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Re: If p is a positive integer, and p ≠ 0, is (p^2 + p)/(2p) an integer? [#permalink]

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23 Aug 2016, 06:41

Bunuel wrote:

If p is a positive integer, and p ≠ 0, is (p^2 + p)/(2p) an integer?

(1) p is an even number. (2) p is a multiple of 3.

Answer will be A as from statement 1 we can definitely say P will not be an integer because and Even number+1 will always be odd and and Odd Number divided by 2 will never be integer hence it is sufficient

and from 2nd statement if P is an multiple of 3 this means it can either be even or either be odd in that case the expression can be or can not be integer hence this statement alone is not sufficient.

Try solving it again. You have missed a couple of things in this question.

buddy it says (P^2+P)/2P lets take value of P=2 4+2=6 6/2*2=6/4=1.5

In this case, simplifying the equation prior to testing numbers is essential.

(P^2+P) / 2P = (P+1)/2.

If P+1=even, P+1/2 is an integer.

The question thus lands in whether P is even or odd.

Let's start with (1) P is even. Look at that, just what we wanted. P+1 won't be even and thus (P+1)/2 won't be an integer. Sufficient.

(2) P is a multiple of 3. Any even multiple of 3 will be even and any odd multiple of 3 will be odd. As such, we can't determine whether p+1 is even or odd and in extension whether the equation gives an integer answer. Insufficient.
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