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by JeetGulia » Fri Aug 20, 2010 11:47 am
If x and y are positive integers and x^2 = 3,150y, what is the minimum possible value of y?

a) 5
b) 6
c) 7
d) 14
e) 15



[spoiler]OA is 14[/spoiler]
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by 4GMAT_Mumbai » Fri Aug 20, 2010 12:12 pm
Hi,

3150 = 7 * 25 * 9 * 2.

To make this a perfect square, it has to be multiplied by a 7 and a 2.

Hence, the answer is 14. Hope this helps.

Thanks.
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by beatthegmatinsept » Fri Aug 20, 2010 12:21 pm
4GMAT_Mumbai wrote:Hi,

3150 = 7 * 25 * 9 * 2.

To make this a perfect square, it has to be multiplied by a 7 and a 2.

Hence, the answer is 14. Hope this helps.

Thanks.
Nice! That's a good approach in solving this question. I took almost 2 mins to solve this question by pugging in each answer choice, and looking for which one gives a perfect square. This definitely makes a lot more sense! Thanks :)
Being defeated is often only a temporary condition. Giving up is what makes it permanent.
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by JeetGulia » Fri Aug 20, 2010 12:46 pm
4GMAT_Mumbai wrote:Hi,

3150 = 7 * 25 * 9 * 2.

To make this a perfect square, it has to be multiplied by a 7 and a 2.

Hence, the answer is 14. Hope this helps.

Thanks.
so the number will be 3,15,014 = 7*7*25*9*4
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by Makushr1 » Fri Aug 20, 2010 2:41 pm
Yeah, I got it right. I've found that for alot of the questions with squares in them require either the quadratic equation or factoring. Looking at this you know it's not going to be the quad eq, so start factoring (another hint is that the number is pretty large).
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