💡Okay, so I have this problem where I need to prove two things about the function Z(a, b) defined as (3a)! times (4b)! divided by (a!)^4 times (b!)^3. The first part is to show that for all a less than or equal to b, Z(a, b) is a non-negative integer. The second part is to show that for any non-negative integer b, there are infinitely many a such that Z(a, b) is not an integer. Let me start with the first part. I need to show that Z(a, b) is an integer when a ≤ b. Hmm, factorials can be tricky, but maybe I can break it down. Let me write out Z(a, b):Z(a, b) = (3a)! * (4b)! / (a!^4 * b!^3)I know that factorials often come up in combinatorics, like binomial coefficients. Maybe I can express this ratio in terms of binomial coefficients or multinomial coefficients. Let me think about how to decompose the numerator and the denominator.First, looking at (3a)! / (a!^3). That reminds me of the multinomial coefficient for dividing 3a items into three groups of a each, which is (3a)! / (a!^3). Similarly, (4b)! / (b!^4) is the multinomial coefficient for dividing 4b items into four groups of b each. So, if I rewrite Z(a, b), it becomes:Z(a, b) = [ (3a)! / (a!^3) ] * [ (4b)! / (b!^4) ] * (a!^3 * b!^4) / (a!^4 * b!^3)Wait, that seems a bit convoluted. Let me simplify that. The denominator is a!^4 * b!^3, and the numerator is (3a)! * (4b)!.So, if I write it as:Z(a, b) = (3a)! / (a!^3) * (4b)! / (b!^4) * (a!^3 * b!^4) / (a!^4 * b!^3)Wait, that's not helpful. Maybe I should think differently. Let me consider the prime factorization of the numerator and denominator. For Z(a, b) to be an integer, the exponents of all primes in the numerator must be greater than or equal to those in the denominator.So, for each prime p, the exponent of p in (3a)! plus the exponent of p in (4b)! must be greater than or equal to 4 times the exponent of p in a! plus 3 times the exponent of p in b!.I recall that the exponent of a prime p in n! is given by the sum of floor(n/p^k) for k from 1 to infinity. So, maybe I can use Legendre's formula here.Let me denote v_p(n!) as the exponent of p in n!. Then, we have:v_p(Z(a, b)) = v_p((3a)!) + v_p((4b)!) - 4*v_p(a!) - 3*v_p(b!)I need to show that this is non-negative for all primes p.Hmm, this might be a bit involved, but maybe I can use Kummer's theorem or Lucas' theorem to analyze the carryover in base p expansions. Alternatively, maybe I can use the fact that a ≤ b to bound the exponents.Wait, another approach: maybe express Z(a, b) as a product of binomial coefficients. For example, (3a)! / (a!^3) is the multinomial coefficient for dividing 3a items into three groups of a each, which is equal to the product of binomial coefficients: C(3a, a) * C(2a, a). Similarly, (4b)! / (b!^4) can be expressed as C(4b, b) * C(3b, b) * C(2b, b).So, Z(a, b) = [C(3a, a) * C(2a, a)] * [C(4b, b) * C(3b, b) * C(2b, b)] / (a! * b!)Wait, no, because the denominator is a!^4 * b!^3, and the numerator is (3a)! * (4b)!. So, if I express (3a)! / (a!^3) as C(3a, a) * C(2a, a), and (4b)! / (b!^4) as C(4b, b) * C(3b, b) * C(2b, b), then Z(a, b) becomes:Z(a, b) = [C(3a, a) * C(2a, a)] * [C(4b, b) * C(3b, b) * C(2b, b)] / (a! * b!)But that still leaves me with a! and b! in the denominator. Maybe I can further decompose those.Alternatively, perhaps I can write Z(a, b) as a product of binomial coefficients and some factorials. Let me try:Z(a, b) = (3a)! / (a!^3) * (4b)! / (b!^4) * (a!^3 * b!^4) / (a!^4 * b!^3)Wait, that's the same as:Z(a, b) = (3a)! / (a!^3) * (4b)! / (b!^4) * (a!^3 * b!^4) / (a!^4 * b!^3) = (3a)! / (a!^3) * (4b)! / (b!^4) * 1 / (a! * b!)Hmm, not helpful. Maybe I need to think about the ratio differently. Let me consider the ratio as:Z(a, b) = (3a)! / (a!^3) * (4b)! / (b!^4) * (a!^3 * b!^4) / (a!^4 * b!^3) = (3a)! / (a!^3) * (4b)! / (b!^4) * 1 / (a! * b!)Wait, that's the same as before. Maybe I should instead think about the ratio as:Z(a, b) = (3a)! * (4b)! / (a!^4 * b!^3) = [(3a)! / (a!^3)] * [(4b)! / (b!^4)] * (a!^3 * b!^4) / (a!^4 * b!^3) = [(3a)! / (a!^3)] * [(4b)! / (b!^4)] / (a! * b!)Still not helpful. Maybe I need to approach it differently. Let me think about specific cases to get an idea.Let me try a = 0. Then Z(0, b) = (0)! * (4b)! / (0!^4 * b!^3) = 1 * (4b)! / (1 * b!^3) = (4b)! / (b!^3). Is this an integer? Yes, because (4b)! / (b!^4) is the multinomial coefficient, which is an integer, and we have an extra b! in the denominator, so (4b)! / (b!^3) is also an integer.Similarly, if a = 1, then Z(1, b) = (3)! * (4b)! / (1!^4 * b!^3) = 6 * (4b)! / (b!^3). Since (4b)! / (b!^4) is an integer, multiplying by 6 and dividing by b! gives an integer as long as b! divides 6*(4b)!.Wait, but for a =1, b must be ≥1 since a ≤ b. For b=1, Z(1,1)=6*24 / (1*1)=144, which is integer. For b=2, Z(1,2)=6*40320 / (1*2^3)=6*40320 /8=6*5040=30240, integer. Seems to hold.Similarly, for a=2, b=2: Z(2,2)= (6)!*(8)! / (2!^4 *2!^3)=720*40320 / (16*8)=720*40320 /128. Let's compute 720/16=45, 40320/8=5040, so 45*5040=226800, which is integer.Okay, so specific cases seem to work. Maybe I can generalize this.I think the key is to express Z(a, b) as a product of binomial coefficients, which are integers, and then show that the remaining terms also result in integers.Let me try to decompose (3a)! / (a!^3). This is the multinomial coefficient for dividing 3a items into three groups of a each, which is equal to C(3a, a) * C(2a, a). Both of these are binomial coefficients and hence integers.Similarly, (4b)! / (b!^4) is the multinomial coefficient for dividing 4b items into four groups of b each, which is equal to C(4b, b) * C(3b, b) * C(2b, b). Again, these are binomial coefficients and hence integers.So, Z(a, b) = [C(3a, a) * C(2a, a)] * [C(4b, b) * C(3b, b) * C(2b, b)] / (a! * b!)Wait, but we still have a! and b! in the denominator. How do they cancel out?Wait, no, because in the numerator, we have (3a)! * (4b)! and in the denominator, we have (a!^4 * b!^3). So, when we write (3a)! / (a!^3) as C(3a, a) * C(2a, a), and (4b)! / (b!^4) as C(4b, b) * C(3b, b) * C(2b, b), then Z(a, b) becomes:Z(a, b) = [C(3a, a) * C(2a, a)] * [C(4b, b) * C(3b, b) * C(2b, b)] * (a!^3 * b!^4) / (a!^4 * b!^3)Wait, that's not correct. Let me clarify:Z(a, b) = (3a)! * (4b)! / (a!^4 * b!^3) = [(3a)! / (a!^3)] * [(4b)! / (b!^4)] * (a!^3 * b!^4) / (a!^4 * b!^3)Which simplifies to:Z(a, b) = [C(3a, a) * C(2a, a)] * [C(4b, b) * C(3b, b) * C(2b, b)] / (a! * b!)Hmm, so we have a product of binomial coefficients divided by a! * b!. I need to show that this is an integer.Wait, but a! and b! are in the denominator, so unless the binomial coefficients are divisible by a! and b!, this might not hold. Maybe I need to think differently.Alternatively, perhaps I can use the fact that a ≤ b to bound the exponents in the prime factorization. Let me consider the exponents of a prime p in Z(a, b). Using Legendre's formula, as I thought earlier.For a given prime p, the exponent in Z(a, b) is:v_p(Z(a, b)) = v_p((3a)!) + v_p((4b)!) - 4*v_p(a!) - 3*v_p(b!)I need to show that this is ≥ 0 for all primes p.Let me recall that v_p(n!) = sum_{k=1}^∞ floor(n/p^k)So, v_p((3a)!) = sum_{k=1}^∞ floor(3a/p^k)v_p((4b)!) = sum_{k=1}^∞ floor(4b/p^k)v_p(a!) = sum_{k=1}^∞ floor(a/p^k)v_p(b!) = sum_{k=1}^∞ floor(b/p^k)Thus, v_p(Z(a, b)) = sum_{k=1}^∞ [floor(3a/p^k) + floor(4b/p^k) - 4*floor(a/p^k) - 3*floor(b/p^k)]I need to show that this sum is non-negative for all primes p.Let me analyze the term inside the sum for each k:floor(3a/p^k) + floor(4b/p^k) - 4*floor(a/p^k) - 3*floor(b/p^k)Let me denote x = a/p^k and y = b/p^k, so floor(3x) + floor(4y) - 4*floor(x) - 3*floor(y)I need to show that this is ≥ 0.Let me consider the fractional parts. Let x = m + f, where m is integer and 0 ≤ f <1. Similarly, y = n + g, with 0 ≤ g <1.Then floor(3x) = floor(3m + 3f) = 3m + floor(3f)floor(4y) = floor(4n + 4g) = 4n + floor(4g)floor(x) = mfloor(y) = nSo, the expression becomes:[3m + floor(3f)] + [4n + floor(4g)] - 4m - 3n = (3m -4m) + (4n -3n) + floor(3f) + floor(4g) = (-m + n) + floor(3f) + floor(4g)Hmm, so it's (-m + n) + floor(3f) + floor(4g)But m = floor(x) = floor(a/p^k), n = floor(y) = floor(b/p^k)Given that a ≤ b, we have m ≤ n because a/p^k ≤ b/p^k, so floor(a/p^k) ≤ floor(b/p^k). Thus, -m + n ≥ 0.Now, floor(3f) can be 0, 1, or 2, since f <1. Similarly, floor(4g) can be 0,1,2,3.So, the total expression is (-m + n) + floor(3f) + floor(4g) ≥ 0 + 0 + 0 =0But wait, is that always true? Let me check.If m = n, then -m + n =0, and floor(3f) + floor(4g) ≥0, which is true.If m < n, then -m + n ≥1, and floor(3f) + floor(4g) ≥0, so total is ≥1.Thus, in all cases, the expression is ≥0.Therefore, v_p(Z(a, b)) is a sum of non-negative terms, hence non-negative. Therefore, Z(a, b) is an integer.Okay, that seems to work. So, for part 1, Z(a, b) is an integer when a ≤ b.Now, part 2: For any non-negative integer b, there are infinitely many a such that Z(a, b) is not an integer.Hmm, so for a fixed b, I need to find infinitely many a where Z(a, b) is not integer.I think I can use the fact that for certain a, the exponents of some prime p in Z(a, b) will be negative, making Z(a, b) not an integer.Let me fix b. Let me choose a prime p such that p > 4b. Then, for a = p, let's see what happens.Compute Z(p, b) = (3p)! * (4b)! / (p!^4 * b!^3)Now, since p >4b, p does not divide (4b)! because all factors in (4b)! are ≤4b <p. So, the exponent of p in (4b)! is zero.Similarly, in the denominator, p!^4 has p^4 in its prime factorization, because p! includes p, and p^4 comes from (p!)^4.In the numerator, (3p)! includes p, 2p, 3p. So, the exponent of p in (3p)! is floor(3p/p) + floor(3p/p^2) + ... = 3 + 0 + ... =3.Thus, the exponent of p in the numerator is 3 (from (3p)!) + 0 (from (4b)!) =3.The exponent of p in the denominator is 4 (from p!^4) + 0 (from b!^3) =4.Thus, the exponent of p in Z(p, b) is 3 -4 = -1, which is negative. Therefore, Z(p, b) is not an integer.Since there are infinitely many primes p >4b, we can choose infinitely many a =p such that Z(a, b) is not integer.Therefore, for any fixed b, there are infinitely many a where Z(a, b) is not integer.I think that works. So, summarizing:1. For a ≤ b, Z(a, b) is an integer because the exponents of all primes in the numerator are at least as large as those in the denominator.2. For any fixed b, choosing a as a prime greater than 4b results in Z(a, b) not being an integer, and since there are infinitely many such primes, there are infinitely many a for each b where Z(a, b) is not integer.

💡Alright, so I've got this problem here about two couples, each with a child, going to the zoo. They need to line up to enter, but there are some specific rules about how they can arrange themselves. The two fathers have to be at the beginning and the end of the line, and the two children have to be together. I need to figure out how many different ways they can arrange themselves in the line following these rules.Okay, let's break this down step by step. First, I need to visualize the line. There are six people in total: two fathers, two mothers, and two children. So, six spots in the line. The fathers have to be at the very front and the very end. That means the first and the sixth positions are reserved for the fathers. The two children have to be together somewhere in the middle.So, let me think about the fathers first. There are two fathers, let's call them Father A and Father B. They need to be at the beginning and the end of the line. How many ways can we arrange them? Well, Father A can be at the front and Father B at the end, or Father B at the front and Father A at the end. So, that's two possible arrangements for the fathers.Now, with the fathers fixed at the ends, we have four people left to arrange in the middle four positions: two mothers and two children. But there's another condition: the two children must be together. So, the children can't be separated; they have to be next to each other somewhere in the middle.Hmm, okay. So, if we think of the two children as a single unit, that means we're effectively arranging three units: the two mothers and the children unit. How many ways can we arrange three units in four positions? Wait, actually, if we're treating the two children as a single unit, that unit takes up two adjacent positions. So, in the four middle positions, we need to place this children unit and the two mothers.Let me think about how many ways we can place the children unit. Since the children have to be together, they can start at position 2, 3, or 4. If they start at position 2, they'll occupy positions 2 and 3. If they start at position 3, they'll occupy positions 3 and 4. If they start at position 4, they'll occupy positions 4 and 5. Wait, but position 5 is the fifth position, which is before the last position where Father B is. So, actually, the children can start at position 2, 3, or 4.So, that's three possible starting positions for the children unit. For each of these starting positions, the children can switch places within their unit. So, for each starting position, there are two arrangements: Child 1 first and Child 2 second, or Child 2 first and Child 1 second.Now, once we've placed the children unit, the remaining two positions will be occupied by the two mothers. How many ways can we arrange the two mothers in those two positions? Well, it's 2 factorial, which is 2 ways. So, for each arrangement of the children unit, there are two ways to arrange the mothers.Putting it all together: we have two choices for the fathers, three starting positions for the children unit, two arrangements within the children unit, and two arrangements for the mothers. So, the total number of arrangements is 2 (fathers) multiplied by 3 (starting positions) multiplied by 2 (children arrangements) multiplied by 2 (mothers arrangements). That gives us 2 * 3 * 2 * 2 = 24.Wait, but let me double-check that. Sometimes when dealing with permutations, it's easy to overcount or undercount. So, let's go through it again.First, fathers: 2 ways.Then, the middle four positions: we have two children who must be together and two mothers. If we treat the children as a single unit, we have three units to arrange: [Mother 1], [Mother 2], and [Child 1, Child 2]. The number of ways to arrange these three units in four positions is... Hmm, actually, when we have a unit that takes up two positions, the number of ways to arrange the units is different.Wait, maybe I should think of it as arranging the three units in the four positions. The number of ways to arrange three units where one unit takes two positions is equal to the number of ways to choose the starting position for the children unit, which we already considered as three possibilities (starting at position 2, 3, or 4). For each of these, the children can be arranged in 2 ways, and the mothers can be arranged in 2 ways. So, 3 * 2 * 2 = 12 ways for the middle four positions.Then, multiplied by the 2 ways for the fathers, gives us 2 * 12 = 24.Yes, that seems consistent. So, the total number of arrangements is 24.But wait, let me think if there's another way to approach this problem. Maybe using permutations with restrictions.We have six people: F1, F2, M1, M2, C1, C2.Constraints:- F1 and F2 must be at the ends.- C1 and C2 must be adjacent.So, first, fix F1 and F2 at the ends. There are 2! ways to arrange F1 and F2 at the ends.Now, we have four positions left: positions 2, 3, 4, 5.We need to arrange M1, M2, C1, C2 in these four positions, with C1 and C2 together.The number of ways to arrange four people with two adjacent is calculated by treating the two children as a single entity. So, we have three entities: [C1C2], M1, M2.The number of ways to arrange these three entities in four positions is equal to the number of ways to choose the starting position for the children unit, which is 3 (as before), multiplied by the number of ways to arrange the three entities, which is 3! = 6, but wait, no, because the children unit takes two positions.Wait, actually, if we treat [C1C2] as a single unit, then we have three units: [C1C2], M1, M2. The number of ways to arrange these three units in four positions is equal to the number of ways to choose the starting position for the children unit, which is 3, and then arrange the three units. But since the children unit is fixed in size, the arrangement of the units is 3! = 6. However, since the children unit takes two positions, we have to consider that the remaining two positions are for the two mothers, which can be arranged in 2! = 2 ways.Wait, I'm getting confused here. Let me try a different approach.The number of ways to arrange four people with two adjacent is equal to the number of ways to choose the two adjacent positions, arrange the children in them, and then arrange the mothers in the remaining positions.There are three possible pairs of adjacent positions in four spots: positions 2-3, 3-4, and 4-5.For each pair, the children can be arranged in 2 ways (C1 first or C2 first). Then, the remaining two positions are for the mothers, which can be arranged in 2! = 2 ways.So, for each of the three adjacent pairs, we have 2 * 2 = 4 arrangements.Therefore, total arrangements for the middle four positions are 3 * 4 = 12.Then, multiplied by the 2 arrangements for the fathers, gives us 2 * 12 = 24.Yes, that matches my earlier calculation.Alternatively, another way to think about it is:Total number of ways without any restrictions: 6! = 720.But we have restrictions, so we need to adjust.First, fix the fathers at the ends. There are 2! ways to arrange the fathers.Then, consider the two children must be together. The number of ways to arrange the remaining four people with the children together is calculated by treating the children as a single unit, so we have three units: [C1C2], M1, M2.The number of ways to arrange these three units in four positions is equal to the number of ways to choose the starting position for the children unit, which is 3, and then arrange the three units. But since the children unit takes two positions, the remaining two positions are for the mothers, which can be arranged in 2! ways. Also, the children can be arranged among themselves in 2! ways.So, total arrangements for the middle four positions: 3 (starting positions) * 2! (children arrangements) * 2! (mothers arrangements) = 3 * 2 * 2 = 12.Then, multiplied by the 2 arrangements for the fathers, gives us 2 * 12 = 24.Yes, that's consistent as well.I think I've approached this problem from a few different angles, and each time I end up with 24 as the total number of arrangements. So, I feel confident that the answer is 24.**Final Answer**boxed{C}

💡Alright, so I have this inequality problem here, and I need to figure out which of the options A, B, C, or D is true given that ( a < b < 0 ). Let me take it step by step because I want to make sure I understand each part correctly.First, let's parse the given information: ( a < b < 0 ). This means both ( a ) and ( b ) are negative numbers, and ( a ) is less than ( b ). So, for example, if ( a = -3 ) and ( b = -2 ), that would satisfy ( a < b < 0 ). Negative numbers can sometimes be tricky when dealing with inequalities because multiplying or dividing by them can reverse the inequality signs. I need to keep that in mind as I go through each option.Let's look at option A: ( frac{1}{a} < frac{1}{b} ). Hmm, so we're comparing the reciprocals of ( a ) and ( b ). Since both ( a ) and ( b ) are negative, their reciprocals will also be negative. Now, when dealing with negative numbers, the number with the smaller absolute value is actually larger. For example, ( -1 ) is greater than ( -2 ) because it's closer to zero.Given that ( a < b < 0 ), ( a ) is more negative than ( b ), which means ( |a| > |b| ). Therefore, ( frac{1}{a} ) will have a smaller absolute value than ( frac{1}{b} ) because dividing 1 by a larger absolute value results in a smaller number. But since both are negative, the one with the smaller absolute value is actually larger. So, ( frac{1}{a} ) is less than ( frac{1}{b} ). Wait, that seems contradictory. Let me think again.If ( a = -3 ) and ( b = -2 ), then ( frac{1}{a} = -frac{1}{3} ) and ( frac{1}{b} = -frac{1}{2} ). Comparing these, ( -frac{1}{3} ) is greater than ( -frac{1}{2} ) because it's closer to zero. So, ( frac{1}{a} > frac{1}{b} ). That means option A is incorrect because it states ( frac{1}{a} < frac{1}{b} ). Okay, so A is false.Moving on to option B: ( b + frac{1}{a} > a + frac{1}{b} ). Let me rearrange this inequality to see if I can simplify it. Subtract ( a ) from both sides: ( b - a + frac{1}{a} - frac{1}{b} > 0 ). We know that ( b - a ) is positive because ( b > a ). Also, from option A, we saw that ( frac{1}{a} > frac{1}{b} ), so ( frac{1}{a} - frac{1}{b} ) is positive. Therefore, the entire expression ( b - a + frac{1}{a} - frac{1}{b} ) is positive, which means ( b + frac{1}{a} > a + frac{1}{b} ). So, option B is correct. That seems solid.But just to be thorough, let me plug in some numbers. Let's take ( a = -3 ) and ( b = -2 ) again. Then, ( b + frac{1}{a} = -2 + (-frac{1}{3}) = -2.333... ) and ( a + frac{1}{b} = -3 + (-frac{1}{2}) = -3.5 ). Comparing these, ( -2.333... > -3.5 ), so indeed, ( b + frac{1}{a} > a + frac{1}{b} ). That confirms option B is correct.Now, option C: ( a + frac{1}{b} > b + frac{1}{a} ). Wait a second, this looks like the opposite of option B. From what I saw earlier, ( b + frac{1}{a} > a + frac{1}{b} ), so this would mean that ( a + frac{1}{b} ) is actually less than ( b + frac{1}{a} ). Let me test it with the same numbers. ( a + frac{1}{b} = -3 + (-frac{1}{2}) = -3.5 ) and ( b + frac{1}{a} = -2 + (-frac{1}{3}) = -2.333... ). So, ( -3.5 < -2.333... ), which means ( a + frac{1}{b} < b + frac{1}{a} ). Therefore, option C is incorrect.Finally, option D: ( frac{b}{a} < frac{b + 1}{a + 1} ). Let me see. Both ( a ) and ( b ) are negative, so ( frac{b}{a} ) is positive because a negative divided by a negative is positive. Similarly, ( frac{b + 1}{a + 1} ) could be positive or negative depending on the values of ( a ) and ( b ). Let me use the same example where ( a = -3 ) and ( b = -2 ). Then, ( frac{b}{a} = frac{-2}{-3} = frac{2}{3} ) and ( frac{b + 1}{a + 1} = frac{-2 + 1}{-3 + 1} = frac{-1}{-2} = frac{1}{2} ). Comparing these, ( frac{2}{3} ) is approximately 0.666..., and ( frac{1}{2} ) is 0.5. So, ( frac{2}{3} > frac{1}{2} ), which means ( frac{b}{a} > frac{b + 1}{a + 1} ). Therefore, option D is incorrect.To double-check, maybe try another set of numbers. Let me pick ( a = -4 ) and ( b = -1 ). Then, ( frac{b}{a} = frac{-1}{-4} = frac{1}{4} ) and ( frac{b + 1}{a + 1} = frac{-1 + 1}{-4 + 1} = frac{0}{-3} = 0 ). So, ( frac{1}{4} > 0 ), which again shows that ( frac{b}{a} > frac{b + 1}{a + 1} ). So, option D is indeed false.So, summarizing my findings:- Option A is incorrect because ( frac{1}{a} > frac{1}{b} ).- Option B is correct because ( b + frac{1}{a} > a + frac{1}{b} ).- Option C is incorrect because ( a + frac{1}{b} < b + frac{1}{a} ).- Option D is incorrect because ( frac{b}{a} > frac{b + 1}{a + 1} ).Therefore, the only correct option is B.**Final Answer**boxed{B}

💡First, I need to determine how many fifty-cent coins each person has. Liam has (3p + 2) fifty-cent coins, and Mia has (2p + 7) fifty-cent coins.Next, I'll find the difference in the number of fifty-cent coins between Liam and Mia by subtracting Mia's coins from Liam's coins:[(3p + 2) - (2p + 7) = 3p + 2 - 2p - 7 = p - 5]This means Liam has (p - 5) more fifty-cent coins than Mia.Since each fifty-cent coin is worth 50 pennies, I'll convert the difference in the number of coins to pennies by multiplying by 50:[50 times (p - 5) = 50p - 250]Therefore, the difference in their total value expressed in pennies is (50p - 250).

💡Alright, so I have this problem about a right triangular prism, ABC-A₁B₁C₁. It says that AB is perpendicular to BC, and the length of AC is 5. I need to find the maximum surface area of an inscribed sphere in this prism. Hmm, okay, let me try to break this down step by step.First, let me visualize the prism. A right triangular prism has two congruent triangular bases connected by rectangles. In this case, the base is triangle ABC, and since AB is perpendicular to BC, triangle ABC must be a right-angled triangle at B. So, AB and BC are the legs, and AC is the hypotenuse, which is given as 5 units.Now, I need to think about the inscribed sphere in this prism. An inscribed sphere, or an insphere, is a sphere that is tangent to all the faces of the prism. For a sphere to be inscribed in a prism, the prism must be such that it's possible for a sphere to touch all its faces. In the case of a right triangular prism, this should be possible, but the size of the sphere will depend on the dimensions of the prism.Since the prism is right, the sides are rectangles, and the height of the prism (let's say it's the length AA₁, BB₁, or CC₁) is perpendicular to the base. Let me denote the height of the prism as h. So, the prism has a right triangle base with legs AB and BC, hypotenuse AC = 5, and height h.To find the inscribed sphere, I think I need to find the radius of the sphere first. The surface area of the sphere will then be 4πr², so if I can find the maximum possible radius, I can compute the maximum surface area.Now, how do I find the radius of the inscribed sphere in a right triangular prism? I recall that for a sphere to be inscribed in a prism, the sphere must be tangent to all the faces. In a right prism, this means the sphere is tangent to the two triangular bases and all three rectangular faces.Let me consider the base triangle ABC first. Since it's a right triangle, it has an incircle. The radius of this incircle will be related to the inradius of the sphere. Wait, is the inradius of the sphere the same as the inradius of the base triangle? Hmm, maybe not directly, because the sphere has to fit within the entire prism, not just the base.Let me think. The sphere must be tangent to the top and bottom triangular faces, as well as the three rectangular faces. So, the diameter of the sphere must be equal to the height of the prism, right? Because the sphere has to fit perfectly between the two triangular bases. So, if the height of the prism is h, then the diameter of the sphere is h, which means the radius r is h/2.But wait, that might not be the only constraint. The sphere also has to fit within the base triangle. So, the inradius of the base triangle must be at least equal to the radius of the sphere. Otherwise, the sphere would protrude outside the base. So, the radius r of the sphere must satisfy r ≤ inradius of triangle ABC.So, to maximize the surface area of the sphere, I need to maximize its radius r. This would happen when r is as large as possible, which would be when r equals the inradius of the base triangle ABC. But wait, is that necessarily true? Because the sphere also has to fit within the height of the prism. So, if the inradius of the base is larger than h/2, then h/2 would be the limiting factor. Conversely, if h/2 is larger than the inradius, then the inradius would limit the sphere's size.Therefore, to maximize r, I need to have both the inradius of the base triangle and h/2 as large as possible. But since the prism's height h is a variable here, perhaps I can adjust h to make both conditions compatible.Wait, but in the problem statement, is the prism fixed? It just says "the right triangular prism ABC-A₁B₁C₁, where AB ⊥ BC and AC=5." So, AC is fixed at 5, but AB and BC can vary as long as they are perpendicular and satisfy AB² + BC² = AC² = 25. So, AB and BC can vary, which affects the inradius of the base triangle. Also, the height h of the prism can vary as well.Therefore, to maximize the radius r, I need to consider both the inradius of the base triangle and the height h. Since r is limited by both h/2 and the inradius of the base, to maximize r, I need to set h such that h/2 equals the inradius of the base triangle. That way, both constraints are satisfied, and r is maximized.So, let me denote AB = a and BC = b. Then, since AB ⊥ BC, AC = √(a² + b²) = 5. So, a² + b² = 25.The inradius of the base triangle ABC can be calculated using the formula for the inradius of a right triangle: r_base = (a + b - c)/2, where c is the hypotenuse. In this case, c = 5, so r_base = (a + b - 5)/2.On the other hand, the radius of the sphere is limited by h/2. So, to have r = h/2 = r_base, we set h = 2*r_base = a + b - 5.But h must also be positive, so a + b - 5 > 0, which implies a + b > 5.But since a² + b² = 25, we can write (a + b)² = a² + 2ab + b² = 25 + 2ab. So, a + b = √(25 + 2ab). Therefore, h = √(25 + 2ab) - 5.But h must be positive, so √(25 + 2ab) > 5, which implies 25 + 2ab > 25, so 2ab > 0, which is always true since a and b are positive lengths.Now, to maximize r, which is equal to r_base = (a + b - 5)/2, we need to maximize (a + b - 5)/2. Since a² + b² = 25, we can express a + b in terms of a single variable.Let me set a = 5 cos θ and b = 5 sin θ, since a² + b² = 25. Then, a + b = 5 (cos θ + sin θ). So, r_base = (5 (cos θ + sin θ) - 5)/2 = (5 (cos θ + sin θ - 1))/2.To maximize r_base, we need to maximize cos θ + sin θ. The maximum value of cos θ + sin θ is √2, which occurs when θ = 45°. So, the maximum value of r_base is (5 (√2 - 1))/2.Therefore, the maximum radius r is (5 (√2 - 1))/2. Then, the surface area of the sphere is 4πr² = 4π * [(5 (√2 - 1))/2]^2.Let me compute that:First, square the radius: [(5 (√2 - 1))/2]^2 = (25 ( (√2 - 1)^2 )) / 4.Compute (√2 - 1)^2: (√2)^2 - 2*√2*1 + 1^2 = 2 - 2√2 + 1 = 3 - 2√2.So, the squared radius becomes (25 (3 - 2√2)) / 4.Then, multiply by 4π: 4π * (25 (3 - 2√2))/4 = 25 (3 - 2√2) π.Wait, but in my earlier thought process, I thought the maximum surface area was 25(3 - 3√2)π, but now I'm getting 25(3 - 2√2)π. Hmm, which one is correct?Wait, let me double-check my calculations.Starting from r = (5 (√2 - 1))/2.So, r² = [5 (√2 - 1)/2]^2 = (25 ( (√2 - 1)^2 )) / 4.(√2 - 1)^2 = 2 - 2√2 + 1 = 3 - 2√2.So, r² = 25 (3 - 2√2)/4.Then, surface area is 4πr² = 4π * 25 (3 - 2√2)/4 = 25 (3 - 2√2) π.So, that's 25(3 - 2√2)π.But in the initial problem, the assistant's answer was 25(3 - 3√2)π. So, which one is correct?Wait, maybe I made a mistake in the earlier step. Let me re-examine.Wait, when I set r = h/2 = r_base, so h = 2 r_base = a + b - 5.But h is also the height of the prism, which is the length of AA₁, etc. So, in the prism, the sphere is inscribed, so it's tangent to all faces. Therefore, the radius of the sphere must satisfy that the sphere is tangent to the top and bottom faces (so h = 2r), and also tangent to the side faces.But in a right triangular prism, the side faces are rectangles. For the sphere to be tangent to these rectangles, the radius must also be less than or equal to the inradius of the base triangle.Wait, perhaps I need to consider that the sphere is tangent to all faces, which includes the two triangular faces and the three rectangular faces. So, the radius must satisfy both the inradius of the base triangle and the height constraint.But in a right triangular prism, the inradius of the sphere is determined by the minimum of the inradius of the base triangle and half the height of the prism.Wait, maybe not exactly. Let me think differently.In a right prism, the inscribed sphere must fit within the prism such that it is tangent to all faces. For the sphere to be tangent to the two triangular bases, the distance between the bases must be twice the radius, so h = 2r.Additionally, the sphere must be tangent to the three rectangular faces. For this, the radius must be less than or equal to the inradius of the base triangle. Because the inradius of the base triangle is the radius of the circle inscribed in the base, which is the largest circle that fits within the base. So, for the sphere to be tangent to the side faces, its radius must be equal to the inradius of the base.Wait, so if h = 2r and r = r_base, then h = 2 r_base.But earlier, I had h = a + b - 5, which is 2 r_base. So, that seems consistent.But then, to maximize r, we need to maximize r_base, which is (a + b - 5)/2.Given that a² + b² = 25, we can express a + b in terms of θ as 5 (cos θ + sin θ), which has a maximum value of 5√2 when θ = 45°, as I did before.Thus, r_base_max = (5√2 - 5)/2 = (5 (√2 - 1))/2.Therefore, h = 2 r_base_max = 5 (√2 - 1).So, the radius r is (5 (√2 - 1))/2, and the surface area is 4πr².Calculating that:r = (5 (√2 - 1))/2r² = [25 ( (√2 - 1)^2 ) ] / 4 = [25 (3 - 2√2)] / 4Surface area = 4π * [25 (3 - 2√2)] / 4 = 25 (3 - 2√2) πSo, that seems correct. But in the initial problem, the assistant's answer was 25(3 - 3√2)π, which is different.Wait, perhaps I made a mistake in interpreting the problem. Let me read it again."In the right triangular prism ABC-A₁B₁C₁, where AB ⊥ BC and AC=5, the maximum surface area of the inscribed sphere is ______."So, the prism is right, meaning that the lateral edges are perpendicular to the base. So, the height h is the length of AA₁, etc., and is perpendicular to the base ABC.Now, for the sphere to be inscribed in the prism, it must be tangent to all faces. So, the sphere must be tangent to the two triangular bases and the three rectangular faces.To be tangent to the two triangular bases, the distance between the bases (the height h) must be equal to 2r, as the sphere must fit exactly between them.To be tangent to the rectangular faces, the sphere must fit within the base triangle. The largest sphere that can fit within the base triangle is the incircle, with radius r_base. So, the radius of the sphere must be equal to r_base, because if it's larger, it would protrude outside the base.Therefore, to have the sphere tangent to all faces, we must have r = r_base and h = 2r.Thus, h = 2 r_base.Given that, we can express h in terms of a and b.From the base triangle, r_base = (a + b - c)/2, where c = 5.So, r_base = (a + b - 5)/2.Therefore, h = 2 * (a + b - 5)/2 = a + b - 5.But h must also satisfy the condition that the sphere is tangent to the rectangular faces. Wait, but h is determined by the inradius of the base. So, perhaps h is dependent on a and b.But in the prism, h is independent of a and b, right? Or is it?Wait, no, in the prism, h is the height, which is independent of the base triangle's dimensions. So, perhaps h can be chosen independently to maximize the sphere's radius.But in reality, for the sphere to be inscribed, h must be equal to 2r, and r must be equal to r_base. So, h = 2 r_base.Therefore, h = a + b - 5.But h is also the height of the prism, which is a separate variable. So, to maximize r, we need to maximize r_base, which is (a + b - 5)/2.Given that a² + b² = 25, we can express a + b in terms of a single variable.Let me set a = 5 cos θ and b = 5 sin θ, as before. Then, a + b = 5 (cos θ + sin θ), which has a maximum value of 5√2 when θ = 45°, as I calculated earlier.Therefore, the maximum value of r_base is (5√2 - 5)/2 = (5 (√2 - 1))/2.Thus, the maximum radius r is (5 (√2 - 1))/2, and the surface area is 4πr².Calculating that:r = (5 (√2 - 1))/2r² = [25 ( (√2 - 1)^2 ) ] / 4 = [25 (3 - 2√2)] / 4Surface area = 4π * [25 (3 - 2√2)] / 4 = 25 (3 - 2√2) πSo, that's the surface area.But wait, in the initial problem, the assistant's answer was 25(3 - 3√2)π. So, why is there a discrepancy?Let me check the assistant's solution again.The assistant wrote:"Let the radius of the inscribed sphere of the prism be r. Then, the inscribed circle of triangle ABC is the great circle of the sphere.Let AB = a and BC = b. Then, we have a² + b² = 25.By equating areas, we get (1/2)ab = (1/2)(a + b + 5)r.Therefore, r = ab / (a + b + 5).Let a = 5 cos α and b = 5 sin α. Then, r = [25 sin α cos α] / [5 cos α + 5 sin α + 5].Let t = cos α + sin α (where |t| ≤ √2), we have r = (5/2)(t - 1).Therefore, the maximum value of r, r_max = (5/2)(√2 - 1).Thus, the maximum surface area of the inscribed sphere in the right triangular prism is 25(3 - 3√2)π."Wait, so the assistant is using a different approach. They are equating the area of the triangle to something involving r.Wait, in their solution, they say "By equating areas, we get (1/2)ab = (1/2)(a + b + 5)r." Hmm, that seems a bit unclear. Let me see.In a right triangle, the area is (1/2)ab. The inradius of the triangle is given by r_base = (a + b - c)/2, where c is the hypotenuse. So, r_base = (a + b - 5)/2.But the assistant is writing (1/2)ab = (1/2)(a + b + 5)r. That seems different. Let me see.Wait, perhaps they are considering the area of the prism's lateral faces or something else. Wait, no, the prism has two triangular bases and three rectangular faces.But the inscribed sphere touches all faces, so maybe the assistant is considering the total surface area of the prism and relating it to the sphere's surface area? That doesn't seem right.Alternatively, perhaps they are considering the area of the base triangle and relating it to the sphere's radius in some way.Wait, let me think. If the inscribed sphere touches the base triangle, then the area of the base triangle can be expressed in terms of the inradius and the semiperimeter.Yes, in a triangle, the area is equal to the inradius times the semiperimeter. So, for triangle ABC, area = r_base * s, where s = (a + b + c)/2.So, (1/2)ab = r_base * (a + b + 5)/2.Therefore, r_base = (ab) / (a + b + 5).Ah, so that's where the assistant got the equation (1/2)ab = (1/2)(a + b + 5)r.So, they are correct in that step.Then, they express a and b in terms of 5 cos α and 5 sin α, so a = 5 cos α, b = 5 sin α.Then, r = [25 sin α cos α] / [5 cos α + 5 sin α + 5] = [25 sin α cos α] / [5 (cos α + sin α + 1)].Simplify numerator and denominator by 5: [5 sin α cos α] / [cos α + sin α + 1].Then, let t = cos α + sin α, so t ranges from -√2 to √2.But since a and b are lengths, cos α and sin α are positive, so t ranges from 1 to √2.Wait, actually, when α is between 0 and π/2, cos α and sin α are positive, so t = cos α + sin α ranges from 1 to √2.So, t ∈ [1, √2].Then, the expression becomes r = [5 sin α cos α] / [t + 1].But sin α cos α can be expressed as (t² - 1)/2, since t = cos α + sin α, so t² = cos² α + 2 sin α cos α + sin² α = 1 + 2 sin α cos α.Therefore, sin α cos α = (t² - 1)/2.So, substituting back, r = [5 * (t² - 1)/2] / (t + 1) = [5 (t² - 1)] / [2 (t + 1)].Factor numerator: t² - 1 = (t - 1)(t + 1).So, r = [5 (t - 1)(t + 1)] / [2 (t + 1)] = [5 (t - 1)] / 2.Therefore, r = (5/2)(t - 1).Since t ∈ [1, √2], the maximum value of r occurs when t is maximum, i.e., t = √2.Thus, r_max = (5/2)(√2 - 1).Therefore, the maximum radius is (5/2)(√2 - 1), and the surface area is 4πr².Calculating that:r = (5/2)(√2 - 1)r² = [25/4]( (√2 - 1)^2 ) = [25/4](3 - 2√2)Surface area = 4π * [25/4](3 - 2√2) = 25(3 - 2√2)π.Wait, so according to this, the surface area is 25(3 - 2√2)π.But in the assistant's solution, they wrote the surface area as 25(3 - 3√2)π.So, which one is correct?Wait, let me check the assistant's final step.They wrote: "Thus, the maximum surface area of the inscribed sphere in the right triangular prism is 25(3 - 3√2)π."But according to my calculation, it should be 25(3 - 2√2)π.Wait, perhaps the assistant made a mistake in their calculation.Let me see:They had r = (5/2)(√2 - 1).Then, surface area is 4πr² = 4π*(25/4)( (√2 - 1)^2 ) = 25*(3 - 2√2)π.Yes, that's correct.But the assistant wrote 25(3 - 3√2)π, which is different.So, perhaps the assistant made an error in their final calculation.Alternatively, maybe I made a mistake in interpreting the problem.Wait, let me think again.The assistant's solution is:r = ab / (a + b + 5)Expressed in terms of t, they got r = (5/2)(t - 1), with t = cos α + sin α.Then, t_max = √2, so r_max = (5/2)(√2 - 1).Then, surface area is 4πr² = 4π*(25/4)( (√2 - 1)^2 ) = 25*(3 - 2√2)π.So, that's correct.But the assistant wrote 25(3 - 3√2)π, which is incorrect.Therefore, the correct surface area should be 25(3 - 2√2)π.But wait, let me check the assistant's final answer again.They wrote: "Hence, the answer is boxed{25(3-3sqrt{2})pi}."So, they have 25(3 - 3√2)π, but according to the correct calculation, it should be 25(3 - 2√2)π.Therefore, the assistant made a mistake in their final step.Alternatively, perhaps I made a mistake in my own calculation.Wait, let me recalculate the surface area.Given r = (5/2)(√2 - 1).Then, r² = [25/4]*( (√2 - 1)^2 ) = [25/4]*(2 - 2√2 + 1) = [25/4]*(3 - 2√2).Therefore, surface area = 4πr² = 4π*(25/4)*(3 - 2√2) = 25*(3 - 2√2)π.Yes, that's correct.Therefore, the correct answer should be 25(3 - 2√2)π.But the assistant's answer was 25(3 - 3√2)π, which is incorrect.So, perhaps the assistant made a mistake in their calculation, or perhaps I'm missing something.Wait, let me think again about the relationship between the sphere and the prism.In the assistant's solution, they considered the inscribed circle of the base triangle as the great circle of the sphere. That might not be accurate.Wait, in a right triangular prism, the inscribed sphere must be tangent to all faces, including the two triangular bases and the three rectangular faces.Therefore, the sphere's center must be equidistant from all these faces.In the base triangle, the inradius is the distance from the incenter to each side. Similarly, in the prism, the center of the sphere must be at a distance r from each face.Therefore, the center of the sphere must be at a distance r from the two triangular bases (so the height h = 2r) and at a distance r from each of the three rectangular faces.But the distance from the center to each rectangular face is determined by the inradius of the base triangle.Wait, no, in the base triangle, the inradius is the distance from the incenter to each side. So, in the prism, the center of the sphere must be at a distance r from each rectangular face, which correspond to the sides of the base triangle.Therefore, the inradius of the base triangle must be equal to r.Wait, so in the base triangle, the inradius r_base = r.But also, the height of the prism h = 2r.Therefore, h = 2r_base.But earlier, we have h = a + b - 5, which is 2 r_base.So, h = 2 r_base.But r_base = (a + b - 5)/2.Therefore, h = a + b - 5.But h is also equal to 2 r_base = a + b - 5.So, that's consistent.But in the assistant's solution, they equated the area of the base triangle to something involving r, which led them to r = ab / (a + b + 5).But in reality, the inradius of the base triangle is r_base = (a + b - 5)/2.So, which one is correct?Wait, in a triangle, the area is equal to r_base * s, where s is the semiperimeter.So, for triangle ABC, area = (1/2)ab = r_base * (a + b + 5)/2.Therefore, (1/2)ab = r_base * (a + b + 5)/2.Therefore, r_base = ab / (a + b + 5).So, the assistant is correct in that step.But earlier, I thought r_base = (a + b - 5)/2.Wait, let's compute both expressions.From the formula, r_base = (a + b - c)/2, where c is the hypotenuse.So, r_base = (a + b - 5)/2.But also, from the area formula, r_base = ab / (a + b + 5).Therefore, equating the two expressions:(a + b - 5)/2 = ab / (a + b + 5)Multiply both sides by 2(a + b + 5):(a + b - 5)(a + b + 5) = 2abExpand the left side:(a + b)^2 - 25 = 2abBut (a + b)^2 = a² + 2ab + b² = 25 + 2ab.Therefore, 25 + 2ab - 25 = 2abSimplify: 2ab = 2ab.So, both expressions are consistent.Therefore, r_base can be expressed as either (a + b - 5)/2 or ab / (a + b + 5).So, the assistant's approach is correct.Therefore, when they express r = ab / (a + b + 5), and then proceed to find its maximum, they are correct.So, in their solution, they set a = 5 cos α, b = 5 sin α, so ab = 25 sin α cos α.Then, a + b = 5 (cos α + sin α).Therefore, r = [25 sin α cos α] / [5 (cos α + sin α) + 5] = [5 sin α cos α] / [cos α + sin α + 1].Then, they let t = cos α + sin α, so t ∈ [1, √2].Then, sin α cos α = (t² - 1)/2.Therefore, r = [5*(t² - 1)/2] / (t + 1) = [5*(t - 1)(t + 1)/2] / (t + 1) = [5*(t - 1)] / 2.Therefore, r = (5/2)(t - 1).Since t ∈ [1, √2], the maximum r occurs at t = √2, so r_max = (5/2)(√2 - 1).Therefore, surface area = 4πr² = 4π*(25/4)*( (√2 - 1)^2 ) = 25*(3 - 2√2)π.So, the correct surface area is 25(3 - 2√2)π.But the assistant wrote 25(3 - 3√2)π, which is incorrect.Therefore, the correct answer should be 25(3 - 2√2)π.But wait, let me check the assistant's final step again.They wrote: "Thus, the maximum surface area of the inscribed sphere in the right triangular prism is 25(3 - 3√2)π."So, they have 25(3 - 3√2)π, which is different from my calculation.Therefore, the assistant made a mistake in their final calculation.Alternatively, perhaps I made a mistake in my own calculation.Wait, let me recalculate the surface area.Given r = (5/2)(√2 - 1).Then, r² = [25/4]*( (√2 - 1)^2 ) = [25/4]*(2 - 2√2 + 1) = [25/4]*(3 - 2√2).Therefore, surface area = 4πr² = 4π*(25/4)*(3 - 2√2) = 25*(3 - 2√2)π.Yes, that's correct.Therefore, the correct answer is 25(3 - 2√2)π.But the assistant's answer was 25(3 - 3√2)π, which is incorrect.So, perhaps the assistant made a mistake in their calculation.Alternatively, maybe the problem is different.Wait, let me think again.Is the sphere inscribed in the prism, or is it inscribed in the base triangle?In the problem statement, it says "the inscribed sphere of the prism."Therefore, the sphere must be tangent to all faces of the prism, including the two triangular bases and the three rectangular faces.Therefore, the radius of the sphere is determined by both the inradius of the base triangle and the height of the prism.As we saw earlier, the radius r must satisfy r = r_base and h = 2r.Therefore, h = 2 r_base.But r_base = (a + b - 5)/2.Therefore, h = a + b - 5.But h is also the height of the prism, which is independent of a and b.Wait, no, in the prism, h is the height, which is a separate variable.Therefore, to maximize r, we need to maximize r_base, which is (a + b - 5)/2, given that a² + b² = 25.As we saw earlier, the maximum value of a + b is 5√2, so the maximum r_base is (5√2 - 5)/2.Therefore, the maximum radius r is (5√2 - 5)/2, and the surface area is 4πr² = 25(3 - 2√2)π.Therefore, the correct answer is 25(3 - 2√2)π.But the assistant's answer was 25(3 - 3√2)π, which is incorrect.Therefore, the correct answer should be 25(3 - 2√2)π.But wait, let me think again.Wait, in the assistant's solution, they wrote:"Let the radius of the inscribed sphere of the prism be r. Then, the inscribed circle of triangle ABC is the great circle of the sphere."Wait, that might not be accurate.In a right triangular prism, the inscribed sphere's great circle would correspond to the base triangle's incircle.But in reality, the sphere's great circle is a circle with radius equal to the sphere's radius, but in the base triangle, the incircle has radius r_base.Therefore, if the sphere's great circle is the incircle of the base triangle, then the sphere's radius would be equal to r_base.But in reality, the sphere's radius is determined by both the inradius of the base and the height of the prism.Therefore, perhaps the assistant's assumption that the inscribed circle of the base triangle is the great circle of the sphere is incorrect.Therefore, their approach might be flawed.Therefore, perhaps the correct approach is to consider that the sphere must be tangent to all faces, leading to r = r_base and h = 2r.Therefore, the maximum surface area is 25(3 - 2√2)π.Therefore, the correct answer is 25(3 - 2√2)π.But the assistant's answer was 25(3 - 3√2)π, which is incorrect.Therefore, the correct answer should be 25(3 - 2√2)π.

💡First, I need to evaluate the expression inside the innermost parentheses: (2 + 3 = 5).Next, I square the result: (5^2 = 25).Then, I substitute this back into the expression: (6 - 5 times (10 - 25) times 2).I calculate the subtraction inside the parentheses: (10 - 25 = -15).Now, the expression becomes: (6 - 5 times (-15) times 2).I perform the multiplication: (5 times (-15) = -75), and then (-75 times 2 = -150).Finally, I subtract this result from 6: (6 - (-150) = 6 + 150 = 156).