<Question>
The product of integers 240 and $k$ is a perfect cube. What is the smallest possible positive value of $k$?
</Question>

<think>
<Plan_1>
Okay, so I need to find the smallest positive integer k such that the product of 240 and k is a perfect cube. Hmm, let me think about how to approach this. I remember that for a number to be a perfect cube, all the exponents in its prime factorization must be multiples of 3. So, first, I should start by prime factorizing 240. Let me do that.

None

<prompt_1.1> Execute the prime factorization of 240 as planned above. </prompt_1.1>
</Plan_1>

<Execution_1>
<execution_1.1>
240 divided by 2 is 120. Divided by 2 again is 60, again by 2 is 30, and once more by 2 is 15. Then 15 divided by 3 is 5, and then 5 divided by 5 is 1. So, the prime factors of 240 are 2^4 * 3^1 * 5^1.
</execution_1.1>
</Execution_1>

<Plan_2>
Based on execution_1.1:
Now, when I multiply 240 by k, the exponents in the combined prime factorization need to all be multiples of 3. Let me think. If 240 is 2^4 * 3^1 * 5^1, then k must provide the necessary exponents to make each of these exponents reach the next multiple of 3.

<prompt_2.1> Determine the exponents needed for k to make 240k a perfect cube, based on the prime factorization of 240. </prompt_2.1>
-----

Alternatively, maybe there's a smaller k? Let me check with k=12.

<prompt_2.2> Check if a smaller k, such as 12, makes 240k a perfect cube. </prompt_2.2>
</Plan_2>

<Execution_2>
<execution_2.1>
Let's break it down prime by prime.

Starting with prime 2: 240 has 2^4. The next multiple of 3 after 4 is 6. So, we need 6 - 4 = 2 more 2s. So, k needs to have 2^2.

Next, prime 3: 240 has 3^1. The next multiple of 3 is 3. So, we need 3 - 1 = 2 more 3s. So, k needs 3^2.

Then, prime 5: 240 has 5^1. Similarly, the next multiple of 3 is 3, so we need 3 - 1 = 2 more 5s. So, k needs 5^2.

Are there any other primes in k's factorization? If k has any other prime factors besides 2, 3, and 5, then those would introduce new primes into the product 240k, which would mean their exponents would have to be multiples of 3 as well. But since we want the smallest k, we should avoid adding unnecessary primes. Therefore, k should only have primes 2, 3, and 5.

Putting that together, k would be 2^2 * 3^2 * 5^2. Let's compute that. 2^2 is 4, 3^2 is 9, 5^2 is 25. Multiplying these together: 4*9=36, 36*25=900. So, k=900.
</execution_2.1>

<execution_2.2>
Wait, if k=12, then 240*12=2880. Let's factorize 2880. 2880 divided by 10 is 288, so 2880=288*10. 288 is 2^5 *3^2, and 10 is 2*5. So overall, 2^6 *3^2 *5^1. The exponents are 6,2,1. 6 is a multiple of 3, but 2 and 1 are not. So 2880 is not a perfect cube.
</execution_2.2>
</Execution_2>

<Plan_3>
Based on execution_2.1:
Wait, is that right? Let me check again.

<prompt_3.1> Verify the calculation of k by recalculating the exponents needed for 240k to be a perfect cube. </prompt_3.1>
-----

Based on execution_2.2:
How about k=15? 240*15=3600.

<prompt_3.2> Verify if k=15 makes 240k a perfect cube. </prompt_3.2>
</Plan_3>

<Execution_3>
<execution_3.1>
240's prime factors: 2^4, 3^1, 5^1. To make each exponent a multiple of 3:

- For 2: 4 + 2 = 6 (which is 3*2)
- For 3: 1 + 2 = 3 (which is 3*1)
- For 5: 1 + 2 = 3 (which is 3*1)
</execution_3.1>

<execution_3.2>
Factorizing 3600: 3600=36*100= (2^2*3^2)*(2^2*5^2)=2^4*3^2*5^2. Exponents 4,2,2. Not multiples of 3.
</execution_3.2>
</Execution_3>

<Plan_4>
Based on execution_3.1:
Therefore, k must supply 2^2, 3^2, and 5^2. So, 2*2=4, 3*3=9, 5*5=25.
4*9=36, 36*25=900. So k=900. But wait, is that the smallest possible?
-----

Based on execution_3.1, execution_3.2:
Wait, maybe if I take a different combination. Let me see.

<prompt_4.1> Re-evaluate the exponents needed for k to ensure 240k is a perfect cube. </prompt_4.1>
</Plan_4>

<Execution_4>
<execution_4.1>
Alternatively, perhaps k doesn't need to have all three primes? Let's see. Suppose k only has 2,3,5. If I take k as 2^a *3^b *5^c, then:

After multiplying by 240, the exponents become:

2: 4 + a

3: 1 + b

5: 1 + c

Each of these must be divisible by 3. So:

4 + a ≡ 0 mod 3 ⇒ a ≡ (-4) mod 3 ⇒ a ≡ 2 mod 3. So the smallest a is 2.

1 + b ≡ 0 mod 3 ⇒ b ≡ 2 mod 3. Smallest b is 2.

1 + c ≡ 0 mod 3 ⇒ c ≡ 2 mod 3. Smallest c is 2.

Therefore, the minimal k is 2^2 *3^2 *5^2 = 4*9*25=900.
</execution_4.1>
</Execution_4>

<Plan_5>
Based on execution_4.1:
So, that seems to be the minimal k. But let me check once more.

<prompt_5.1> Confirm the smallest k by verifying the calculations and ensuring no smaller k exists. </prompt_5.1>
</Plan_5>

<Execution_5>
<execution_5.1>
Suppose there is a smaller k. Let's see, if we take k as 2^2*3^2*5^2=900. If we tried to reduce any exponent in k, say, make a=1, then 4 +1=5, which is not a multiple of 3. Similarly, a=0 gives 4, which is not. So a must be 2. Same logic for b and c. Therefore, k must have exponents 2 on each prime. So 2^2*3^2*5^2=900 is indeed minimal.
</execution_5.1>
</Execution_5>

<Final_answer>
Therefore, the smallest possible positive value of $k$ is $\boxed{900}$.
</Final_answer>
</think>