Is k the square of an integer?

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Gmat_mission: Legendary Member; Posts: 1622; Joined: Thu Mar 01, 2018 7:22 amFollowed by: 2 members

by Gmat_mission » Sat Apr 14, 2018 3:04 am

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Is k the square of an integer? $$(1)\ \ \ k=t^2*q^6*r^{10}\ ,\ where\ t,\ q\ and\ r\ are\ integers.$$ $$(2)\ \ k=t^2*m^{15}\ ,\ where\ \ t\ and\ m\ are\ integers.$$ [spoiler]OA=A[/spoiler].

Why is A the correct answer? Can someone give me an explanation? Please.

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Source: Beat The GMAT — Data Sufficiency | Sat Apr 14, 2018 3:04 am

Vincen: Legendary Member; Posts: 2898; Joined: Thu Sep 07, 2017 2:49 pmFollowed by: 5 members; Thanked: 6 times

by Vincen » Sat Apr 14, 2018 6:10 am

Hello.

I'd solve it like this: $$\left(1\right)\ \ \ k=t^2\cdot q^6\cdot r^{10}$$ If we rewrite this expression we get: $$k=t^2\cdot\left(q^3\right)^2\cdot\left(r^5\right)^2\ =\ \ \left(t\cdot q^3\cdot r^5\right)^2.$$ Since t, q and r are integers, then t*q^3 * r^5 is an integer. Therefore, this statement is SUFFICIENT. $$\left(2\right)\ \ \ k=t^2\cdot m^{15}$$ If we rewrite this expression we will get $$k=t^2\cdot\left(m^{\frac{15}{2}}\right)^2=\left(t\cdot m^{\frac{15}{2}}\right)^2\ .$$ Now, since we don't know the value of m, we can't sy that m^(15/2) is integer. Therefore, this statement is NOT SUFFICIENT.

This is why the correct answer is the option A.

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