2002 CEMC Gauss (Grade 8) Problems/Problem 5

Problem

Which of the following expressions is equal to an odd integer?

$\text{ (A) }\ 3(5) + 1 \qquad\text{ (B) }\ 2(3 + 5) \qquad\text{ (C) }\ 3(3 + 5) \qquad\text{ (D) }\ 3 + 5 + 1 \qquad\text{ (E) }\ \frac{3 + 5}{2}$

Solution 1

We can evaluate each of the answer choices until we get an odd integer.

$3(5) + 1 = 15 + 1 = 16$ . $16$ is even, so A cannot be the answer.

$2(3 + 5) = 2 \times 8 = 16$ . We got the same number, so B also cannot be the answer.

$3(3 + 5) = 3 \times 8 = 24$ . $24$ is even, so C cannot be the answer.

$\frac{3 + 5}{2} = \frac{8}{2} = 4$ , $4$ is even, so E cannot be the answer.

$3 + 5 + 1 = 8 + 1 = 9$ . $9$ is odd, so D must be the answer.

Combining all of these or by seeing all of the answer choices are eliminated, we see that $\boxed {\textbf {(D) } 3 + 5 + 1}$ .

~anabel.disher

Solution 2 (patterns)

We can eliminate answer choice A because an odd multiplied by an odd is an odd. Adding an odd to another odd number gives an even number, so the answer cannot be A.

We also see that $3 + 5$ is even due to two odds being added together in both B and C. Since we have an even number multiplied by an integer in both answer choices, both B and C are even. Thus, we can eliminate both B and C.

With answer choice D, we see $3 + 5$ is even because two odds added together gives an even number. However, adding $1$ makes an odd number because an even number added to an odd number is odd.

Thus, the answer is $\boxed {\textbf {(D) } 3 + 5 + 1}$ .

~anabel.disher

2002 CEMC Gauss (Grade 8) (Problems • Answer Key • Resources)
Preceded by Problem 4	Followed by Problem 6
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CEMC Gauss (Grade 8)

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2002 CEMC Gauss (Grade 8) Problems/Problem 5

Problem

Solution 1

Solution 2 (patterns)