Problem of the Day


Which of the following expressions yields an even integer for any integer n?

A. n^2-10n+21
B. n^2-2n-24
C. n^2+8n+7
D. n^2+11n+18
E. n^2-4n-60

Reveal Answer

Answer

[latexpage]

D. $n^2+11n+18$

See the Solution

Solution

[latexpage]

First, factor all of the expressions into a product of binomials:

A. $n^2-10n+21=(n-7)(n-3)$
B. $n^2-2n-24=(n-6)(n+4)$
C. $n^2+8n+7=(n+7)(n+1)$
D. $n^2+11n+18=(n+9)(n+2)$
E. $n^2-4n-60=(n-10)(n+6)$

In choice D, we have $n$ plus an odd number times $n$ plus an even number. Thus, if $n$ is even, $n+2$ will be even, and if $n$ is odd, $n+9$ will be odd, thus guaranteeing that the product will be even whether $n$ is odd or even. The other expressions have either both odd numbers or both even numbers and thus would not necessarily produce an even product.


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