If n is any integer, which of the following expressions must be an even integer? — GMAT

If $n$ is any integer, which of the following expressions must be an even integer?

$3n^2 + 5n + 2$ Answer
B
$n^2 + n + 1$
C
$2n^2 + 3n$
D
$n^3 + 2n$
E
$n^2 + 3$

Answer

The expression

3n^2 + 5n + 2

must be an even integer for any integer

n

The expression

3n^2 + 5n + 2

can be factored as

3n(n+1) + 2

. Since

n

and

n+1

are consecutive integers, one of them must be even, making

n(n+1)

an even integer. Multiplying an even integer by 3 produces an even integer, and adding 2 maintains even parity. Therefore, this expression is guaranteed to be even for all integer values of

n

Step-by-Step Solution

Analyze the parity property of consecutive integers.

For any integer

n

, one of the terms in the pair

\{n, n+1\}

is even. Hence, the product

n(n+1)

is always an even integer.

The product of an even integer and any integer is always even.

Rewrite the target expression

3n^2 + 5n + 2

to isolate the consecutive integer product.

3n^2 + 5n + 2 = 3n^2 + 3n + 2 = 3n(n+1) + 2

Algebraic manipulation isolates known parity components.

Determine the parity of

3n(n+1) + 2

Since

n(n+1)

is even,

3 \times \text{even} = \text{even}

. Then

\text{even} + 2 = \text{even}

Multiplying an even integer by an odd integer yields an even integer, and adding an even integer preserves even parity.

Key Concept

Product of consecutive integers

n(n+1)

is always even; zero is an even integer; basic parity rules under addition and multiplication.

Estimated Time:1m 30s

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