Number Properties Q

czechchamp wrote:
PS:
If Y and N are positive integers and 450Y = N cubed, which of the following must be an integer?

I. Y / 3 x 2 squared x 5
II. Y / 3 squared x 2 x 5
III. Y / 3 x 2 x 5 squared

A) None
B) I only
C) II only
D) III only
E) I,II, III
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For what it's worth, I think this is among the most difficult number theory questions you can find on the GMAT (along with others that test a similar concept). I'm sure you have a good command of prime factorizations. To illustrate the important concept in this question (because you might see it again, in different guises), let's take three different primes, p, q and r, and say (p^4)*(q^2)*(r^5) is the prime factorization of some number n. What will the prime factorization of n^3 look like?

n = (p^4)*(q^2)*(r^5)
n^3 = [(p^4)*(q^2)*(r^5)]^3
n^3 = (p^12)*(q^6)*(r^15)

That is, we just triple every exponent in the prime factorization of n. This is how you can recognize a perfect cube from a prime factorization: every exponent is divisible by 3. Similarly, the exponents are all even in prime factorizations of perfect squares, are all divisible by 4 in prime factorizations of perfect fourth powers, and so on.

Okay, back to the original question:

450Y = N^3

Prime factorize, of course!

(2^1)*(3^2)*(5^2)*Y = N^3

But the left side is equal to a perfect cube- the exponents on 2, 3 and 5 must be divisible by 3 after we multiply by Y. So Y must be divisible at least by (2^2)*(3)*(5). So only I) must be an integer (the others could be integers if Y is large enough, but don't need to be).

ildude02 wrote:
We know that reducing 450 into it's prime factors gives, 2 * 3 ^2 * 5^2.
So, from equation 1, we can write y/ 2 ^2 * 3 * 5 as n ^3 / 2 ^3 * 3^3 * 5 ^3 => (n/2 * 3* 5) ^ 3.

I think you're doing the following:

n^3 = 450y
n^3 = 2 * 3 ^2 * 5^2 * y
n^3/(2 * 3 ^2 * 5^2) = y
n^3/(2^3 * 3^3 * 5^3) = y/(2^2 * 3 * 5)
(n/(2*3*5))^3 = y/(2^2 * 3 * 5)

Now we know that n must be divisible by 2, 3 and 5 (from the equation n^3 = 450y), so the left side of the above must be an integer. That is, y must be divisible by (2^2 * 3 * 5). But that's the best we can say here.

ildude02 wrote: We know that n is an integer, so it could be that n can be an integer with prime factors of 2, 3 and 5 , meaning the result will always give us an integer. But if n were 4 or 8, the n/2 * 3 * 5 is NOT an integer. So how can we say it 's always an integer with my approach ? Appreciate your response.

From the original question, if n^3 = 450y, n *must* be divisible by 2, 3 and 5. The primes that divide the right side of the equation must also divide the left side (assuming we're dealing with integers).

aditikedia wrote:III. Y / 3 x 2 x 5 squared

Is this Y/ (3x2x5^2) OR is it (Y/3)X2X(5^2)

If it's the latter.. could someone explain why this can't be an integer?

Thanks

It's the former- all of the numbers are in the denominator.