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If n and k are integers and n^2 – kn is even, which of the following m
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17 Mar 2021, 02:59
17 Mar 2021, 02:59
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If n and k are integers and n^2 – kn is even, which of the following must be even?
A) n^2
B) k^2
C) 2n + k^2
D) n(k + 1)
E) k(k + n)
A) n^2
B) k^2
C) 2n + k^2
D) n(k + 1)
E) k(k + n)
Retired Moderator
Joined: 10 Apr 2015
Posts: 6218
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Re: If n and k are integers and n^2 – kn is even, which of the following m
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08 Jun 2021, 14:43
08 Jun 2021, 14:43
3
GeminiHeat wrote:
If n and k are integers and n^2 – kn is even, which of the following must be even?
A) n^2
B) k^2
C) 2n + k^2
D) n(k + 1)
E) k(k + n)
A) n^2
B) k^2
C) 2n + k^2
D) n(k + 1)
E) k(k + n)
A lot of Integer Properties questions can be answered quickly by testing values that satisfy the given information.
So, for example, if n² - kn is EVEN, it COULD be the case that n = 1 and k = 1
Now use these values to check the answer choices
A) n² = 1² = 1 (not even - ELIMINATE)
B) k² = 1² = 1 (not even - ELIMINATE)
C) 2n + k² = 2(1) + 1² = 3 (not even - ELIMINATE)
D) n(k + 1) = 1(1 + 1) = 2 (EVEN - keep!!)
E) k(k + n) = 1(1 + 1) = 2 (EVEN - keep!!)
We are now down to answer choices D and E.
So let's test a second pair of values.
If n² - kn is EVEN, it could also be the case that n = 2 and k = 1
Now use these values to check the remaining answer choices
D) n(k + 1) = 2(1 + 1) = 4 (EVEN - keep!!)
E) k(k + n) = 1(1 + 2) = 3 (not even - ELIMINATE)
By the process of elimination the answer must be D
Cheers,
Brent
General Discussion
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Re: If n and k are integers and n^2 – kn is even, which of the following m
[#permalink]
17 Mar 2021, 11:03
17 Mar 2021, 11:03
3
1
GeminiHeat wrote:
If n and k are integers and n^2 – kn is even, which of the following must be even?
A) n^2
B) k^2
C) 2n + k^2
D) n(k + 1)
E) k(k + n)
A) n^2
B) k^2
C) 2n + k^2
D) n(k + 1)
E) k(k + n)
\(n^2 - kn = even\)
Cases:
I: (even) - (even) = even
i.e. n = even, k = odd or even
II: (odd) - (odd) = even
i.e. n = odd, k = odd
Let's check the option choices:
A. \(n^2\) - NO
When \(n\) is odd, \(n^2\) is odd
B. \(k^2\) - NO
When \(k\) is odd, \(k^2\) is odd
C. \(2n + k^2\) - NO
\(2n\) will always be even and since \(k\) can be odd - even + odd = odd
D. \(n(k + 1)\) - YES
When \(n\) is even - the expression would always be even
When \(n\) is odd - \(k\) is odd i.e. odd(odd + odd) = even
E. k(k + n) - NO
When \(n\) is even, \(k\) can be odd or even
even(even + odd) = even
But, odd(odd + even) = odd
Hence, option D
