answer：To solve for ( A ), I start by translating the given conditions into mathematical equations.First, since 1047 leaves a remainder of 23 when divided by ( A ), I can write:[1047 = A times k + 23]This implies:[A times k = 1047 - 23 = 1024]So, ( A ) must be a divisor of 1024.Next, considering the second condition, 1047 leaves a remainder of 7 when divided by ( A + 1 ):[1047 = (A + 1) times m + 7]This simplifies to:[(A + 1) times m = 1047 - 7 = 1040]Therefore, ( A + 1 ) must be a divisor of 1040.To find ( A ), I list the divisors of 1024 and 1040 and look for a pair where one is exactly one less than the other.The divisors of 1024 are:[1, 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024]The divisors of 1040 include:[1, 2, 4, 5, 8, 10, 13, 16, 20, 26, 40, 52, 65, 80, 104, 130, 208, 260, 520, 1040]Looking for a pair where one divisor is one less than the other, I find that 32 and 33 fit this criterion.Thus, ( A = 32 ) and ( A + 1 = 33 ) satisfy both conditions.

answer：First, I need to determine the total amount of water required to fill the pool. Since the pool takes 50 hours to fill and the hose runs at 100 gallons per hour, the total water needed is 50 hours multiplied by 100 gallons per hour, which equals 5,000 gallons.Next, I calculate the cost of this water. Water costs 1 cent for every 10 gallons. Therefore, for 5,000 gallons, the cost is 5,000 divided by 10, resulting in 500 cents.Finally, I convert the cost from cents to dollars. Since 100 cents equal 1 dollar, 500 cents is equivalent to 5 dollars.So, the total cost to fill the pool is 5 dollars.

answer：Okay, so I have this problem: Find the maximum value of ( k ) such that ( 3^{11} ) can be expressed as the sum of ( k ) consecutive positive integers. Hmm, interesting. I remember that there's a formula for the sum of consecutive integers, so maybe I can use that.Let me recall the formula. If we have ( k ) consecutive integers starting from ( n ), the sum ( S ) is given by:[S = frac{k}{2} times [2n + (k - 1)]]Right? Because it's an arithmetic series where the first term is ( n ) and the common difference is 1. So, the sum is the number of terms times the average of the first and last term. The last term would be ( n + (k - 1) ), so the average is ( frac{n + (n + k - 1)}{2} = frac{2n + k - 1}{2} ). Multiply that by ( k ) terms, and you get the formula above.So, in this problem, ( S = 3^{11} ). Let me write that down:[3^{11} = frac{k}{2} times [2n + (k - 1)]]I need to find the maximum ( k ) such that this equation holds for some positive integer ( n ). So, let me rearrange this equation to solve for ( n ) in terms of ( k ):First, multiply both sides by 2:[2 times 3^{11} = k times [2n + k - 1]]Let me denote ( 2 times 3^{11} ) as a number. Calculating that, ( 3^{11} ) is 177147, so ( 2 times 177147 = 354294 ). So,[354294 = k times (2n + k - 1)]So, ( k ) must be a divisor of 354294. Since we're looking for the maximum ( k ), I should find the largest divisor ( k ) such that ( n ) is a positive integer.Let me express this as:[2n + k - 1 = frac{354294}{k}]Then, solving for ( n ):[2n = frac{354294}{k} - k + 1][n = frac{ frac{354294}{k} - k + 1 }{2}]Since ( n ) must be a positive integer, the expression ( frac{354294}{k} - k + 1 ) must be even and positive.So, two conditions:1. ( frac{354294}{k} - k + 1 ) is even.2. ( frac{354294}{k} - k + 1 > 0 ).Let me first consider the second condition:[frac{354294}{k} - k + 1 > 0][frac{354294}{k} > k - 1][354294 > k(k - 1)]So, ( k(k - 1) < 354294 ). Since ( k ) is a positive integer, this gives an upper bound on ( k ). Let me approximate ( k^2 ) is roughly 354294, so ( k ) is approximately ( sqrt{354294} ). Calculating that, ( sqrt{354294} ) is approximately 595.2, so ( k ) must be less than 595.2. So, the maximum possible ( k ) is 595, but we need to check if ( k=595 ) satisfies the conditions.But before that, let me note that 354294 is equal to ( 2 times 3^{11} ). So, the prime factorization of 354294 is ( 2 times 3^{11} ). Therefore, the divisors of 354294 are of the form ( 2^a times 3^b ), where ( a = 0 ) or ( 1 ), and ( b ) ranges from 0 to 11.So, all divisors can be written as ( 2 times 3^b ) or ( 3^b ), for ( b ) from 0 to 11.Therefore, the divisors are:- ( 1, 2, 3, 6, 9, 18, 27, 54, 81, 162, 243, 486, 729, 1458, 2187, 4374, 6561, 13122, 19683, 39366, 59049, 118098, 177147, 354294 ).Wait, hold on. Let me list them properly.Since 354294 is ( 2 times 3^{11} ), the number of divisors is (1+1)(11+1) = 24. So, there are 24 divisors. Let me list them:Starting from the smallest:1. ( 1 )2. ( 2 )3. ( 3 )4. ( 6 )5. ( 9 )6. ( 18 )7. ( 27 )8. ( 54 )9. ( 81 )10. ( 162 )11. ( 243 )12. ( 486 )13. ( 729 )14. ( 1458 )15. ( 2187 )16. ( 4374 )17. ( 6561 )18. ( 13122 )19. ( 19683 )20. ( 39366 )21. ( 59049 )22. ( 118098 )23. ( 177147 )24. ( 354294 )So, these are all the divisors. Now, since we need the maximum ( k ), we can start checking from the largest divisor and see if it satisfies the conditions.But wait, the largest divisor is 354294, but let's check if ( k=354294 ) is possible.Plugging into the equation:[n = frac{ frac{354294}{354294} - 354294 + 1 }{2} = frac{1 - 354294 + 1}{2} = frac{-354292}{2} = -177146]Negative, which is not a positive integer. So, ( k=354294 ) is invalid.Next, ( k=177147 ):[n = frac{ frac{354294}{177147} - 177147 + 1 }{2} = frac{2 - 177147 + 1}{2} = frac{-177144}{2} = -88572]Still negative. Not good.Next, ( k=118098 ):[n = frac{ frac{354294}{118098} - 118098 + 1 }{2} = frac{3 - 118098 + 1}{2} = frac{-118094}{2} = -59047]Still negative. Hmm.Continuing this way, but maybe it's too time-consuming. Perhaps there's a smarter way.Wait, let me think. Since ( n ) must be positive, the expression ( frac{354294}{k} - k + 1 ) must be positive and even.So, ( frac{354294}{k} - k + 1 > 0 implies frac{354294}{k} > k - 1 implies 354294 > k(k - 1) ).So, ( k^2 - k - 354294 < 0 ). Let me solve for ( k ):The quadratic equation ( k^2 - k - 354294 = 0 ).Using the quadratic formula:[k = frac{1 pm sqrt{1 + 4 times 354294}}{2} = frac{1 pm sqrt{1 + 1417176}}{2} = frac{1 pm sqrt{1417177}}{2}]Calculating ( sqrt{1417177} ). Let me see, 1190^2 is 1,416,100, which is close. 1190^2 = 1,416,100. Then, 1,417,177 - 1,416,100 = 1,077. So, 1190 + x squared is 1,417,177.Approximate x:(1190 + x)^2 = 1190^2 + 2*1190*x + x^2 ≈ 1,416,100 + 2380x.Set equal to 1,417,177:1,416,100 + 2380x ≈ 1,417,1772380x ≈ 1,077x ≈ 1,077 / 2380 ≈ 0.452.So, sqrt(1,417,177) ≈ 1190.452.Thus,k ≈ (1 + 1190.452)/2 ≈ 1191.452 / 2 ≈ 595.726.So, the positive root is approximately 595.726. Therefore, the inequality ( k^2 - k - 354294 < 0 ) holds for ( k < 595.726 ). So, the maximum integer ( k ) can be is 595.But wait, 595 is not a divisor of 354294, right? Let me check.Looking back at the list of divisors:1, 2, 3, 6, 9, 18, 27, 54, 81, 162, 243, 486, 729, 1458, 2187, 4374, 6561, 13122, 19683, 39366, 59049, 118098, 177147, 354294.So, 595 is not in the list. Therefore, the next lower divisor is 486.Wait, 486 is a divisor. Let me check ( k=486 ).Compute ( n ):[n = frac{ frac{354294}{486} - 486 + 1 }{2}]Calculate ( 354294 / 486 ):Divide 354294 by 486:First, 486 * 700 = 340,200.354,294 - 340,200 = 14,094.486 * 29 = 14,094 (since 486*30=14,580, which is too much, so 14,580 - 486=14,094).So, 700 + 29 = 729.So, ( 354294 / 486 = 729 ).Therefore,[n = frac{729 - 486 + 1}{2} = frac{244}{2} = 122]122 is a positive integer. So, ( k=486 ) is valid. So, that's a candidate.But wait, is 486 the maximum? Because 595 is not a divisor, so the next lower divisor is 486. But maybe there is a larger divisor between 486 and 595 that is a divisor of 354294.Wait, looking back at the list of divisors, after 486 comes 729, which is higher than 595, so 729 is too big because ( k=729 ) would give:Compute ( n ):[n = frac{354294 / 729 - 729 + 1}{2}]354294 / 729: Let's compute that.729 * 486 = 354,294. Wait, that's the same as 354294.Wait, 729 * 486 = 354,294. So, 354,294 / 729 = 486.Therefore,[n = frac{486 - 729 + 1}{2} = frac{-242}{2} = -121]Negative, so invalid. So, ( k=729 ) is invalid.So, the next lower divisor after 486 is 243.Wait, 243 is lower than 486, so 486 is higher. So, 486 is the highest divisor less than 595.726.Wait, but wait, 486 is 486, which is less than 595.726, but is there a higher divisor between 486 and 595.726?Looking at the list of divisors:After 486, the next higher divisor is 729, which is 729, which is higher than 595.726, so it's too big.Therefore, 486 is the highest divisor less than 595.726.But wait, 486 is 486, which is 486. So, is 486 the maximum k?But let me check if there are any other divisors between 486 and 595.726.Looking at the list:Divisors are 1, 2, 3, 6, 9, 18, 27, 54, 81, 162, 243, 486, 729, 1458, etc.So, between 486 and 729, there are no other divisors. So, 486 is indeed the next lower divisor.Therefore, ( k=486 ) is the maximum possible.Wait, but let me check ( k=243 ):Compute ( n ):[n = frac{354294 / 243 - 243 + 1}{2}]Calculate ( 354294 / 243 ):243 * 1000 = 243,000354,294 - 243,000 = 111,294243 * 400 = 97,200111,294 - 97,200 = 14,094243 * 58 = 14,094 (since 243*50=12,150; 243*8=1,944; 12,150+1,944=14,094)So, total is 1000 + 400 + 58 = 1458.Thus,[n = frac{1458 - 243 + 1}{2} = frac{1216}{2} = 608]Positive integer. So, ( k=243 ) is also valid, but it's smaller than 486, so 486 is still the maximum.Wait, but hold on. Let me think again. Maybe I can have a larger ( k ) if the expression ( frac{354294}{k} - k + 1 ) is even. So, perhaps even if ( k ) is not a divisor, but wait, no, ( k ) must divide 354294 because ( k times (2n + k - 1) = 354294 ). So, ( k ) must be a divisor.Therefore, ( k ) must be one of the divisors listed above.So, since 486 is the largest divisor less than 595.726, and it gives a positive integer ( n=122 ), which is valid, so 486 is the maximum ( k ).Wait, but let me check another divisor just above 486, but in the list, the next one is 729, which is too big, as we saw earlier.Wait, but 486 is 486, which is 486, and 486 is equal to ( 2 times 3^5 ). So, 486 is a divisor.But just to make sure, let me check ( k=486 ) gives ( n=122 ), which is positive, so that's good.But wait, is there a way to get a larger ( k ) beyond 486?Wait, if ( k ) is larger than 486, say, 487, but 487 is not a divisor of 354294, so ( k=487 ) won't work because ( 354294 ) divided by 487 is not an integer, so ( 2n + k - 1 ) would not be an integer, which is required because ( n ) must be integer.Therefore, ( k ) must be a divisor of 354294. So, the maximum ( k ) is 486.Wait, but hold on. Let me think again. Because 354294 is ( 2 times 3^{11} ), so all its divisors are of the form ( 2 times 3^b ) or ( 3^b ). So, the largest divisor less than 595.726 is 486, as 486 is ( 2 times 3^5 ), and the next one is 729, which is too big.Therefore, 486 is the maximum ( k ).But wait, let me check ( k=243 times 2=486 ). So, 486 is double of 243.Wait, but maybe I can find a larger ( k ) by considering the expression ( 2n + k - 1 ). Since ( k ) is a divisor, but perhaps if ( k ) is odd, then ( 2n + k -1 ) is even or odd?Wait, let me think about the parity.Looking back at the equation:[354294 = k times (2n + k - 1)]So, ( k ) and ( 2n + k -1 ) are two factors of 354294. Since 354294 is even, one of the factors must be even.So, either ( k ) is even, or ( 2n + k -1 ) is even.But ( 2n + k -1 ) is equal to ( 2n + (k -1) ). So, if ( k ) is odd, then ( k -1 ) is even, so ( 2n + (k -1) ) is even + even = even. If ( k ) is even, then ( k -1 ) is odd, so ( 2n + (k -1) ) is even + odd = odd.Thus, if ( k ) is even, then ( 2n + k -1 ) is odd, so ( k ) must be even and ( 2n + k -1 ) must be odd. If ( k ) is odd, then ( 2n + k -1 ) is even.So, in either case, one factor is even and the other is odd.Given that, since 354294 is even, one factor must be even, which is already satisfied.But how does this affect ( n )?Well, since ( n ) must be integer, ( frac{354294}{k} - k + 1 ) must be even. So, let's consider:Case 1: ( k ) is even.Then, ( frac{354294}{k} ) is an integer, since ( k ) divides 354294. Let me denote ( m = frac{354294}{k} ). So, ( m ) is an integer.Then, ( m - k + 1 ) must be even.Since ( k ) is even, ( m ) is ( frac{354294}{k} ). Since 354294 is even and ( k ) is even, ( m ) is an integer.So, ( m - k + 1 ) is even.Which implies ( m - k ) is odd.So, ( m ) and ( k ) have opposite parity.But ( k ) is even, so ( m ) must be odd.Therefore, ( m ) is odd.But ( m = frac{354294}{k} ). Since 354294 is even, and ( k ) is even, ( m ) is ( frac{even}{even} ), which could be integer or not, but in our case, ( k ) is a divisor, so ( m ) is integer.But for ( m ) to be odd, ( k ) must contain all the factors of 2 in 354294.Wait, 354294 is ( 2 times 3^{11} ). So, the exponent of 2 is 1. So, if ( k ) is even, it must contain exactly one factor of 2, because if ( k ) had more than one factor of 2, then ( m = frac{354294}{k} ) would have less than one factor of 2, which is not possible because 354294 only has one factor of 2.Therefore, if ( k ) is even, it must be of the form ( 2 times 3^b ), where ( b ) is from 0 to 11.Similarly, if ( k ) is odd, it must be of the form ( 3^b ), since 354294 has only one factor of 2.So, that gives us the list of divisors as before.So, for ( k ) even, ( k = 2 times 3^b ), and for ( k ) odd, ( k = 3^b ).So, in the case where ( k ) is even, ( m = frac{354294}{k} = frac{2 times 3^{11}}{2 times 3^b} = 3^{11 - b} ), which is odd, as expected.So, ( m - k + 1 = 3^{11 - b} - 2 times 3^b + 1 ) must be even.Which is:[3^{11 - b} - 2 times 3^b + 1]We need this to be even.Since ( 3^{11 - b} ) is odd (as any power of 3 is odd), ( 2 times 3^b ) is even, and 1 is odd.So, odd - even + odd = odd - even is odd, odd + odd is even. So, the entire expression is even.Therefore, for any even ( k ), the expression ( m - k + 1 ) is even, so ( n ) is integer.Therefore, for all even ( k ), ( n ) is integer as long as ( n ) is positive.Similarly, for odd ( k ), let's see.If ( k ) is odd, then ( m = frac{354294}{k} ) is even, because 354294 is even and ( k ) is odd, so ( m ) must be even.Then, ( m - k + 1 ) must be even.Since ( m ) is even, ( k ) is odd, so:Even - odd + 1 = even - odd is odd, odd + 1 is even. So, the expression is even.Therefore, for odd ( k ), ( n ) is also integer as long as ( n ) is positive.Therefore, regardless of whether ( k ) is odd or even, as long as ( k ) is a divisor of 354294, ( n ) will be integer. So, the only condition is that ( n ) is positive.Therefore, the problem reduces to finding the largest divisor ( k ) of 354294 such that:[frac{354294}{k} - k + 1 > 0]Which simplifies to:[frac{354294}{k} > k - 1][354294 > k(k - 1)]As before, which gives ( k < 595.726 ). So, the maximum integer ( k ) is 595, but since 595 is not a divisor, we take the largest divisor less than 595.726, which is 486.Therefore, ( k=486 ) is the maximum value.But wait, let me check ( k=486 ):Compute ( n ):[n = frac{354294 / 486 - 486 + 1}{2} = frac{729 - 486 + 1}{2} = frac{244}{2} = 122]Which is positive, so that's good.Is there a larger ( k ) that is a divisor of 354294 and less than 595.726? The next divisor after 486 is 729, which is too big, as we saw earlier.Therefore, 486 is indeed the maximum.But just to make sure, let me see if there are any other divisors between 486 and 595.726.Looking back at the list of divisors:1, 2, 3, 6, 9, 18, 27, 54, 81, 162, 243, 486, 729, 1458, 2187, 4374, 6561, 13122, 19683, 39366, 59049, 118098, 177147, 354294.So, between 486 and 729, there are no other divisors. So, 486 is the next one before 729, which is too big.Therefore, 486 is the maximum ( k ).Wait, but hold on. Let me think about the formula again.The sum of ( k ) consecutive integers starting from ( n ) is:[S = frac{k}{2} times [2n + (k - 1)]]Which is equal to ( 3^{11} ).So, ( frac{k}{2} times (2n + k - 1) = 3^{11} ).So, ( k times (2n + k - 1) = 2 times 3^{11} ).So, ( k ) must divide ( 2 times 3^{11} ), which is 354294.So, ( k ) is a divisor of 354294, which is ( 2 times 3^{11} ).Therefore, ( k ) must be of the form ( 2 times 3^b ) or ( 3^b ), as I thought earlier.So, the maximum ( k ) is 486, as it's the largest divisor less than 595.726.Therefore, the answer is ( boxed{486} ).**Final Answer**The maximum value of ( k ) is boxed{486}.

answer：To compute 98 multiplied by 102, I can use the difference of squares formula. First, I recognize that 98 is 100 minus 2, and 102 is 100 plus 2. This allows me to express the product as (100 - 2)(100 + 2).Applying the difference of squares formula, which states that (a - b)(a + b) equals a squared minus b squared, I can rewrite the expression as 100 squared minus 2 squared.Calculating 100 squared gives me 10,000, and 2 squared is 4. Subtracting these values, I find that 10,000 minus 4 equals 9,996.Therefore, the product of 98 and 102 is 9,996.

answer：Okay, so I have this problem where I need to distribute 4 college graduates to intern at three units: A, B, and C. Each unit has to receive at least one person. Also, there's a condition that student A will only go to unit A. I need to figure out how many different distribution plans there are. Hmm, let me think through this step by step.First, let me parse the problem. There are 4 students, and each needs to be assigned to one of three units: A, B, or C. But each unit must have at least one intern. So, no unit can be empty. Additionally, one of the students, specifically student A, can only go to unit A. So, student A is fixed in unit A, and the other three students can go to any of the units, but with the constraint that each unit must have at least one person.Wait, so student A is fixed in unit A, so unit A already has at least one person. That means units B and C each need to have at least one person as well. So, the other three students (let's call them B, C, D for simplicity) need to be assigned to units A, B, or C, but units B and C must each have at least one person. So, we can't have all three students going to unit A or all going to unit B or all going to unit C. Also, we can't have all three going to units A and B, leaving unit C empty, or all going to units A and C, leaving unit B empty. So, the assignments must cover all three units.But wait, since student A is already in unit A, if the other three students are assigned in such a way that units B and C each get at least one, then we're good. So, the problem reduces to assigning three students (B, C, D) to three units (A, B, C), with the condition that units B and C each have at least one student. So, we need to count the number of ways to assign these three students such that units B and C are non-empty.Alternatively, since each unit must have at least one person, and unit A already has student A, we need to ensure that units B and C each have at least one of the remaining three students. So, the number of ways is equal to the number of onto functions from the set of three students to the three units, but with the constraint that units B and C are non-empty. Wait, but since unit A can have more than one, but units B and C must each have at least one.Alternatively, maybe it's better to model this as distributing the three students into three units, with each unit being able to have any number, but units B and C must have at least one each. So, the number of distributions is equal to the total number of distributions without restrictions minus the distributions where unit B is empty or unit C is empty.But actually, since we have to ensure both units B and C are non-empty, it's similar to inclusion-exclusion. The total number of ways to assign the three students to three units is 3^3 = 27. Then, subtract the cases where unit B is empty and the cases where unit C is empty. But then, we have to add back the case where both unit B and unit C are empty because we subtracted it twice.Wait, but in this case, if both unit B and unit C are empty, that would mean all three students are assigned to unit A. But since student A is already in unit A, that's allowed, but in our problem, we need each unit to have at least one person. So, if all three students are assigned to unit A, then units B and C would be empty, which violates the condition. So, we need to exclude that case.So, using inclusion-exclusion, the number of valid assignments is:Total assignments: 3^3 = 27Subtract assignments where unit B is empty: 2^3 = 8 (since each student can go to A or C)Subtract assignments where unit C is empty: 2^3 = 8 (since each student can go to A or B)But now, we've subtracted too much because the case where both unit B and unit C are empty (all students go to A) has been subtracted twice. So, we need to add that back once.Number of assignments where both unit B and unit C are empty: 1^3 = 1 (all students go to A)So, applying inclusion-exclusion, the number of valid assignments is:27 - 8 - 8 + 1 = 12So, there are 12 ways to assign the three students such that units B and C are each non-empty.Therefore, the total number of distribution plans is 12.Wait, let me double-check that. So, total assignments: 3^3 = 27.Subtract assignments where unit B is empty: 2^3 = 8.Subtract assignments where unit C is empty: 2^3 = 8.Add back assignments where both unit B and C are empty: 1.So, 27 - 8 - 8 + 1 = 12.Yes, that seems correct.Alternatively, another way to think about it is that since each unit must have at least one person, and unit A already has student A, we need to distribute the remaining three students such that units B and C each have at least one. So, this is equivalent to the number of onto functions from three elements to two elements, but with the twist that unit A can also receive more students.Wait, no, actually, it's not exactly onto functions because unit A can have multiple students, but units B and C must have at least one each. So, it's similar to surjective functions from the three students to the three units, but with the condition that units B and C are non-empty, regardless of unit A.Wait, maybe another approach is to consider the possible distributions of the three students into units A, B, and C, with the constraints that B and C have at least one each.So, the number of ways is equal to the number of ways to assign the three students to the three units, minus the assignments where B is empty or C is empty, which is exactly what I did earlier.Alternatively, we can think of it as the number of ways to distribute the three students into units A, B, and C, with B and C each getting at least one. So, the number of such distributions is equal to the total number of distributions minus the distributions where B is empty minus the distributions where C is empty plus the distributions where both B and C are empty.Which is the same as inclusion-exclusion.So, 3^3 - 2*2^3 + 1 = 27 - 16 + 1 = 12.Yes, that's consistent.Alternatively, we can think of it as the number of ways to assign the three students to units A, B, and C, with each of B and C getting at least one. So, for each student, they can go to A, B, or C, but we need at least one in B and at least one in C.So, the number of such assignments is equal to the total number of assignments minus the assignments where all are in A or C (excluding B) minus the assignments where all are in A or B (excluding C) plus the assignments where all are in A (since we subtracted those twice).Which again is 3^3 - 2^3 - 2^3 + 1^3 = 27 - 8 - 8 + 1 = 12.So, that seems consistent.Alternatively, another way to think about it is to consider that each of the three students must be assigned to A, B, or C, with B and C each having at least one student. So, for each student, there are three choices, but we have constraints on the total counts.But that might complicate things, so inclusion-exclusion seems the straightforward approach.Alternatively, maybe using Stirling numbers of the second kind. The number of ways to partition n elements into k non-empty subsets is given by S(n, k), and then multiplied by k! for labeled boxes.But in this case, the units are labeled (A, B, C), so it's similar.But wait, in our case, we have three students and three units, but unit A can have any number, including zero, but in our problem, unit A already has student A, so we don't need to worry about unit A being empty. However, units B and C must have at least one each.Wait, so the problem is equivalent to distributing three distinguishable objects (students) into three distinguishable boxes (units A, B, C), with the condition that boxes B and C are non-empty. So, the number of such distributions is equal to the total number of distributions minus the distributions where B is empty minus the distributions where C is empty plus the distributions where both B and C are empty.Which is exactly what I did earlier, leading to 12.Alternatively, if I think in terms of surjective functions, the number of onto functions from three elements to three elements is 3! * S(3,3) = 6 * 1 = 6. But that's the number of ways to assign three students to three units such that each unit gets exactly one student. But in our case, we allow units to have more than one student, as long as B and C have at least one each.Wait, so that's different. So, the number of onto functions is 6, but that's only when each unit has exactly one student. But in our problem, we can have multiple students in a unit, as long as B and C have at least one each.So, that approach might not directly apply.Alternatively, maybe using multiplication principle.We have three students: B, C, D.Each can go to A, B, or C.But we need at least one in B and at least one in C.So, let's count the number of assignments where at least one goes to B and at least one goes to C.So, for each student, they have three choices, but we need to subtract the cases where all go to A and C, or all go to A and B.Wait, that's similar to inclusion-exclusion.Alternatively, think of it as:Total assignments: 3^3 = 27.Subtract assignments where no one goes to B: 2^3 = 8.Subtract assignments where no one goes to C: 2^3 = 8.Add back assignments where no one goes to B and no one goes to C: 1^3 = 1.So, 27 - 8 - 8 + 1 = 12.Same result.Alternatively, maybe using constructive counting.We need to assign the three students such that at least one goes to B and at least one goes to C.So, we can break it down into cases based on how many students go to B and C.Case 1: One student goes to B, one student goes to C, and the remaining one goes to A.Case 2: Two students go to B, one goes to C, and the remaining one goes to A.Case 3: One student goes to B, two go to C, and the remaining one goes to A.Case 4: All three students go to B or C, but ensuring at least one in each.Wait, but actually, in this approach, we have to consider all possible distributions where B and C each have at least one student, and the rest can go to A.So, let's enumerate the possible distributions.Each distribution can be characterized by the number of students in B and C, with the rest in A.So, possible distributions:- 1 in B, 1 in C, 1 in A.- 2 in B, 1 in C, 0 in A.- 1 in B, 2 in C, 0 in A.- 2 in B, 1 in C, 1 in A.- 1 in B, 2 in C, 1 in A.- 3 in B, 0 in C, but this is invalid because C must have at least one.- 0 in B, 3 in C, invalid.- 3 in B, 1 in C, but that's covered in the above.Wait, perhaps it's better to think in terms of the number of students in B and C.Since each must have at least one, the possible pairs (number in B, number in C) are:(1,1), (1,2), (2,1), (2,2), (1,3), (3,1). But since we have only three students, the maximum in B or C is 3.But wait, let's list all possible distributions where B and C each have at least one, and the rest go to A.So, for each possible number of students in B and C:1. B=1, C=1, A=1: This is one distribution where each unit has one student.2. B=1, C=2, A=0: All three students are distributed as 1 in B, 2 in C.3. B=2, C=1, A=0: Similarly, 2 in B, 1 in C.4. B=1, C=2, A=1: Wait, but that would require 1 in B, 2 in C, and 1 in A, but we only have three students. So, 1+2+1=4, which is more than 3. So, that's not possible.Wait, no, actually, if we have three students, and we assign 1 to B, 2 to C, then the remaining 0 go to A. Similarly, 2 to B, 1 to C, 0 to A.Alternatively, 1 to B, 1 to C, and 1 to A.Wait, so the possible distributions are:- 1 in B, 1 in C, 1 in A.- 1 in B, 2 in C, 0 in A.- 2 in B, 1 in C, 0 in A.So, these are the three possible distributions.Now, for each distribution, calculate the number of assignments.1. For 1 in B, 1 in C, 1 in A:We need to choose which student goes to B, which goes to C, and which goes to A.Number of ways: 3! / (1!1!1!) = 6.But since the students are distinguishable, it's 3! = 6.2. For 1 in B, 2 in C, 0 in A:We need to choose 1 student out of 3 to go to B, and the remaining 2 go to C.Number of ways: C(3,1) = 3.3. For 2 in B, 1 in C, 0 in A:Similarly, choose 2 students out of 3 to go to B, and the remaining 1 goes to C.Number of ways: C(3,2) = 3.So, total number of assignments is 6 + 3 + 3 = 12.Yes, that's consistent with the inclusion-exclusion result.Therefore, the number of different distribution plans is 12.Wait, just to make sure, let me think if there's another way to approach this.Another approach could be using the principle of multiplication.Since each student can choose among A, B, or C, but with constraints.But since the constraints are on the units, it's a bit more involved.Alternatively, since we have three students and three units, with B and C needing at least one each, we can model this as assigning each student independently, but ensuring that B and C are covered.But that's essentially what inclusion-exclusion does.Alternatively, think of it as arranging the students into the units, considering the constraints.But I think the constructive counting approach I did earlier is solid, breaking it down into cases where we have 1,1,1; 1,2,0; and 2,1,0 distributions.Each case gives 6, 3, and 3 ways respectively, totaling 12.Therefore, I'm confident that the number of different distribution plans is 12.**Final Answer**The number of different distribution plans is boxed{12}.

answer：Alright, so I have this function here, f(x) = 1 + x/(x² + 1). The problem is asking me to find its maximum value M and its minimum value N, and then compute M + N. Hmm, okay. Let me think about how to approach this.First, I remember that to find the maximum and minimum values of a function, I can use calculus. Specifically, I need to find the critical points by taking the derivative of f(x) and setting it equal to zero. Then, I can determine whether those critical points are maxima or minima by using the second derivative test or analyzing the sign changes of the first derivative.So, let me start by finding the derivative of f(x). The function is f(x) = 1 + x/(x² + 1). The derivative of 1 is 0, so I just need to differentiate the second term, which is a quotient. To differentiate x/(x² + 1), I'll use the quotient rule. The quotient rule says that if I have a function u(x)/v(x), its derivative is (u’v - uv’)/v².Let me assign u = x and v = x² + 1. Then, u’ is 1, and v’ is 2x. Plugging into the quotient rule, the derivative of the second term is (1*(x² + 1) - x*(2x))/(x² + 1)². Let me compute that numerator:1*(x² + 1) is x² + 1, and x*(2x) is 2x². So, subtracting, we get (x² + 1 - 2x²) = (-x² + 1). Therefore, the derivative of the second term is (-x² + 1)/(x² + 1)².So, putting it all together, the derivative of f(x) is f’(x) = 0 + (-x² + 1)/(x² + 1)². Simplifying, that's f’(x) = (1 - x²)/(x² + 1)².Now, to find the critical points, I need to set f’(x) equal to zero and solve for x. So, (1 - x²)/(x² + 1)² = 0. The denominator is always positive because it's squared, so the only way for the fraction to be zero is if the numerator is zero. Therefore, 1 - x² = 0, which gives x² = 1, so x = ±1.Okay, so the critical points are at x = 1 and x = -1. Now, I need to determine whether these points are maxima or minima. I can use the second derivative test or analyze the sign of the first derivative around these points.Let me try the second derivative test. First, I need to find the second derivative f''(x). To do that, I can differentiate f’(x) = (1 - x²)/(x² + 1)².This looks a bit complicated, but let's take it step by step. Let me denote the numerator as u = 1 - x² and the denominator as v = (x² + 1)². Then, f’(x) = u/v, so the derivative f''(x) will be (u’v - uv’)/v².First, compute u’ and v’:u = 1 - x², so u’ = -2x.v = (x² + 1)², so v’ is 2*(x² + 1)*(2x) = 4x(x² + 1). Wait, hold on. Let me double-check that.Actually, v = (x² + 1)², so by the chain rule, v’ = 2*(x² + 1)*(d/dx)(x² + 1) = 2*(x² + 1)*(2x) = 4x(x² + 1). Yes, that's correct.So, now, plug into the formula:f''(x) = [u’v - uv’]/v² = [(-2x)*(x² + 1)² - (1 - x²)*(4x(x² + 1))]/(x² + 1)^4.Hmm, that's a bit messy, but let's factor out common terms in the numerator. I notice that both terms in the numerator have a factor of (x² + 1) and x. Let me factor that out.First term: (-2x)*(x² + 1)² = (-2x)(x² + 1)(x² + 1)Second term: -(1 - x²)*(4x(x² + 1)) = -4x(1 - x²)(x² + 1)So, factoring out (-2x)(x² + 1) from both terms:Numerator = (-2x)(x² + 1)[(x² + 1) + 2(1 - x²)]Wait, let me see:Wait, actually, let me factor out (-2x)(x² + 1) from both terms:First term: (-2x)(x² + 1)(x² + 1)Second term: -4x(1 - x²)(x² + 1) = (-2x)(x² + 1)*[2(1 - x²)]So, factoring out (-2x)(x² + 1):Numerator = (-2x)(x² + 1)[(x² + 1) + 2(1 - x²)]Let me compute the bracket:(x² + 1) + 2(1 - x²) = x² + 1 + 2 - 2x² = (-x²) + 3So, numerator becomes (-2x)(x² + 1)(-x² + 3)Therefore, f''(x) = [(-2x)(x² + 1)(-x² + 3)] / (x² + 1)^4Simplify numerator and denominator:We can cancel (x² + 1) in numerator and denominator:f''(x) = [(-2x)(-x² + 3)] / (x² + 1)^3Simplify the numerator:(-2x)(-x² + 3) = 2x(x² - 3)So, f''(x) = [2x(x² - 3)] / (x² + 1)^3Okay, so f''(x) is equal to 2x(x² - 3)/(x² + 1)^3.Now, let's evaluate f''(x) at the critical points x = 1 and x = -1.First, at x = 1:f''(1) = 2*1*(1² - 3)/(1² + 1)^3 = 2*(1 - 3)/(2)^3 = 2*(-2)/8 = (-4)/8 = -0.5.Since f''(1) is negative, the function is concave down at x = 1, which means this point is a local maximum.Next, at x = -1:f''(-1) = 2*(-1)*((-1)^2 - 3)/((-1)^2 + 1)^3 = 2*(-1)*(1 - 3)/(2)^3 = 2*(-1)*(-2)/8 = (4)/8 = 0.5.Since f''(-1) is positive, the function is concave up at x = -1, which means this point is a local minimum.Great, so we have a local maximum at x = 1 and a local minimum at x = -1. Now, we need to compute the function values at these points to find M and N.First, compute f(1):f(1) = 1 + (1)/(1² + 1) = 1 + 1/2 = 3/2.Then, compute f(-1):f(-1) = 1 + (-1)/((-1)^2 + 1) = 1 + (-1)/(1 + 1) = 1 - 1/2 = 1/2.So, M, the maximum value, is 3/2, and N, the minimum value, is 1/2. Therefore, M + N is 3/2 + 1/2 = 2.Wait, that seems straightforward, but let me double-check my calculations to make sure I didn't make any mistakes.Starting with f(1):f(1) = 1 + (1)/(1 + 1) = 1 + 1/2 = 3/2. Correct.f(-1):f(-1) = 1 + (-1)/(1 + 1) = 1 - 1/2 = 1/2. Correct.So, M + N is indeed 3/2 + 1/2 = 2.Just to be thorough, let me also consider the behavior of the function as x approaches infinity and negative infinity to ensure that these are indeed the global maxima and minima.As x approaches infinity, the term x/(x² + 1) behaves like x/x² = 1/x, which approaches 0. Therefore, f(x) approaches 1 + 0 = 1.Similarly, as x approaches negative infinity, x/(x² + 1) behaves like x/x² = 1/x, which approaches 0. So, f(x) approaches 1 + 0 = 1.Therefore, the function approaches 1 at both ends, and since we have a maximum of 3/2 and a minimum of 1/2, which are both greater than and less than 1, respectively, these are indeed the global maximum and minimum.Hence, M + N is 2.**Final Answer**The value of ( M + N ) is boxed{2}.