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]
a) 5
b) 6
c) 7
d) 14
e) 15
[spoiler]OA is 14[/spoiler]
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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! Thanks4GMAT_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*44GMAT_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.
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