plane. That seems straightforward. So the first part is proven by choosing the coordinate system appropriately and noting that rotation around the z-axis doesn't change the z-coordinate, hence keeping the line in the xy-plane.Now, for the angle between a and a' being equal to the angle of rotation. Let's think about a specific example. Suppose line a is along the x-axis in the xy-plane. If we rotate it by an angle θ around the z-axis, the image a' would make an angle θ with the original x-axis. The angle between a and a' would then be θ. That works for this case. But how to generalize this?Let me consider any line a perpendicular to the z-axis. Let’s parametrize line a. Since it's perpendicular to the z-axis, it can be represented in the xy-plane. Suppose line a is given by some direction vector (l, m, 0) passing through a point (x0, y0, z0), but since it's perpendicular to the z-axis, all points on a must have the same z-coordinate. Wait, but if the axis is the z-axis, then any line perpendicular to the z-axis must lie in a plane z = constant. However, the axis of rotation is the z-axis, so rotating around the z-axis doesn't change the z-coordinate. Therefore, the entire line a, lying in a plane z = k, when rotated about the z-axis, will remain in z = k, hence the image a' is also in z = k. Therefore, a and a' lie in the same plane z = k. So that covers the first part.Now, the angle between a and a'. Let’s consider two points on line a. Suppose line a is in the plane z = k, and let's take two points P and Q on line a. After rotation by angle θ around the z-axis, these points become P' and Q' on line a'. The angle between vectors OP and OP' (assuming rotation about the origin) would be θ. But since line a is a straight line, the entire line is rotated, so the angle between the original line and the rotated line should be θ. Wait, maybe a better approach is to consider the direction vectors of the lines. If line a has a direction vector v in the xy-plane, then rotating v by θ around the z-axis would result in a direction vector v' which is v rotated by θ. The angle between v and v' is θ. Therefore, the angle between the lines a and a' is θ. Since lines are infinite in both directions, the angle between two lines in the same plane is defined as the smallest angle between their direction vectors, which in this case is θ. That seems right.But let me check for a non-axis-aligned line. Suppose line a is in the xy-plane but not along the x or y-axis. For example, a line with direction vector (1, 1, 0). Rotating this by θ around the z-axis would result in a direction vector (cosθ - sinθ, sinθ + cosθ, 0) if we apply a rotation matrix. Wait, actually, applying the rotation matrix to the direction vector (1,1,0) would give (cosθ - sinθ, sinθ + cosθ, 0). The angle between the original vector (1,1,0) and the rotated vector can be found using the dot product. The cosine of the angle between them is [(1)(cosθ - sinθ) + (1)(sinθ + cosθ)] divided by the product of their magnitudes. Let's compute:Numerator: cosθ - sinθ + sinθ + cosθ = 2cosθOriginal vector magnitude: sqrt(1^2 + 1^2) = sqrt(2)Rotated vector magnitude: sqrt( (cosθ - sinθ)^2 + (sinθ + cosθ)^2 ) = sqrt( cos²θ - 2cosθsinθ + sin²θ + sin²θ + 2sinθcosθ + cos²θ ) = sqrt(2cos²θ + 2sin²θ) = sqrt(2(cos²θ + sin²θ)) = sqrt(2)Therefore, cosine of the angle is 2cosθ / (sqrt(2)*sqrt(2)) ) = 2cosθ / 2 = cosθ. Therefore, the angle between the direction vectors is θ. Hence, the angle between the lines is θ. That's exactly what we needed to prove.So putting it all together: since the line a is perpendicular to the axis of rotation (z-axis), it lies in a plane perpendicular to the axis (like the xy-plane or a plane parallel to it). Rotating a around the z-axis by angle θ keeps it in the same plane (since rotation around z doesn't affect the z-coordinate) and rotates its direction vector by θ, resulting in the angle between a and a' being θ. Therefore, both conditions are satisfied.I should also consider if there are any edge cases or other possibilities. For example, if the line a is not passing through the origin, does that affect the angle? Well, the angle between two lines is determined by their direction vectors, not their positions. So even if the line is translated in the plane (but still perpendicular to the axis), the direction vector remains the same after rotation, so the angle between the lines is still θ. Another thought: the axis of rotation could be any line in space, not necessarily the z-axis. But the problem states that the line a is perpendicular to the axis of rotation. So if the axis is an arbitrary line, then line a must lie in a plane perpendicular to that axis. However, rotating around the axis would still keep line a' in the same plane because all points on a rotate in circles perpendicular to the axis, hence within the plane. The angle between the lines would still be equal to the rotation angle θ. To formalize this without relying on a specific coordinate system, maybe using vectors. Let’s denote the axis of rotation as a line L. Let line a be perpendicular to L, so they meet at a point O (since a is perpendicular to L, they must intersect at some point; otherwise, if they are skew, the perpendicularity might not hold). Let’s take O as the origin. Then line L can be considered as the z-axis, and line a lies in the xy-plane passing through O. Rotating line a around L (z-axis) by angle θ will map it to line a' in the same xy-plane, and the angle between a and a' is θ. Alternatively, if line a does not pass through the axis of rotation but is still perpendicular to it, then it must lie in a plane parallel to some plane perpendicular to the axis. For example, if the axis is the z-axis, line a could be in the plane z = k, but displaced from the origin. However, rotation around the z-axis would still keep it in z = k, and the direction vector would rotate by θ, hence the angle between a and a' is θ. The displacement doesn't affect the angle between the lines because angle is determined by direction, not position.Therefore, regardless of whether the line a intersects the axis or is parallel but offset, the angle between a and its image a' after rotation is equal to the angle of rotation θ, and they lie in the same plane. So the proof holds in general.

✅:To prove that a line ( a ), perpendicular to the axis of rotation, maps under rotation to a line ( a' ) in the same plane as ( a ), with the angle between ( a ) and ( a' ) equal to the angle of rotation ( theta ), follow these steps: 1. Coordinate System Setup- Let the axis of rotation be the ( z )-axis.- Since ( a ) is perpendicular to the ( z )-axis, it lies in a plane parallel to the ( xy )-plane (e.g., ( z = k ), where ( k ) is constant). 2. Effect of Rotation on Plane and Lines- Rotation about the ( z )-axis by ( theta ) preserves the ( z )-coordinate of all points. Thus, the entire plane ( z = k ) (containing ( a )) is invariant under rotation. This ensures ( a' ) also lies in ( z = k ), so ( a ) and ( a' ) are coplanar. 3. Direction Vectors and Angle Between Lines- Let ( mathbf{v} ) be the direction vector of ( a ). Since ( a ) is in ( z = k ), ( mathbf{v} = (l, m, 0) ).- Under rotation by ( theta ), ( mathbf{v} ) transforms to ( mathbf{v}' = (lcostheta - msintheta, lsintheta + mcostheta, 0) ).- The angle between ( mathbf{v} ) and ( mathbf{v}' ) is computed via the dot product: [ cosphi = frac{mathbf{v} cdot mathbf{v}'}{|mathbf{v}| |mathbf{v}'|} = frac{l(lcostheta - msintheta) + m(lsintheta + mcostheta)}{l^2 + m^2}. ] Simplifying: [ cosphi = costheta implies phi = theta. ]- Since lines ( a ) and ( a' ) share the same plane and their direction vectors differ by ( theta ), the angle between ( a ) and ( a' ) is ( theta ). 4. Generalization to Arbitrary Axis- For a general axis ( L ), decompose space into planes perpendicular to ( L ). Line ( a ), being perpendicular to ( L ), lies in one such plane. Rotation about ( L ) preserves these planes and rotates direction vectors by ( theta ), ensuring ( a' ) remains in the same plane and the angle between ( a ) and ( a' ) is ( theta ). ConclusionAfter rotation:1. ( a' ) lies in the same plane as ( a ).2. The angle between ( a ) and ( a' ) equals the rotation angle ( theta ).Final Answer: A line ( a ) perpendicular to the axis of rotation maps to a line ( a' ) in the same plane, and the angle between ( a ) and ( a' ) is exactly the angle of rotation. This follows from the invariance of the plane under rotation and the transformation of direction vectors by ( theta ). (boxed{text{The angle between } a text{ and } a' text{ equals the rotation angle, and they lie in the same plane.}})

🔑:# 1. Define the Setup:Let ( l ) be the axis of rotation and let us consider a plane ( alpha ) which is perpendicular to ( l ) and intersects ( l ) at some point ( O ). Assume ( a ) to be a line perpendicular to the axis ( l ) and lying on the plane ( alpha ).# 2. Consider the Point and its Image under Rotation:Let ( A ) be a point on the line ( a ) such that ( (O A) perp a ). When we perform the rotation around the axis ( l ), the point ( A ) will map to ( A' ) in the plane ( alpha ), maintaining the perpendicularity to ( l ).# 3. Determine Relationship of ( A ) and ( A' ):In the plane ( alpha ), consider a right-angle triangle ( triangle OAM ), where ( M ) is another point on ( a ). Upon rotation, the point ( M ) maps to ( M' ), and ( M' ) lies on the image line ( a' ), which is also perpendicular to ( l ).# 4. Relationships of Rotated Points:- Since ( |OM'| = |OM| ) due to the properties of the rotation preserving distances.- The rotation angle ( varphi ) between ( OA ) and ( OA' ) means that ( angle MOA = angle A'OM' = varphi ).# 5. Prove Isoceles Triangles:Triangles ( triangle OAM ) and ( triangle OA'M' ) are congruent because they share ( |OA| = |OA'| ) and ( |OM| = |OM'| ), and the angle between them remains ( varphi ).# 6. Prove Lines ( a ) and ( a' ) Maintain Angle:Given ( angle AOM = angle A'OM' ) and both triangles being congruent, the angle between lines ( a ) and ( a' ) is preserved and equal to the rotation angle ( varphi ).# Conclusion:The line ( a ) after the rotation maps onto the line ( a' ) in such a way that they lie on the same plane (( alpha )) and the angle between ( a ) and ( a' ) is exactly the rotation angle ( varphi ). Hence, it is proven that any line perpendicular to the rotation axis maintains the correct angular relationship after rotation.[boxed{text{The angle between } a text{ and } a' text{ is equal to the rotation angle } varphi.}](blacksquare)

🧠:Okay, so I need to show that the sum of s(1) + s(2) + ... + s(n-1) is equal to n times s(n) minus n. Let me start by understanding what s(n) is. It's the nth harmonic number, right? So s(n) = 1 + 1/2 + 1/3 + ... + 1/n. The problem wants me to sum these harmonic numbers from s(1) up to s(n-1) and show that this sum equals n*s(n) - n. Hmm. Let me write that down more formally. Let's denote the left-hand side (LHS) as S = s(1) + s(2) + ... + s(n-1). The right-hand side (RHS) is n*s(n) - n. I need to show S = RHS.First, maybe I can compute both sides for a small n to see if the equality holds. Let's take n=2. Then LHS is s(1) = 1. RHS is 2*s(2) - 2. s(2) = 1 + 1/2 = 3/2. So RHS = 2*(3/2) - 2 = 3 - 2 = 1. Okay, it works for n=2.Try n=3. LHS is s(1) + s(2) = 1 + (1 + 1/2) = 2.5. RHS is 3*s(3) - 3. s(3) = 1 + 1/2 + 1/3 ≈ 1.8333. So RHS = 3*1.8333 - 3 ≈ 5.5 - 3 = 2.5. That matches too. Let me check n=4. LHS: s(1)+s(2)+s(3) = 1 + 1.5 + 1.8333 ≈ 4.3333. RHS: 4*s(4) -4. s(4) = 1 + 1/2 + 1/3 + 1/4 ≈ 2.0833. So 4*2.0833 ≈ 8.3333 -4 ≈ 4.3333. Okay, so the examples check out. That's reassuring.Now, to prove it in general. Let's think about how to approach this. Maybe induction? Or perhaps manipulating the sum directly. Let's try both approaches and see which one works.First, induction. Let's see. For n=2, we already checked the base case. Assume that for some k ≥2, the statement holds: s(1) + ... + s(k-1) = k*s(k) -k. Then, for n = k+1, we need to show that s(1) + ... + s(k) = (k+1)*s(k+1) - (k+1).Using the induction hypothesis, the left-hand side would be [k*s(k) -k] + s(k). So that's k*s(k) -k + s(k) = (k+1)*s(k) -k. We need to show this equals (k+1)*s(k+1) - (k+1). Let's see:(k+1)*s(k+1) - (k+1) = (k+1)(s(k) + 1/(k+1)) - (k+1) = (k+1)s(k) + (k+1)*(1/(k+1)) - (k+1) = (k+1)s(k) +1 - (k+1) = (k+1)s(k) -k. Which matches the left-hand side. So by induction, the formula holds. Therefore, the statement is true for all n ≥2. But since we checked n=2, and induction step works, that's a valid proof.Alternatively, maybe there's a direct algebraic way. Let's consider the sum S = s(1) + s(2) + ... + s(n-1). Each s(k) is the sum of reciprocals up to 1/k. So if we write out S, it would be:s(1) = 1s(2) = 1 + 1/2s(3) = 1 + 1/2 + 1/3...s(n-1) = 1 + 1/2 + 1/3 + ... + 1/(n-1)So when we sum all these, how many times does each term 1/m (for m from 1 to n-1) appear? For each m, the term 1/m appears in s(m), s(m+1), ..., s(n-1). So the number of times 1/m appears is (n-1) - m +1 = n - m. Therefore, the total sum S can be written as the sum over m=1 to n-1 of (n - m)/m.So S = Σ_{m=1}^{n-1} (n - m)/m = Σ_{m=1}^{n-1} (n/m - 1) = n Σ_{m=1}^{n-1} 1/m - Σ_{m=1}^{n-1} 1.But Σ_{m=1}^{n-1} 1/m is s(n-1), and Σ_{m=1}^{n-1} 1 = (n-1). So S = n*s(n-1) - (n-1). Wait, but the RHS we need is n*s(n) - n. So how does this relate?Wait, let's see. If S = n*s(n-1) - (n-1), and we need to show S = n*s(n) -n. Therefore, is n*s(n) -n equal to n*s(n-1) - (n-1)? Let's check:n*s(n) -n = n*(s(n-1) + 1/n) -n = n*s(n-1) + n*(1/n) -n = n*s(n-1) +1 -n = n*s(n-1) - (n-1). Which is exactly S. So that works. Therefore, S = n*s(n) -n.So this direct method also confirms the formula. Therefore, the identity holds.Alternatively, another way to think about it: Let's consider the difference between n*s(n) and the sum S. Let's write n*s(n) as n*(1 + 1/2 + ... + 1/n). Then, n*s(n) = n*1 + n*(1/2) + ... + n*(1/n). The sum S is the sum of s(1) to s(n-1), which as we saw earlier, can be represented as Σ_{k=1}^{n-1} Σ_{m=1}^k 1/m. Changing the order of summation, this becomes Σ_{m=1}^{n-1} Σ_{k=m}^{n-1} 1/m = Σ_{m=1}^{n-1} (n - m)/m. Which is the same as before.But perhaps integrating these two expressions. Let's see:n*s(n) = n*(1 + 1/2 + ... + 1/n) = sum_{m=1}^n n/mS = sum_{k=1}^{n-1} s(k) = sum_{k=1}^{n-1} sum_{m=1}^k 1/m = sum_{m=1}^{n-1} sum_{k=m}^{n-1} 1/m = sum_{m=1}^{n-1} (n - m)/mTherefore, n*s(n) - S = sum_{m=1}^n n/m - sum_{m=1}^{n-1} (n - m)/mLet's compute this difference:For m from 1 to n-1, the term is n/m - (n - m)/m = [n - (n - m)]/m = m/m = 1. So each term from m=1 to n-1 contributes 1, and there's an additional term from m=n in n*s(n), which is n/n = 1. Therefore, n*s(n) - S = sum_{m=1}^{n-1} 1 + 1 = (n-1) +1 = n. Therefore, n*s(n) - S = n => S = n*s(n) -n. Which is exactly the identity we needed to prove.So that's another way to look at it by considering the difference between n*s(n) and S, and realizing that the difference is n, leading to the conclusion that S = n*s(n) -n.All these methods converge to the same result, so the identity holds. The key idea seems to be switching the order of summation or using induction. Both approaches validate the formula. It's interesting how different methods can lead to the same conclusion. The induction approach is straightforward once you see the inductive step, but the direct algebraic manipulation by changing the order of summation also works neatly. Another perspective is computing the difference between n*s(n) and the sum S, which directly gives the desired result by telescoping or term-wise comparison.Therefore, after exploring different methods, the proof holds through multiple valid approaches, reinforcing the correctness of the identity.

✅:To prove that ( s(1) + s(2) + ldots + s(n-1) = n s(n) - n ), where ( s(n) = sum_{k=1}^n frac{1}{k} ), we use the following steps:Step 1: Expand the Sum ( S = sum_{k=1}^{n-1} s(k) )[S = sum_{k=1}^{n-1} sum_{m=1}^k frac{1}{m}]Step 2: Swap the Order of Summation[S = sum_{m=1}^{n-1} sum_{k=m}^{n-1} frac{1}{m} = sum_{m=1}^{n-1} frac{n - m}{m}]Step 3: Split the Fraction[S = sum_{m=1}^{n-1} left( frac{n}{m} - 1 right) = n sum_{m=1}^{n-1} frac{1}{m} - sum_{m=1}^{n-1} 1]Step 4: Simplify Using ( s(n) )[S = n left( s(n) - frac{1}{n} right) - (n - 1) = n s(n) - 1 - n + 1 = n s(n) - n]Conclusion:[s(1) + s(2) + ldots + s(n-1) = n s(n) - n](boxed{n s(n) - n})

🔑:To prove the statement [s(1) + s(2) + cdots + s(n-1) = n cdot s(n) - n,] where (s(n) = 1 + frac{1}{2} + frac{1}{3} + cdots + frac{1}{n},) we will use mathematical induction.1. Base Case: - For (n = 1): [ text{LHS: } s(1) = 1 ] Since the left-hand side (LHS) is considered empty (sum of no terms), the base case is trivially true. - For (n = 2): [ s(1) = 1 ] [ text{LHS: } s(1) = 1 ] [ text{RHS: } 2 cdot s(2) - 2 = 2 left( 1 + frac{1}{2} right) - 2 = 2 cdot frac{3}{2} - 2 = 3 - 2 = 1 ] Hence, the equation holds for (n = 2). 2. Inductive Step: - Assume the statement is true for some (k geq 2): [ s(1) + s(2) + cdots + s(k-1) = k cdot s(k) - k ] - We need to prove it for (k + 1): [ s(1) + s(2) + cdots + s(k-1) + s(k) ] By the inductive hypothesis: [ s(1) + s(2) + cdots + s(k-1) = k cdot s(k) - k ] So: [ s(1) + s(2) + cdots + s(k-1) + s(k) = k cdot s(k) - k + s(k) ] [ = k cdot s(k) - k + left( s(k) right) ] [ = k cdot s(k) + s(k) - k ] [ = (k + 1) cdot s(k) - k ] - Now, using the definition of (s(n)): [ s(k + 1) = s(k) + frac{1}{k+1} ] Therefore: [ (k+1) cdot s(k+1) - (k+1) = (k+1) left(s(k) + frac{1}{k+1}right) - (k+1) ] [ = (k+1) cdot s(k) + (k+1) cdot frac{1}{k+1} - (k+1) ] [ = (k + 1) cdot s(k) + 1 - (k + 1) ] [ = (k + 1) cdot s(k) - k ] - Since both conditions match: [ s(1) + s(2) + cdots + s(k) = (k + 1) cdot s(k + 1) - (k + 1) ] 3. Conclusion: By mathematical induction, the given statement holds for all (n). [ boxed{s(1) + s(2) + cdots + s(n-1) = n cdot s(n) - n} ]

🧠:Okay, so I need to find the last 5 digits of the product 99 × 10101 × 111 × 1001001. Hmm, let's see. Since the question is about the last five digits, that means I need to compute the product modulo 100000. Because when you take a number modulo 10^n, you get its last n digits. So, modulo 100000 in this case. Alright, so maybe instead of multiplying all these big numbers together and then taking modulo 100000, which could be cumbersome, I can take each number modulo 100000 first and then multiply them step by step, taking modulo 100000 at each step to keep the numbers small. That should work because of the property that (a × b) mod m = [(a mod m) × (b mod m)] mod m. Yeah, that should make the calculations manageable.Let me start by breaking down each number modulo 100000.First, 99 mod 100000 is just 99. That's straightforward.Next, 10101 mod 100000. Well, 10101 is less than 100000, right? Wait, 100000 is 10^5 which is 100,000. So 10101 is 10,101. So yes, less than 100,000. So 10101 mod 100000 is 10101.Then, 111 mod 100000 is 111. Again, straightforward.Finally, 1001001 mod 100000. Let's compute that. 1001001 divided by 100000 is 10 with a remainder of 1001. Because 100000 × 10 = 1,000,000. Then 1,001,001 - 1,000,000 = 1,001. So 1001001 mod 100000 is 1001.So now, the problem reduces to calculating (99 × 10101 × 111 × 1001) mod 100000.Hmm, let me write this step by step. Let's compute the product incrementally, taking modulo 100000 at each step.First, multiply 99 and 10101:99 × 10101. Let me compute that. Let's do 99 × 10101.Breaking it down: 10101 × 100 = 1,010,100. Subtract 10101 from that: 1,010,100 - 10,101 = 999,999. Wait, no, that's if it's 99 × 10101. Wait, actually, 99 × 10101 is the same as (100 - 1) × 10101 = 100×10101 - 10101 = 1,010,100 - 10,101 = 999,999. Oh, that's clever. So 99 × 10101 = 999,999.Now, 999,999 mod 100,000. Since 999,999 divided by 100,000 is 9 with a remainder of 99,999. So 999,999 mod 100000 = 99,999.Okay, so after the first multiplication step, we have 99,999.Next, multiply this by 111. So 99,999 × 111. Let me compute that.Again, maybe break it down. 99,999 × 100 = 9,999,900. Then 99,999 × 10 = 999,990. Then 99,999 × 1 = 99,999. Adding those together: 9,999,900 + 999,990 = 10,999,890 + 99,999 = 11,099,889. So 99,999 × 111 = 11,099,889.Now, take this modulo 100,000. So 11,099,889 divided by 100,000 is 110 with remainder 99,889. Therefore, 11,099,889 mod 100000 = 99,889.So now, we have 99,889. Next, multiply by 1001. So 99,889 × 1001. Hmm, that's a big number. Let's see if I can find a smarter way to compute this modulo 100,000.Alternatively, note that 1001 mod 100000 is 1001, so we have to compute 99,889 × 1001 mod 100000.But perhaps we can break it down:99,889 × 1001 = 99,889 × (1000 + 1) = 99,889 × 1000 + 99,889 × 1.Compute each part:99,889 × 1000 = 99,889,000.99,889 × 1 = 99,889.Now, add these together: 99,889,000 + 99,889 = 99,988,889.Now, take this modulo 100,000. So 99,988,889 divided by 100,000 is 999 with a remainder of 88,889. Therefore, 99,988,889 mod 100,000 = 88,889.So putting it all together, the result of 99 × 10101 × 111 × 1001001 modulo 100,000 is 88,889. Therefore, the last five digits are 88889.Wait, but let me double-check my steps because that seems straightforward, but maybe I made an arithmetic error somewhere.Let me verify each step again.First, 99 × 10101:Yes, 99 × 10101 is 999,999. Then modulo 100,000 is 99,999. That's correct.Next, 99,999 × 111:Breaking down 99,999 × 100 = 9,999,90099,999 × 10 = 999,99099,999 × 1 = 99,999Adding up:9,999,900 + 999,990 = 10,999,89010,999,890 + 99,999 = 11,099,889. Correct.11,099,889 mod 100,000 is indeed 99,889. Because 11,000,000 mod 100,000 is 00,000, then 99,889 remains.Then 99,889 × 1001:Breaking into 99,889 × 1000 = 99,889,00099,889 × 1 = 99,889Adding gives 99,889,000 + 99,889 = 99,988,889Modulo 100,000: 99,988,889 divided by 100,000 is 999, remainder 88,889. So yes, 88,889.Therefore, the last five digits are 88889.Wait, but let me cross-validate using another approach. Maybe compute the product step by step but using modulo 100,000 at each intermediate step.Alternatively, perhaps factor each number into its prime factors and compute modulo 100,000. But that might complicate. Alternatively, notice that 100000 = 10^5 = 2^5 × 5^5. Maybe compute the product modulo 2^5 and modulo 5^5, then apply the Chinese Remainder Theorem. Hmm, that might be a more advanced method but could be a good check.But since the numbers involved are manageable, maybe not necessary. However, to ensure accuracy, let's try this approach.First, compute the product modulo 32 (since 2^5 = 32) and modulo 3125 (since 5^5 = 3125). Then combine the results using Chinese Remainder Theorem.First, modulo 32:Compute each term:99 mod 32: 99 ÷32=3×32=96, remainder 3. So 99 ≡3 mod32.10101 mod32: Let's divide 10101 by32.32×315=10080, 10101-10080=21. So 10101≡21 mod32.111 mod32: 32×3=96, 111-96=15. So 111≡15 mod32.1001001 mod32: Let's compute 1001001 ÷32.But 32×31281=1,000,992. 1001001 -1,000,992= 9. So 1001001≡9 mod32.Therefore, the product modulo32 is 3×21×15×9.Compute step by step:3×21=63; 63 mod32=63-2×32=63-64= -1≡31 mod32.31×15=465; 465 mod32: 32×14=448, 465-448=17. So 17 mod32.17×9=153; 153 mod32: 32×4=128, 153-128=25. So 25 mod32.So the product ≡25 mod32.Now, modulo3125:Compute each term:99 mod3125=99.10101 mod3125: Let's divide 10101 by3125.3125×3=9375, 10101-9375=726. So 10101≡726 mod3125.111 mod3125=111.1001001 mod3125: Let's compute 1001001 ÷3125.3125×320=1,000,000. So 1001001 -1,000,000=1001. So 1001001≡1001 mod3125.Therefore, the product modulo3125 is 99×726×111×1001 mod3125.This seems complicated. Let's compute step by step.First, compute 99×726 mod3125.Compute 99×726:Calculate 100×726=72,600. Subtract 1×726=726. So 72,600 -726=71,874.71,874 mod3125: Let's divide 71,874 by3125.3125×22=68,750. 71,874 -68,750=3,124. So 71,874≡3,124 mod3125.Wait, 3125×22=68,750. 71,874-68,750=3,124. So 3,124 is less than 3125? Wait, no: 3,124 is just 1 less than 3,125. So 3,124 mod3125=3,124.So 99×726≡3,124 mod3125.Next, multiply by 111:3,124×111 mod3125.Compute 3,124×100=312,4003,124×10=31,2403,124×1=3,124Total: 312,400 +31,240=343,640 +3,124=346,764.Now, 346,764 mod3125. Let's divide 346,764 by3125.3125×100=312,500. 346,764 -312,500=34,264.3125×10=31,250. 34,264 -31,250=3,014.So total quotient is 100+10=110, remainder 3,014. So 346,764≡3,014 mod3125.Then multiply by 1001:3,014×1001 mod3125.Again, break into 3,014×1000 +3,014×1.3,014×1000=3,014,0003,014×1=3,014Total:3,014,000 +3,014=3,017,014.Now, 3,017,014 mod3125. Let's compute.3125×965=3125×(900+65)=3125×900=2,812,500; 3125×65=203,125. Total 2,812,500+203,125=3,015,625.3,017,014 -3,015,625=1,389.So 3,017,014≡1,389 mod3125.Therefore, the product ≡1,389 mod3125.So now, we have the product ≡25 mod32 and ≡1,389 mod3125.We need to find a number x such that:x ≡25 mod32x ≡1389 mod3125We can write x =3125k +1389 for some integer k. Then substitute into the first equation:3125k +1389 ≡25 mod32Compute 3125 mod32 and 1389 mod32.3125 ÷32: 32×97=3,104. 3125-3,104=21. So 3125≡21 mod32.1389 ÷32: 32×43=1,376. 1389-1,376=13. So 1389≡13 mod32.Thus:21k +13 ≡25 mod3221k ≡25 -13=12 mod32So 21k ≡12 mod32We need to solve for k: 21k ≡12 mod32First, find the modular inverse of 21 mod32.21 and32 are coprime (gcd(21,32)=1), so inverse exists.Find an integer m such that21m≡1 mod32.Trying m= 21*21=441≡441-13*32=441-416=25 mod32.Not 1. Try m= 21*5=105≡105-3*32=105-96=9 mod32.21*13=273≡273-8*32=273-256=17 mod32.21* 21*29=21*(32-3)=21*29=609≡609-19*32=609-608=1 mod32. Wait, 21*29 mod32=?Wait, 21*29: 20*29=580, 1*29=29. Total 609. 609 divided by32:32*19=608, so 609-608=1. So 21*29≡1 mod32. Therefore, the inverse of21 mod32 is29.Thus, k ≡12*29 mod32.Compute 12*29=348. 348 mod32: 32*10=320, 348-320=28. So k≡28 mod32.Therefore, k=32m +28 for some integer m.Then x=3125k +1389=3125*(32m +28)+1389=3125*32m +3125*28 +1389.Compute 3125*28:3125*28=3125*(20+8)=62,500 +25,000=87,500. Wait, no:Wait, 3125*20=62,5003125*8=25,000So total 62,500 +25,000=87,500. Then 87,500 +1,389=88,889.Therefore, x=3125*32m +88,889. Thus, the solution is x≡88,889 mod100,000.Therefore, the last five digits are 88889. Which matches the previous result.So both methods confirm that the last five digits are 88889. Hence, the answer should be boxed{88889}.Final Answerboxed{88889}

✅:To find the last 5 digits of (99 times 10101 times 111 times 1001001), we need to compute the product modulo (100000). 1. Modulo Reduction of Each Factor: - (99 mod 100000 = 99) - (10101 mod 100000 = 10101) - (111 mod 100000 = 111) - (1001001 mod 100000 = 1001)2. Step-by-Step Multiplication with Modulo 100000: - First multiplication: (99 times 10101 = 999999) - (999999 mod 100000 = 99999) - Second multiplication: (99999 times 111 = 11099889) - (11099889 mod 100000 = 99889) - Third multiplication: (99889 times 1001 = 99988889) - (99988889 mod 100000 = 88889)3. Verification Using Chinese Remainder Theorem: - Compute modulo (32) and (3125) separately, then combine results. - Modulo (32): The product is (25 mod 32). - Modulo (3125): The product is (1389 mod 3125). - Combining results using Chinese Remainder Theorem confirms the result (88889 mod 100000).Thus, the last five digits of (99 times 10101 times 111 times 1001001) are (boxed{88889}).

🔑:1. Consider the given expression: (99 times 10101 times 111 times 1001001).2. To find the last 5 digits, we will compute the given product modulo ( 10^5 ).3. First, observe the factors modulo (10^5): - (99) mod (10^5 = 99) - (10101) mod (10^5 = 10101) - (111) mod (10^5 = 111) - (1001001) mod (10^5 = 1001) 4. We now compute the product (99 times 10101 times 111 times 1001) mod (10^5).5. Break this computation into parts and repeatedly reduce modulo (10^5): [ begin{align*} 99 times 10101 &equiv 99 times 10101 mod 10^5 99 times 10101 &= 999999, 999999 mod 10^5 &equiv 99999. end{align*} ]6. Next step: [ begin{align*} 99999 times 111 &equiv 11100000 - 111 mod 10^5 &= 11110889 mod 10^5, &equiv 8889. end{align*} ]8. Now for the final multiplication: [ begin{align*} 8889 times 1001 &equiv (8889 times 1000 + 8889) mod 10^5 &= (8889000 + 8889) mod 10^5, &= 8897889 mod 10^5 &equiv 88889. end{align*} ] Conclusion:The last 5 digits of (99 times 10101 times 111 times 1001001 ) are ( boxed{88889} ).

🧠:Okay, so I need to find a positive integer n that satisfies the equation 1/(2^10) + 1/(2^9) + 1/(2^8) = n/(2^10). Let me think about how to approach this. First, all the terms on the left side have denominators that are powers of 2, and the right side also has a denominator of 2^10. Maybe I can combine the left side into a single fraction with denominator 2^10 and then compare the numerators. That seems like a good plan because the denominators would be the same, making it easier to solve for n.Let me write down each term with denominator 2^10. The first term is already 1/(2^10), so that's easy. The second term is 1/(2^9). To convert that to denominator 2^10, I need to multiply both numerator and denominator by 2. So, 1/(2^9) = 2/(2^10). Similarly, the third term is 1/(2^8). To convert that to denominator 2^10, I have to multiply numerator and denominator by 2^2, which is 4. Therefore, 1/(2^8) = 4/(2^10).Now, substituting these back into the equation, we get:1/(2^10) + 2/(2^10) + 4/(2^10) = n/(2^10).Since all the denominators are the same, I can add the numerators:(1 + 2 + 4)/2^10 = n/2^10.Adding those numbers together: 1 + 2 is 3, and 3 + 4 is 7. So the left side becomes 7/(2^10). Therefore, the equation simplifies to 7/(2^10) = n/(2^10).Since the denominators are equal and non-zero, the numerators must be equal. Therefore, n = 7.Wait, let me verify that again. If I add 1, 2, and 4, that's 7. Yeah, that's correct. So n is 7. Let me check if there's another way to approach this in case I missed something.Alternatively, I could factor out 1/(2^10) from each term on the left side. Let's see:1/(2^10) + 1/(2^9) + 1/(2^8) = 1/(2^10) + (1/2)/(2^9 * 1/2) + (1/4)/(2^8 * 1/4). Hmm, maybe that's complicating things. Wait, another way: express each term in terms of 2^10.Alternatively, maybe convert all terms to a common denominator step by step. Let's see. Start with the first two terms: 1/(2^10) + 1/(2^9). The common denominator here would be 2^10. So as before, 1/(2^10) + 2/(2^10) = 3/(2^10). Then add the third term, which is 1/(2^8). The common denominator between 2^10 and 2^8 is 2^10, so 3/(2^10) + 4/(2^10) = 7/(2^10). So same result. Another approach: note that 1/(2^8) is equal to 4/(2^10). So, the third term is 4/(2^10), the second term is 2/(2^10), and the first term is 1/(2^10). Adding them gives 4 + 2 + 1 = 7. So n is 7. Wait, maybe I can check using decimal values. Let's compute each term:1/(2^10) = 1/1024 ≈ 0.00097656251/(2^9) = 1/512 ≈ 0.0019531251/(2^8) = 1/256 ≈ 0.00390625Adding these decimals: 0.0009765625 + 0.001953125 = 0.0029296875; then add 0.00390625 gives 0.0068359375.Now, n/(2^10) would be n/1024. If n = 7, then 7/1024 ≈ 0.0068359375, which matches the sum. So that confirms it.Therefore, n must be 7. I don't see any mistakes in the reasoning. The key steps were converting all terms to the same denominator, adding the numerators, and verifying the result both by fraction addition and decimal conversion. Everything checks out. So the answer should be 7.

✅:To determine the positive integer ( n ) that satisfies the equation:frac{1}{2^{10}} + frac{1}{2^{9}} + frac{1}{2^{8}} = frac{n}{2^{10}},we proceed as follows:1. Convert all terms to the common denominator ( 2^{10} ): - ( frac{1}{2^{10}} ) remains ( frac{1}{2^{10}} ). - ( frac{1}{2^{9}} = frac{2}{2^{10}} ) (multiply numerator and denominator by 2). - ( frac{1}{2^{8}} = frac{4}{2^{10}} ) (multiply numerator and denominator by ( 2^2 = 4 )).2. Sum the numerators: [ frac{1}{2^{10}} + frac{2}{2^{10}} + frac{4}{2^{10}} = frac{1 + 2 + 4}{2^{10}} = frac{7}{2^{10}}. ]3. Equate to the right-hand side: [ frac{7}{2^{10}} = frac{n}{2^{10}}. ] Since the denominators are equal, the numerators must be equal. Thus, ( n = 7 ).Verification via decimal conversion:- ( frac{1}{2^{10}} approx 0.0009765625 )- ( frac{1}{2^{9}} approx 0.001953125 )- ( frac{1}{2^{8}} approx 0.00390625 )- Sum: ( 0.0009765625 + 0.001953125 + 0.00390625 = 0.0068359375 )- ( frac{7}{2^{10}} = frac{7}{1024} approx 0.0068359375 ), confirming consistency.Final Answer: (boxed{7})

🔑:Given the equation:[frac{1}{2^{10}}+frac{1}{2^{9}}+frac{1}{2^{8}} = frac{n}{2^{10}}]1. Express each term with a common denominator: We need to make the denominators the same so that we can combine the fractions. Notice that the common denominator here is (2^{10}). [ frac{1}{2^{10}} + frac{1}{2^{9}} + frac{1}{2^{8}} ]2. Rewrite each term with the common denominator: [ frac{1}{2^{10}} + frac{1}{2^{9}} + frac{1}{2^{8}} = frac{1}{2^{10}} + frac{2}{2^{10}} + frac{2^{2}}{2^{10}} ] Here, (frac{1}{2^9}) can be written as (frac{2}{2^{10}}) because (2^9 = 2 cdot 2^{8}), and (frac{1}{2^8}) can be written as (frac{2^2}{2^{10}}) because (2^8 = 4 cdot 2^6).3. Combine the numerators: Now, we add the numerators: [ frac{1}{2^{10}} + frac{2}{2^{10}} + frac{4}{2^{10}} = frac{1 + 2 + 4}{2^{10}} ]4. Calculate the sum in the numerator: [ 1 + 2 + 4 = 7 ] Therefore, we have: [ frac{1 + 2 + 4}{2^{10}} = frac{7}{2^{10}} ]5. Match with the right-hand side: According to the given equation: [ frac{7}{2^{10}} = frac{n}{2^{10}} ]6. Conclude the value of (n): Therefore, (n = 7). Conclusion[boxed{7}]