doudou wrote:If n and y are positive integers and 450y = n^3, which of the following must be an integer?

1 - y/(3*2^2*5)
2 - y/(3^2*2*5)
3 - y/(3*2*5^2)

We don't need exact values for n and y (it's impossible to find exact values), but we do need to calculate the minimum value for y.

To do so, start by isolating y:

y = n^3/450

Next, let's break 450 down into it's primes:

450 = 5 * 90 = 5 * 5 * 18 = 5 * 5 * 2 * 9 = 5 * 5 * 2 * 3 * 3

Now, since we know that n is an integer, n^3 must have triplets of all of it's primes (e.g. if n=5, then n^3 = 5*5*5).

If n^3/450 is an integer, then all of the prime factors of 450 must be prime factors of n^3 as well.

Therefore, 5, 5, 2, 3 and 3 are prime factors of n^3. Since all primes of n^3 will appear in triplets, to round out the value of n^3 we're going to need, at a minimum, one more 5, two more 2s and one more 3.

So, the minimum value of N^3/450 = 5*2*2*3.

Since this minimum value is also the minimum value of y (since n^3/450=y), all values of y must be divisible by 5*2*2*3, which is answer choice (1).

I will choose the first choice.

Answer: I only

I. For this expression to be an integer, y must be at least 60. (3x2x2x5 so 60/60 = 1)

450x60 = 27000 (we can easily see this as 30x30x30)

Use the same methodology on II and III. Only "I" can do a square root three successfully. Thus, I only.

Stuart Kovinsky wrote:

doudou wrote:If n and y are positive integers and 450y = n^3, which of the following must be an integer?

1 - y/(3*2^2*5)
2 - y/(3^2*2*5)
3 - y/(3*2*5^2)

We don't need exact values for n and y (it's impossible to find exact values), but we do need to calculate the minimum value for y.

To do so, start by isolating y:

y = n^3/450

Next, let's break 450 down into it's primes:

450 = 5 * 90 = 5 * 5 * 18 = 5 * 5 * 2 * 9 = 5 * 5 * 2 * 3 * 3

Now, since we know that n is an integer, n^3 must have triplets of all of it's primes (e.g. if n=5, then n^3 = 5*5*5).

If n^3/450 is an integer, then all of the prime factors of 450 must be prime factors of n^3 as well.

Therefore, 5, 5, 2, 3 and 3 are prime factors of n^3. Since all primes of n^3 will appear in triplets, to round out the value of n^3 we're going to need, at a minimum, one more 5, two more 2s and one more 3.

So, the minimum value of N^3/450 = 5*2*2*3.

Since this minimum value is also the minimum value of y (since n^3/450=y), all values of y must be divisible by 5*2*2*3, which is answer choice (1).

awesome breakdown Stuart. thanks!