answer：Okay, so I have this problem here: I need to find a natural number ( n ) such that the expression ( 2n(n+1)(n+2)(n+3) + 12 ) can be written as the sum of squares of two positive integers. The options are whether such an ( n ) doesn't exist, exists once, exists twice, or exists infinitely many times. Hmm, interesting.First, let me try to understand the expression ( 2n(n+1)(n+2)(n+3) + 12 ). It looks like it's a product of four consecutive integers multiplied by 2 and then adding 12. Maybe I can simplify this expression to make it easier to handle.Let me expand ( n(n+1)(n+2)(n+3) ). I know that ( n(n+3) = n^2 + 3n ) and ( (n+1)(n+2) = n^2 + 3n + 2 ). So, multiplying these two together, I get:[(n^2 + 3n)(n^2 + 3n + 2)]Let me set ( t = n^2 + 3n ). Then the expression becomes:[t(t + 2) = t^2 + 2t]So, the original expression is:[2(t^2 + 2t) + 12 = 2t^2 + 4t + 12]Hmm, that's a quadratic in terms of ( t ). Maybe I can complete the square or factor it somehow. Let me see:[2t^2 + 4t + 12 = 2(t^2 + 2t) + 12]Completing the square inside the parentheses:[t^2 + 2t = (t + 1)^2 - 1]So, substituting back:[2((t + 1)^2 - 1) + 12 = 2(t + 1)^2 - 2 + 12 = 2(t + 1)^2 + 10]Hmm, so the expression simplifies to ( 2(t + 1)^2 + 10 ). But ( t = n^2 + 3n ), so:[2(n^2 + 3n + 1)^2 + 10]Wait, that might not be the most helpful form. Let me think again.Alternatively, maybe I can factor the original expression differently. Let me go back to ( 2n(n+1)(n+2)(n+3) + 12 ). Notice that ( n(n+1)(n+2)(n+3) ) is the product of four consecutive integers, which is equal to ( (n^2 + 3n)(n^2 + 3n + 2) ). So, as I had before, ( t(t + 2) ) where ( t = n^2 + 3n ).So, ( 2t(t + 2) + 12 = 2t^2 + 4t + 12 ). Maybe I can factor this quadratic:Looking for factors of ( 2t^2 + 4t + 12 ). Hmm, discriminant is ( 16 - 96 = -80 ), so it doesn't factor over the reals. Maybe I can factor it as ( 2(t^2 + 2t + 6) ). Not sure if that helps.Wait, maybe I can write the entire expression as something squared plus something else. Let me see:We have ( 2t^2 + 4t + 12 ). Let me factor out a 2:[2(t^2 + 2t + 6)]Hmm, ( t^2 + 2t + 6 ) can be written as ( (t + 1)^2 + 5 ). So,[2((t + 1)^2 + 5) = 2(t + 1)^2 + 10]So, the original expression is ( 2(t + 1)^2 + 10 ). Since ( t = n^2 + 3n ), then ( t + 1 = n^2 + 3n + 1 ). So, the expression is:[2(n^2 + 3n + 1)^2 + 10]Hmm, not sure if that helps. Maybe I need a different approach.The problem states that this expression should be equal to the sum of squares of two positive integers. So, ( 2n(n+1)(n+2)(n+3) + 12 = a^2 + b^2 ) for some positive integers ( a ) and ( b ).I remember that numbers expressible as the sum of two squares have certain properties related to their prime factorization. Specifically, a number can be expressed as the sum of two squares if and only if in its prime factorization, every prime congruent to 3 modulo 4 appears an even number of times.So, maybe I can analyze the expression ( 2n(n+1)(n+2)(n+3) + 12 ) modulo 4 or modulo some other number to see if it can be a sum of two squares.Let me compute the expression modulo 4. Since squares modulo 4 are either 0 or 1, the sum of two squares modulo 4 can be 0, 1, or 2.Let me compute ( 2n(n+1)(n+2)(n+3) + 12 ) modulo 4.First, note that ( n(n+1)(n+2)(n+3) ) is the product of four consecutive integers. Among any four consecutive integers, there are two even numbers, one of which is divisible by 4. So, the product is divisible by ( 2 times 4 = 8 ). Therefore, ( n(n+1)(n+2)(n+3) ) is divisible by 8.Thus, ( 2n(n+1)(n+2)(n+3) ) is divisible by 16. So, ( 2n(n+1)(n+2)(n+3) equiv 0 mod 16 ).Adding 12, we get:[2n(n+1)(n+2)(n+3) + 12 equiv 12 mod 16]So, the expression is congruent to 12 modulo 16. Now, let's see what the sum of two squares can be modulo 16.Squares modulo 16 are 0, 1, 4, 9. So, possible sums are:0+0=00+1=10+4=40+9=91+1=21+4=51+9=104+4=84+9=139+9=18≡2 mod16So, the possible residues are 0,1,2,4,5,8,9,10,13.But our expression is 12 mod16, which is not among these. Therefore, the expression cannot be written as the sum of two squares because it's congruent to 12 mod16, which is not achievable by the sum of two squares.Wait, does that mean there are no such natural numbers ( n )? So, the answer is A: Does not exist.But let me double-check my reasoning because sometimes I might make a mistake.First, I concluded that ( n(n+1)(n+2)(n+3) ) is divisible by 8 because in four consecutive numbers, there are two even numbers, one divisible by 4. So, 2*4=8. Then, multiplying by 2, it's divisible by 16. So, ( 2n(n+1)(n+2)(n+3) equiv 0 mod16 ). Adding 12, we get 12 mod16.Then, I considered the possible residues of the sum of two squares modulo16. I listed them as 0,1,2,4,5,8,9,10,13. 12 is not among them, so the expression cannot be expressed as the sum of two squares.Therefore, there are no natural numbers ( n ) satisfying the condition. So, the correct answer is A.But wait, let me test for small values of ( n ) to see if maybe for some small ( n ), the expression is a sum of two squares.Let me compute the expression for ( n = 1 ):( 2*1*2*3*4 + 12 = 2*24 +12 = 48 +12=60 ). Is 60 expressible as the sum of two squares?Let me see: 60. The squares less than 60 are 1,4,9,16,25,36,49.Check:49 + 11=60, 11 is not a square.36 +24=60, 24 not a square.25 +35=60, 35 not a square.16 +44=60, 44 not a square.9 +51=60, nope.4 +56=60, nope.1 +59=60, nope.So, 60 cannot be expressed as the sum of two squares.Next, ( n=2 ):( 2*2*3*4*5 +12 = 2*120 +12=240 +12=252 ).Is 252 expressible as the sum of two squares?Squares less than 252: up to 225.225 +27=252, 27 not square.196 +56=252, 56 not square.169 +83=252, nope.144 +108=252, nope.121 +131=252, nope.100 +152=252, nope.81 +171=252, nope.64 +188=252, nope.49 +203=252, nope.36 +216=252, nope.25 +227=252, nope.16 +236=252, nope.9 +243=252, nope.4 +248=252, nope.1 +251=252, nope.So, 252 cannot be expressed as the sum of two squares.Next, ( n=3 ):( 2*3*4*5*6 +12 = 2*360 +12=720 +12=732 ).Is 732 expressible as the sum of two squares?Squares up to 729.729 +3=732, nope.676 +56=732, 56 not square.625 +107=732, nope.576 +156=732, nope.529 +203=732, nope.484 +248=732, nope.441 +291=732, nope.400 +332=732, nope.361 +371=732, nope.324 +408=732, nope.289 +443=732, nope.256 +476=732, nope.225 +507=732, nope.196 +536=732, nope.169 +563=732, nope.144 +588=732, nope.121 +611=732, nope.100 +632=732, nope.81 +651=732, nope.64 +668=732, nope.49 +683=732, nope.36 +696=732, nope.25 +707=732, nope.16 +716=732, nope.9 +723=732, nope.4 +728=732, nope.1 +731=732, nope.So, 732 also cannot be expressed as the sum of two squares.Hmm, seems like for n=1,2,3, it's not possible. Maybe n=0? Although n is a natural number, which usually starts at 1, but sometimes 0 is included.If n=0: ( 2*0*1*2*3 +12=0 +12=12 ). Is 12 expressible as sum of two squares? 12=9+3, 3 not square; 4+8, 8 not square; 1+11, nope. So, no.n=4:( 2*4*5*6*7 +12=2*840 +12=1680 +12=1692 ).Is 1692 a sum of two squares?Squares up to 1681.1681 +11=1692, nope.1600 +92=1692, 92 not square.1521 +171=1692, nope.1444 +248=1692, nope.1369 +323=1692, nope.1296 +396=1692, nope.1225 +467=1692, nope.1156 +536=1692, nope.1089 +603=1692, nope.1024 +668=1692, nope.961 +731=1692, nope.900 +792=1692, nope.841 +851=1692, nope.784 +908=1692, nope.729 +963=1692, nope.676 +1016=1692, nope.625 +1067=1692, nope.576 +1116=1692, nope.529 +1163=1692, nope.484 +1208=1692, nope.441 +1251=1692, nope.400 +1292=1692, nope.361 +1331=1692, nope.324 +1368=1692, nope.289 +1403=1692, nope.256 +1436=1692, nope.225 +1467=1692, nope.196 +1496=1692, nope.169 +1523=1692, nope.144 +1548=1692, nope.121 +1571=1692, nope.100 +1592=1692, nope.81 +1611=1692, nope.64 +1628=1692, nope.49 +1643=1692, nope.36 +1656=1692, nope.25 +1667=1692, nope.16 +1676=1692, nope.9 +1683=1692, nope.4 +1688=1692, nope.1 +1691=1692, nope.So, 1692 also cannot be expressed as the sum of two squares.Hmm, seems like for n=1,2,3,4, it's not possible. Maybe this pattern continues? Or perhaps there's a mathematical reason why it can't be expressed as such.Earlier, I considered the expression modulo16 and saw that it's congruent to 12, which is not a possible residue for the sum of two squares. That seems like a solid argument because the sum of two squares can only be 0,1,2,4,5,8,9,10,13 modulo16, and 12 isn't among them.Therefore, regardless of the value of ( n ), the expression ( 2n(n+1)(n+2)(n+3) + 12 ) will always be congruent to 12 modulo16, which cannot be expressed as the sum of two squares. Hence, there are no natural numbers ( n ) satisfying the condition.So, the answer should be A: Does not exist.**Final Answer**boxed{A}

answer：Okay, so I have this problem about grass growing in a pasture and how many cows it can feed for a certain number of days. Let me try to understand it step by step.The problem says: "The grass grows at a constant rate every day. This pasture of grass can feed 20 cows for 40 days, or it can feed 35 cows for 10 days. How many days can it feed 25 cows?"Hmm, okay. So, grass grows every day, and depending on how many cows are eating it, the pasture will last a certain number of days. I need to find out how many days 25 cows can be fed.First, I think I need to figure out the rate at which the grass grows and the rate at which the cows eat the grass. Maybe I can set up some equations to model this.Let me denote:- Let G be the initial amount of grass in the pasture.- Let R be the rate at which grass grows per day.- Let C be the rate at which one cow eats grass per day.So, if there are 20 cows, each cow eats C amount of grass per day. Therefore, 20 cows eat 20C amount of grass per day. The pasture can feed them for 40 days, so the total grass consumed by the cows in 40 days is 20C * 40.But during these 40 days, the grass is also growing. So, the total grass available after 40 days is the initial grass G plus the grass grown over 40 days, which is R * 40.Since the pasture is just enough to feed the cows for 40 days, the total grass consumed by the cows should equal the total grass available. So, I can write the equation:20C * 40 = G + 40RSimilarly, for the second scenario where 35 cows can be fed for 10 days, the total grass consumed is 35C * 10, and the total grass available is G + 10R. So, the equation is:35C * 10 = G + 10RNow, I have two equations:1. 800C = G + 40R2. 350C = G + 10RI can solve these two equations to find G and R in terms of C. Maybe I can subtract the second equation from the first to eliminate G.Subtracting equation 2 from equation 1:800C - 350C = (G + 40R) - (G + 10R)450C = 30RSo, 450C = 30RDividing both sides by 30:15C = RSo, R = 15CThat means the grass grows at a rate of 15C per day, where C is the amount one cow eats per day.Now, I can substitute R back into one of the original equations to find G. Let's use equation 2:350C = G + 10R350C = G + 10*(15C)350C = G + 150CSubtracting 150C from both sides:200C = GSo, G = 200CAlright, so the initial amount of grass is 200C, and the grass grows at 15C per day.Now, the question is, how many days can 25 cows be fed?Let me denote D as the number of days. So, 25 cows will eat 25C amount of grass per day. The total grass consumed in D days is 25C * D.The total grass available after D days is the initial grass G plus the grass grown over D days, which is R * D.So, setting up the equation:25C * D = G + R * DSubstituting G = 200C and R = 15C:25C * D = 200C + 15C * DLet me simplify this equation. First, I can divide both sides by C to eliminate C:25D = 200 + 15DNow, subtract 15D from both sides:10D = 200Divide both sides by 10:D = 20So, the pasture can feed 25 cows for 20 days.Wait, let me double-check my calculations to make sure I didn't make a mistake.Starting with the two equations:1. 800C = G + 40R2. 350C = G + 10RSubtracting equation 2 from equation 1:450C = 30R => R = 15CThen, substituting R into equation 2:350C = G + 150C => G = 200CThen, for 25 cows:25C * D = 200C + 15C * DDivide by C:25D = 200 + 15D25D - 15D = 20010D = 200 => D = 20Yes, that seems correct. So, the answer is 20 days.

answer：Okay, so I have this problem where I'm given a function f(x), and its derivative f'(x). The equation provided is f(x) = 2x f'(1) + ln(x). I need to find the value of f'(1). Hmm, let's see how I can approach this.First, I know that f'(x) is the derivative of f(x). So, maybe I can differentiate both sides of the equation with respect to x. That should give me an expression involving f'(x) and f'(1). Let me try that.Differentiating the left side, I get f'(x). Now, differentiating the right side: 2x f'(1) is a term where f'(1) is a constant because it's the derivative at a specific point, x=1. So, the derivative of 2x f'(1) with respect to x is just 2 f'(1). Then, the derivative of ln(x) is 1/x. So putting it all together, I have:f'(x) = 2 f'(1) + 1/x.Okay, so now I have an expression for f'(x) in terms of f'(1) and x. But I need to find f'(1). Maybe I can plug in x=1 into this equation because f'(1) is the derivative at x=1.Let me substitute x=1 into the equation:f'(1) = 2 f'(1) + 1/1.Simplifying that, I get:f'(1) = 2 f'(1) + 1.Hmm, now I have an equation where f'(1) is expressed in terms of itself. Let me solve for f'(1). Subtract 2 f'(1) from both sides:f'(1) - 2 f'(1) = 1.This simplifies to:- f'(1) = 1.So, multiplying both sides by -1 gives:f'(1) = -1.Wait, that seems straightforward. Let me double-check to make sure I didn't make any mistakes.Starting from the original equation: f(x) = 2x f'(1) + ln(x). Differentiating both sides, I get f'(x) = 2 f'(1) + 1/x. Plugging in x=1, f'(1) = 2 f'(1) + 1. Solving for f'(1), I subtract 2 f'(1) from both sides: -f'(1) = 1, so f'(1) = -1.Yes, that seems correct. I don't see any errors in my calculations. So, the value of f'(1) is -1.

answer：Okay, so I have this problem here: I need to find the coefficient of ( x^{-3} ) in the expansion of ( left(2x + frac{1}{x^2}right)^6 ). Hmm, binomial expansion, right? I remember that when you have something like ( (a + b)^n ), you can expand it using the binomial theorem, which involves combinations and powers of ( a ) and ( b ).Let me recall the binomial theorem formula. It says that:[(a + b)^n = sum_{r=0}^{n} binom{n}{r} a^{n - r} b^r]So, each term in the expansion is given by ( binom{n}{r} a^{n - r} b^r ). In this case, ( a ) is ( 2x ) and ( b ) is ( frac{1}{x^2} ), and ( n ) is 6. So, applying the formula, each term ( T_{r+1} ) in the expansion would be:[T_{r+1} = binom{6}{r} (2x)^{6 - r} left(frac{1}{x^2}right)^r]Let me simplify this term. First, ( (2x)^{6 - r} ) is ( 2^{6 - r} x^{6 - r} ). Then, ( left(frac{1}{x^2}right)^r ) is ( x^{-2r} ). So, combining these, the term becomes:[T_{r+1} = binom{6}{r} cdot 2^{6 - r} cdot x^{6 - r} cdot x^{-2r}]Simplifying the exponents of ( x ), we add them together:[x^{6 - r - 2r} = x^{6 - 3r}]So, the term is:[T_{r+1} = binom{6}{r} cdot 2^{6 - r} cdot x^{6 - 3r}]Now, I need to find the term where the exponent of ( x ) is ( -3 ). That means I need to solve for ( r ) in the equation:[6 - 3r = -3]Let me solve this step by step. Subtract 6 from both sides:[-3r = -3 - 6]Which simplifies to:[-3r = -9]Now, divide both sides by -3:[r = frac{-9}{-3} = 3]So, ( r = 3 ). That means the term corresponding to ( r = 3 ) will have ( x^{-3} ). Let me write out that term.First, compute the combination ( binom{6}{3} ). I remember that ( binom{n}{r} = frac{n!}{r!(n - r)!} ). So,[binom{6}{3} = frac{6!}{3! cdot (6 - 3)!} = frac{720}{6 cdot 6} = frac{720}{36} = 20]Wait, hold on. Let me double-check that calculation. ( 6! ) is 720, ( 3! ) is 6, and ( (6 - 3)! = 3! ) which is also 6. So, ( 720 ) divided by ( 6 times 6 ) is indeed ( 20 ). Okay, that seems right.Next, compute ( 2^{6 - r} ). Since ( r = 3 ), this becomes ( 2^{6 - 3} = 2^3 = 8 ).So, putting it all together, the term is:[binom{6}{3} cdot 2^{3} cdot x^{-3} = 20 cdot 8 cdot x^{-3}]Multiplying 20 and 8 gives:[20 times 8 = 160]Therefore, the coefficient of ( x^{-3} ) is 160.Wait, let me just make sure I didn't make any mistakes. Let me go through the steps again.1. Start with the binomial expansion formula.2. Identify ( a = 2x ), ( b = frac{1}{x^2} ), ( n = 6 ).3. General term: ( binom{6}{r} (2x)^{6 - r} left(frac{1}{x^2}right)^r ).4. Simplify exponents: ( x^{6 - 3r} ).5. Set exponent equal to -3: ( 6 - 3r = -3 ) gives ( r = 3 ).6. Compute combination: ( binom{6}{3} = 20 ).7. Compute ( 2^{6 - 3} = 8 ).8. Multiply together: ( 20 times 8 = 160 ).Everything seems to check out. I don't think I made any calculation errors. So, I'm confident that the coefficient is 160.**Final Answer**The coefficient of ( x^{-3} ) is boxed{160}.

answer：Alright, so I have this problem about arithmetic sequences, and I need to figure out parts (a), (b), and (c). Let me start with part (a).**Part (a):** The sequence is 2, b, 8, and it's supposed to be an arithmetic sequence. I remember that in an arithmetic sequence, the difference between consecutive terms is constant. So, the difference between the first and second term should be the same as the difference between the second and third term.Let me write that down:b - 2 = 8 - bHmm, okay, so if I solve for b, I can find its value. Let me do that step by step.First, I'll add b to both sides to get rid of the negative b on the right:b - 2 + b = 8 - b + bThat simplifies to:2b - 2 = 8Now, I'll add 2 to both sides to isolate the term with b:2b - 2 + 2 = 8 + 2Which simplifies to:2b = 10Finally, I'll divide both sides by 2 to solve for b:2b / 2 = 10 / 2So, b = 5.Let me check if that makes sense. If b is 5, then the sequence is 2, 5, 8. The difference between 2 and 5 is 3, and the difference between 5 and 8 is also 3. Yep, that works. So, part (a) seems done.**Part (b):** Now, this part is a bit more complex. We have a sequence a, b, c, and we need to find d1 and d2. From the problem statement, d1 is the non-negative number we need to add or subtract from b to make the sequence arithmetic without changing a or c. Similarly, d2 is the positive number we need to add or subtract from c to make the sequence arithmetic without changing a or b.They gave an example with the sequence 3, 10, 13. They adjusted b to 8 (so d1 = 2) and adjusted c to 17 (so d2 = 4). They noticed that 2d1 = d2, which is 2*2 = 4.Now, our sequence is 1, 13, 17. I need to find d1 and d2 for this sequence.Let me first find d1. To find d1, I need to adjust b (which is 13) so that the sequence becomes arithmetic. The first term is 1, and the third term is 17. So, what should b be to make it arithmetic?In an arithmetic sequence, the middle term is the average of the first and third terms. So, the new b should be (1 + 17)/2.Let me calculate that:(1 + 17)/2 = 18/2 = 9So, the new b should be 9. But originally, b was 13. So, how much do we need to change b? That's d1.d1 = |13 - 9| = 4So, d1 is 4. That means we need to decrease b by 4 to make the sequence arithmetic.Now, let's find d2. To find d2, we need to adjust c (which is 17) so that the sequence becomes arithmetic. We keep a and b the same, which are 1 and 13.In an arithmetic sequence, the difference between consecutive terms should be the same. So, the difference between b and a is 13 - 1 = 12. Therefore, the difference between c and b should also be 12.So, the new c should be b + 12 = 13 + 12 = 25.Originally, c was 17. So, how much do we need to change c? That's d2.d2 = |25 - 17| = 8So, d2 is 8. That means we need to increase c by 8 to make the sequence arithmetic.Let me verify if these values make sense. If we adjust b to 9, the sequence becomes 1, 9, 17. The differences are 8 and 8, which is arithmetic. If we adjust c to 25, the sequence becomes 1, 13, 25. The differences are 12 and 12, which is also arithmetic. So, d1 = 4 and d2 = 8 seem correct.**Part (c):** Now, I need to prove that for all three-term sequences, 2d1 = d2. From part (b), we saw that in the example given, 2d1 = d2, and in our specific case, 2*4 = 8, which also holds. So, it seems like this relationship is general.Let me think about how to approach this proof. Maybe I can express d1 and d2 in terms of a, b, and c, and then show that 2d1 = d2.First, let's recall that in an arithmetic sequence, the middle term is the average of the first and third terms. So, if we have a sequence a, b, c, to make it arithmetic, the middle term should be (a + c)/2.So, if we adjust b to make the sequence arithmetic, the new b would be (a + c)/2. Therefore, the change in b, which is d1, is:d1 = |b - (a + c)/2|Similarly, if we adjust c to make the sequence arithmetic, the new c should be 2b - a. Therefore, the change in c, which is d2, is:d2 = |c - (2b - a)|Let me simplify d2:d2 = |c - 2b + a| = |(a + c) - 2b|But notice that d1 is |b - (a + c)/2|, which can be rewritten as |(2b - a - c)/2|.So, d1 = |(2b - a - c)/2| = (|2b - a - c|)/2Similarly, d2 = |(a + c - 2b)| = |2b - a - c|So, d2 = 2 * |(2b - a - c)/2| = 2 * d1Therefore, 2d1 = d2.Wait, let me make sure I did that correctly.Starting from d1:d1 = |b - (a + c)/2| = |(2b - a - c)/2| = (|2b - a - c|)/2And d2 = |c - (2b - a)| = |c - 2b + a| = |a + c - 2b| = |2b - a - c|So, d2 = |2b - a - c| = 2 * (|2b - a - c|)/2 = 2d1Yes, that makes sense. So, 2d1 = d2.Alternatively, I can think about the differences. If we adjust b, the required change is half the difference needed for c. So, changing b by d1 affects the sequence in a way that only requires half the adjustment compared to changing c.Either way, the relationship holds because d1 is essentially half the discrepancy between the current sequence and the desired arithmetic sequence, while d2 is the full discrepancy.So, I think this proves that 2d1 = d2 for all three-term sequences.**Final Answer**(a) boxed{5}(b) boxed{d_1 = 4} and boxed{d_2 = 8}(c) Proved that (2d_1 = d_2).

answer：First, identify the integers between 1 and 500 that are multiples of both 4 and 6. The least common multiple of 4 and 6 is 12, so we look for multiples of 12 within this range.The multiples of 12 between 1 and 500 are: 12, 24, 36, ..., 492. These can be generated by the arithmetic sequence 12n, where n ranges from 1 to 41 (since 500 ÷ 12 ≈ 41.67). This gives us 41 multiples of 12.Next, we need to exclude the multiples of 5 and 9 from these 41 multiples of 12.1. **Multiples of 5 among the multiples of 12:** - The least common multiple of 12 and 5 is 60. - The multiples of 60 between 1 and 500 are: 60, 120, 180, 240, 300, 360, 420, 480. - There are 8 such numbers.2. **Multiples of 9 among the multiples of 12:** - The least common multiple of 12 and 9 is 36. - The multiples of 36 between 1 and 500 are: 36, 72, 108, 144, 180, 216, 252, 288, 324, 360, 396, 432, 468. - There are 13 such numbers.However, some numbers are multiples of both 60 and 36, which means they are multiples of the least common multiple of 60 and 36, which is 180. The multiples of 180 between 1 and 500 are: 180, 360. There are 2 such numbers.Using the inclusion-exclusion principle, we calculate the total number of valid multiples:Total valid multiples = Total multiples of 12 - Multiples of 5 - Multiples of 9 + Multiples of both 180Total valid multiples = 41 - 8 - 13 + 2 = 22Therefore, there are 22 integers between 1 and 500 that are multiples of both 4 and 6 but not of either 5 or 9.