Perfect Square Question OG 142

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Perfect Square Question OG 142

by EMAN » Sat Sep 19, 2009 10:03 am
If y is the smallest positive integer such that 3,150 multiplied by y is the square on an integer, then y must be

(A) 2
(B) 5
(C) 6
(D) 7
(E) 14

[spoiler]

3,150 = 2 x 3^2 x 5^2 x 7

so, (2)(7) = 14

Okay, I get the first step where you have to solve this question by breaking the number down to its prime factors. I just need to understand why we multiply 2 x 7. Is this because 2 and 7 are not even powers like 3 and 5 (in context of a perfect square)?

OG Explanation: To be a perfect square, 3,150 y must have an even number of prime factors (Agreed). At a minimum, y must have one factor of 2 and one factor of 7 so that 3,150y has two factors of each of the primes 2, 3, 5, and 7. The smallest integer value of y is then (2)(7) = 14.

[/spoiler]
Source: — Problem Solving |

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Follow Up

by EMAN » Sat Sep 19, 2009 10:04 am
Please see my question and comments after trying out this question. Thank you.

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by sanjeevgogi » Wed Sep 23, 2009 10:27 pm
Eman,

The powers of the prime factorization of any perfect square must be even.

3150 = 5^2*3^2*7*2

so we need another 7 and another 2 make the powers even,

hence y = 7*2, so

3150 y = 5^2*3^2*7^2*2^2 which satisfies the requirement.

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Re: Follow Up

by andrew.ng » Wed Sep 23, 2009 11:15 pm
EMAN wrote:Please see my question and comments after trying out this question. Thank you.
Thanks for the explanation.
I think it could be faster in this case if we just simply plug in the number. It's easy and quick to find out 14 is the only choice.