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If m and n are integers, is n even? (1) m^2-n^2=2n-m (2) m [#permalink]
01 Jul 2008, 07:51
00:00
A
B
C
D
E
Difficulty:
(N/A)
Question Stats:
0% (00:00) correct
0% (00:00) wrong based on 0 sessions
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If m and n are integers, is n even? (1) m^2-n^2=2n-m (2) m is an even number.
A. Statement (1) ALONE is sufficient but Statement (2) ALONE is not sufficient. B. Statement (2) ALONE is sufficient but Statement (1) ALONE is not sufficient. C. BOTH Statements TOGETHER are sufficient, but NEITHER Statement alone is sufficient. D. Each Statement ALONE is sufficient. E. Statements (1) and (2) TOGETHER are NOT sufficient.
I agree with Oski till n = (m-n)(m+n+1) then my argument is ... (m-n) and (m+n) are both either odd or even. And (m+n+1) is always going to be opposite of that.
hence n = (m-n)(m+n+1) = odd * even or n = (m-n)(m+n+1) = even * odd In any case, as one of its factors is even n is even.
If m and n are integers, is n even? (1) m^2-n^2=2n-m (2) m is an even number.
A. Statement (1) ALONE is sufficient but Statement (2) ALONE is not sufficient. B. Statement (2) ALONE is sufficient but Statement (1) ALONE is not sufficient. C. BOTH Statements TOGETHER are sufficient, but NEITHER Statement alone is sufficient. D. Each Statement ALONE is sufficient. E. Statements (1) and (2) TOGETHER are NOT sufficient.
I am getting A...though initially i thought the ans is C..but as i began to write out my work..it became clear to me A is sufficient
(m+n)(m-n)=n-m+n divide by (m+n)
(m-n)=n/(m+n) -1
now we know m and n are integers... therefore n/(m+n)=integer too..
m^2 - n^2 = 2n - m, can be written as m^2 + m = 2n + n ^2 => m (m + 1) = n (n+2)
m (m+1) need to be even becos, either m or (m+1) need to be even, as m is an integer. So the product n (n + 2) need to be even. n & (n+2) can be the same either even or odd. That leaves the option that n is even.
This is pretty gud question, First I went wrong on this, checked again and then came back with A.