Unsure of the "Odds and Ms" question I came across on CAT 1

[sahil.kumar1]

Hi,
Question on the CAT is:

If m, n, and p are integers, is m + n odd?
(1) m = p^2 + 4p + 4
(2) n = p^2 + 2m + 1

Now, when I solve (1), then I get that

m = (p+2)^2

Thus, if m is Odd, then p is Odd, and if m is even, then p is Even. But the answer explanation says that it cannot be determined. Please help! I'm sure I'm missing out on something.

Thanks,
S

[biswas_sandeep]

Hi Sahil,
I think the question is asking whether m+n is odd and not about m in particular.
I hope this clarifies your doubt.

Sandeep

[jnelson0612]

Sandeep is correct. Let us know if you are still confused.

[aagar2003]

If m, n, and p are integers, is m + n odd?
(1) m = p^2 + 4p + 4
(2) n = p^2 + 2m + 1
But the answer explanation says that it cannot be determined.

Are you sure the answer is E and not C. See my working below;

1. 4p+4 = even, p^2 can be either even or odd => m cannot be said either even or odd. n is not even in picture. So A and D are out.

2. 2m does not make a difference to status of n. p^2 + 1 is even when p is odd and odd when p is even. Status of m is not in picture. Hence B goes out.

Now solving together to check C.
Adding both equations, m+n = 2p^2+4p+2m+5 = even + 5(odd) = odd always

Hence answer is C and not E, unless either the posted question or intended answer is incorrect.

[hoxlay]

Answer should be 'C' since combining statements 1 and 2 gives--->

====> m+n = (2p^2) + (sum of a bunch of even integers) + (1)
====> m+n = (even) + (even) + odd = ALWAYS odd

Hence m+n is ALWAYS odd and question can be answered.

Thus, C.

[jnelson0612]

Yeah, I not only understand your logic but I've even plugged odd and even numbers in for the values and I keep getting the same result: m+n is always odd.

I'm going to flag this to see if Ron or Tim would like to weigh in. There may be something easy that we're all missing.