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If n and k are integers and n^2 – kn is even, which of the following

Expert replies
by BTGModeratorVI » Tue Feb 18, 2020 11:10 am
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)

Answer: D
Source: Manhattan Prep
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Source: — Problem Solving |

BTGModeratorVI wrote: ↑
Tue Feb 18, 2020 11:10 am
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)

Answer: D
Source: Manhattan Prep
So, we have n^2 – kn = n(n – k) is even.

Case 1: Say n = even, then (n – k) can be even or odd. Thus, k can be even or odd.
Case 2: Say n = odd, then (n – k) must be even. Since n is odd, we have k = odd.

Let's see each option one by one.

A) n^2: Since n can be even or odd, n^2 can also be even or odd.

B) k^2: From Case 1, k can be even or odd, so k^2 can also be even or odd.

C) 2n + k^2:

Whether n is even or odd, 2n is even. We have already seen in Option B that k^2 can also be even or odd; thus, 2n + k^2 can be even or odd.

D) n(k + 1):

Case 1: If n is even, it is immaterail whether (k + 1) is even or odd; we see that n(k + 1) is even.
Case 2: If n is odd, we know that k is also odd; thus, (k + 1) would be even. Thus, n(k + 1) = Odd*Even = Even.

This option is must be true.

E) k(k + n):

Case 1: If k is even, it is immaterial whether (k + n) is even or odd; we see that k(k + n) is even.
Case 2: If k is odd and n is even, then k(k + n) = Odd*(Odd + Even) = Odd*Odd = Odd.

No unique answer.

The correct answer: D

Hope this helps!

-Jay
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BTGModeratorVI wrote: ↑
Tue Feb 18, 2020 11:10 am
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)

Answer: D
Source: Manhattan Prep
You can easily test cases by assuming if n and k are even or odd

n*n - k*n = Even

Case #1

n = Odd
k = Odd

Odd * Odd - Odd * Odd = Even

This holds since we know that:

Odd*Odd = Odd
Even + Odd = Odd

Now you can eliminate
(A)
(B)
(C)
Since those would be all odds by the same conditions

So you are left with (D) and (E), which are both Even

Now you can assume

n = Even
k = Odd

Even*Even - Even*Odd = Even ?

Yes by:

Even*Even =Even
Even*Odd = Even
Even + Even = Even

Therefore we test again our two answer choices left:

(D) Even(Odd+Odd) = Even
(E) Odd(Odd+Even) = Odd

So D is the correct answer.
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BTGModeratorVI wrote: ↑
Tue Feb 18, 2020 11:10 am
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)

Answer: D
Source: Manhattan Prep
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
Brent Hanneson - Creator of GMATPrepNow.com
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BTGModeratorVI wrote: ↑
Tue Feb 18, 2020 11:10 am
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)

Answer: D
Source: Manhattan Prep
If we choose n = 3 and k = 1, we see that n^2 = 9 and k^2 = 1 are odd; thus, we can eliminate answer choices A and B.

If we choose n = 2 and k = 1, we see that 2n + k^2 = 4 + 1 = 5 and k(k + n) = 1(1 + 2) = 3 are odd; thus, we eliminate answer choices C and E as well.

The only remaining answer choice is D.

Answer: D

Scott Woodbury-Stewart
Founder and CEO
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