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Re: Is k the square of an integer? [#permalink]
it's easy to miss that m can be zero or a squre.
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Re: Is k the square of an integer? [#permalink]
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A square of a number is derived when the number is multiplied by itself-> it has 2 copies of a number in a product

For K to be a square of an integer, it has to be an integer^2

1. k=t^2∗q^6∗r^10 where t, q and r are integers

sqrt(k) = sqrt(t^2∗q^6∗r^10) = t*q^3*r^5-> hence sqrt(K) is an integer -> sufficient

2.k=t^2*m^15 , where t and m are integers
sqrt(k)=sqrt(t^2*m^15) = t*m^7*sqrt(m)-> now in this case we cannot say whether k is a square of an integer , if m=3, then K is not the square of an integer, if m=4 then K is a square of an integer-> hence insufficient
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Re: Is k the square of an integer? [#permalink]
Solved this question using no of factors concept. No of factors of a perfect square is odd. But the reverse is not true right?
Can someone please confirm. Because I got option b wrong.
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Re: Is k the square of an integer? [#permalink]
Can someone clarify the doubt above?
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Re: Is k the square of an integer? [#permalink]
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TBT wrote:
Solved this question using no of factors concept. No of factors of a perfect square is odd. But the reverse is not true right?
Can someone please confirm. Because I got option b wrong.


The number of distinct factors of a perfect square is ALWAYS ODD. The reverse is also true: if a number has the odd number of distinct factors then it's a perfect square.


From (2) we cannot say whether the number of factors of k is even or odd because we don't know the number of factors of m. Check the solution here: https://gmatclub.com/forum/is-k-the-squ ... l#p1158051
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Re: Is k the square of an integer? [#permalink]
Quote:
bunuel
I got my mistake. I assumed m cannot be a perfect square itself. Thankyou!
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Re: Is k the square of an integer? [#permalink]
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Re: Is k the square of an integer? [#permalink]
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