Please correct the typo in statement 1...

1) we have\(k={(tq^3r^5)}^2\). Since t,q and r are intergers, k is the square of an integer. Sufficient.

2)m can take several values for k to not be the square of an integer. If m =1, k is the square of an integer. Insufficient.

Answer is A.

Is k the square of an integer?

We need to determine whether k is a perfect square.

(1) k = t^2*q^6*r^10, where t, q, and r are integers --> \(k=(tq^3r^5)^2=integer^2\). Sufficient.

(2) k = t^2*m^15, where t and m are integers --> if m is a perfect square itself (say if m=n^2 for some integer n), then k will also be a perfect square but if m is NOT a perfect square, then k also will NOT be a perfect square. For example consider m=1 and m=2. Not sufficient.

Answer: A.
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it's easy to miss that m can be zero or a squre.

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