The integer v is greater than 1

Adam

If v = 4, then 81v = 81(4) = 324. And 324 is the square of an integer. It is 18*18. So choice A is not 18, it is 18 *squared*. What we do is take the square root of the choice. If that's an integer, then we know that the choice is the square of an integer. Because the square root of choice A is an integer, 18, we know that choice A is the square of an integer.
Hope that helps!

Feb 12, 2020 • Reply

Sam Kinsman

Hi Rayssa,

You're right that v^2 would also be an integer. But note that the question is not asking us for answer choices that are an integer. The question is asking us for numbers that "must also be the square of an integer." And although C is an integer, it's not necessarily the square of an integer.

Aug 9, 2018 • Reply

Jonathan , Magoosh Tutor

Hi Jorge,
We prove algebraically that A and B must be true. 4 is just an example.
We know v is an integer from the question.
A) 81v = (9v)^2 and 9v must be an integer. So (A) must be correct.

B) We should know our FOIL identities cold, including (a + b)^2 = a^2 + 2ab + b^2

So We can see that (B) factors to

(5v + 1)^2 and we know 5v is an integer so therefore 5v + 1 is also an integer.

So (B) is correct.

For (C), we could see that we can't factor the expression so therefore there is no algebraic reason that it must be a square . To be sure, we can find one example for which it's not a square. If v = 4 The expression equals 75 which is not the square of an integer. So we know (C) is not correct.

Oct 30, 2017 • Reply

Bethany D Zacharias

Thank you for all of these explanation videos. The GRE answer key is not nearly as easy to understand.

Oct 14, 2013 • Reply