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Bunuel
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Bunuel
If k is an odd integer, which of the following must be an even integer?


A. \(k^2 – 4\)

B. \(3k + 2\)

C. \(2k + 1\)

D. \(\frac{12k}{8}\)

E. \(\frac{6k}{3}\)

Plug in any value of k and check, let k = 3

(A) 5
(B) 11
(C) 7
(D) 4.5
(E) 6 , Thus Answer must be (E)
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Bunuel
If k is an odd integer, which of the following must be an even integer?


A. \(k^2 – 4\)

B. \(3k + 2\)

C. \(2k + 1\)

D. \(\frac{12k}{8}\)

E. \(\frac{6k}{3}\)

\(\frac{6k}{3}\) --- Will always be even coz 6k/3 is 2K
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Bunuel
If k is an odd integer, which of the following must be an even integer?


A. \(k^2 – 4\)

B. \(3k + 2\)

C. \(2k + 1\)

D. \(\frac{12k}{8}\)

E. \(\frac{6k}{3}\)

A, B, C. Always odd
D. Fraction.
E. 2k. Even.

E!
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Bunuel
If k is an odd integer, which of the following must be an even integer?


A. \(k^2 – 4\)

B. \(3k + 2\)

C. \(2k + 1\)

D. \(\frac{12k}{8}\)

E. \(\frac{6k}{3}\)

Since 6k/3 = 2k and 2k is a multiple of 2, then it must be even.

Answer: E
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Bunuel
If k is an odd integer, which of the following must be an even integer?


A. \(k^2 – 4\)

B. \(3k + 2\)

C. \(2k + 1\)

D. \(\frac{12k}{8}\)

E. \(\frac{6k}{3}\)

The key word in must, which means the correct answer will yield an EVEN value for ALL odd values of k.
So, if we find an odd value for k that does NOT yield an even output, then we can eliminate that answer choice.

Let's see what happens when k = 1 (1 is a nice odd integer to work with)
Plug k = 1 to get...
A. 1² – 4 = -3. -3 is NOT even. ELIMINATE A
B. 3(1) + 2 = 5. 5 is NOT even. ELIMINATE B
C. 2(1) + 1 = 3. 3 is NOT even. ELIMINATE C
D. 12(1)/8 = 3/2. 3/2 is NOT even. ELIMINATE D

By the process of elimination, the correct answer must be E

Cheers,
Brent
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