Is v^2 an even integer? (1) v is an even integer (2) √v is an even

Even X Even = Even

1) If v is even then\( v^2 \)is Even X Even = Even. Sufficient
2) \(\sqrt{v}\) is an even integer -> v^2 = Even X Even X Even X Even = Even. Sufficient

Ans D

The pre-thinking isn't quite right. For this reason, a reformed version of this question would be significantly harder:

Is v an even integer?

1). v^2 is an even integer

2). √v is an even integer


I do this question in my classes sometimes. In my years of teaching, I honestly think fewer than 15% of my students get this question right. Statement 1 is very, very tempting--but it's not sufficient.

Instead of picking v's and seeing which follow the rule, pick v^2s.

v^2 must be an even integer, so v^2 could be:

2
4
6
8
10
12... etc.

It becomes immediately clear why this statement is not sufficient.

v cold be √2 (not an even integer) or 2 (an even integer).

But it's so easy to assume 'integers' and, since 'fractions' don't work in this instance, think any non-integers aren't in play.

Very interesting question, and I must admit I did fall into the trap. Thank you for posting. Also, just to confirm is the answer to this B? I can't think of any case that would make B insufficient, but I wouldn't be surprised if there was one.

Posted from my mobile device

Yes the correct answer would be B. If √v is an even integer, √v could be:

2
4
6
8
10...

Meaning v is

4
16
36
64
100...

(any even perfect square, which are all even integers).

Is v^2 an even integer?

Stat1: v is an even integer, then v^2 must be an integer. Sufficient

Stat2: √v is an even integer, then v^2 must be an integer. Sufficient

So, I think D.