Okay, so I need to find the smallest positive integer k such that when I multiply 240 by k, the result is a perfect cube. Hmm, let me think about how to approach this.

First, I remember that a perfect cube is a number that can be written as some integer raised to the power of 3. For example, 8 is a perfect cube because it's 2^3, and 27 is 3^3. So, if 240 * k is a perfect cube, then all the exponents in its prime factorization must be multiples of 3. That makes sense because when you multiply prime factors, you add their exponents.

So, the plan is probably to factorize 240 into its prime components, see what each exponent is, and then figure out what additional factors are needed to make each exponent a multiple of 3. Then, multiply those additional factors together to get k. Let me try that.

Let's start by factorizing 240. I know that 240 divided by 2 is 120, divided by 2 again is 60, again by 2 is 30, and again by 2 is 15. Then 15 divided by 3 is 5, and 5 divided by 5 is 1. So the prime factors of 240 are 2^4 * 3^1 * 5^1. Let me confirm that:

2^4 is 16, 3^1 is 3, 5^1 is 5. Multiplying them together: 16 * 3 = 48, 48 * 5 = 240. Yep, that's correct.

Now, for 240 * k to be a perfect cube, each prime in the factorization must have an exponent that's a multiple of 3. Let's look at each prime:

- The exponent of 2 is 4. To make this a multiple of 3, we need to add 2 more 2s because 4 + 2 = 6, which is divisible by 3. Wait, 4 divided by 3 leaves a remainder of 1, so actually, the next multiple of 3 after 4 is 6. So, 6 - 4 = 2. So we need 2 more 2s.

- The exponent of 3 is 1. The next multiple of 3 is 3, so we need 3 - 1 = 2 more 3s.

- The exponent of 5 is 1. Similarly, we need 3 - 1 = 2 more 5s.

Therefore, k must be 2^2 * 3^2 * 5^2. Let me compute that:

2^2 is 4, 3^2 is 9, 5^2 is 25. Multiplying them together: 4 * 9 = 36, 36 * 25 = 900. So k is 900. Let me check if 240 * 900 is a perfect cube.

Calculating 240 * 900: 240 * 900. Let's compute that. 240 * 900 = 240 * 9 * 100 = (240 * 9) * 100. 240 * 9: 240 * 10 = 2400, subtract 240 gives 2160. Then 2160 * 100 = 216000. Now, is 216000 a perfect cube?

Let's factorize 216000. Let's break it down step by step. 216000 is 216 * 1000. 216 is 6^3, which is 2^3 * 3^3, and 1000 is 10^3, which is 2^3 * 5^3. So multiplying those together: (2^3 * 3^3) * (2^3 * 5^3) = 2^(3+3) * 3^3 * 5^3 = 2^6 * 3^3 * 5^3. Wait, but 2^6 is (2^2)^3, so the entire thing is (2^2 * 3 * 5)^3. Let's compute that: 2^2 is 4, 4 * 3 = 12, 12 * 5 = 60. So 60^3 is 216000. Yes, that's correct. So 216000 is indeed a perfect cube, specifically 60^3. Therefore, k=900 is a valid solution.

But wait, the question asks for the smallest possible positive k. Is there a smaller k that could work? Let me think. Since we have to make each exponent in the prime factorization of 240 * k a multiple of 3. The exponents of 2, 3, and 5 in 240 are 4, 1, and 1 respectively. To get the next multiple of 3, we need to add 2, 2, and 2 respectively. Therefore, the minimal k is 2^2 * 3^2 * 5^2, which is 4*9*25=900. There's no way to have a smaller k without reducing the exponents below the required multiples of 3, which would make the product not a perfect cube. Therefore, 900 is indeed the smallest possible positive k.

Wait, but let me double-check. Suppose I tried to use exponents less than 2 for any prime. For example, if I tried to use 2^1 * 3^1 * 5^1, which is 30, but then 240*30=7200. Let's factorize 7200. 7200 divided by 100 is 72. 72 is 8*9, so 7200 is 72*100= (8*9)*(10^2) = 2^3 * 3^2 * 2^2 * 5^2 = 2^(3+2) * 3^2 * 5^2 = 2^5 * 3^2 * 5^2. The exponents are 5, 2, 2. 5 is not a multiple of 3, so 7200 is not a perfect cube. Therefore, 30 doesn't work.

Alternatively, if I tried using k= 2^2 *3^2 *5^1= 4*9*5=180. Then 240*180=43200. Let's factorize 43200. 43200 divided by 100 is 432. 432 is 16*27, which is 2^4 * 3^3. So 43200=2^4 *3^3 *2^2 *5^2=2^(4+2)*3^3 *5^2=2^6 *3^3 *5^2. The exponents of 2 and 5 are 6 and 2. 6 is a multiple of 3, but 2 is not. So 43200 is not a perfect cube. Therefore, 180 doesn't work.

Similarly, if I tried k=2^2*3^1*5^2=4*3*25=300. Then 240*300=72000. Factorizing 72000: 72000=72*1000= (8*9)*10^3=2^3*3^2*2^3*5^3=2^(3+3)*3^2*5^3=2^6*3^2*5^3. Exponents of 3 and 2 are 2 and 6. 2 is not a multiple of 3, so 72000 is not a perfect cube. Therefore, 300 doesn't work.

Therefore, the minimal k is indeed 900. Since any smaller k would result in at least one exponent in the prime factorization not being a multiple of 3, making the product not a perfect cube. Therefore, 900 is the smallest possible k.

Another way to think about it is that k must provide the missing exponents to each prime factor in 240 to make them all multiples of 3. Since 240 has 2^4, 3^1, and 5^1, the exponents needed are 2 more 2s, 2 more 3s, and 2 more 5s. Multiplying those together gives 2^2*3^2*5^2=4*9*25=900. Yep, that's the same result.

So I think that's solid. I can't see any mistakes in the reasoning here. The key steps are prime factorization of 240, figuring out the required exponents, and then multiplying the necessary primes raised to those required exponents. Since each step is verified, and testing smaller k's shows they don't work, I'm confident that 900 is the correct answer.
Final Answer
The smallest possible positive value of $k$ is $\boxed{900}$.
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