1951 AMC 12 Problem 18

Attempt Problem 18 of the 1951 AMC 12 below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 1951 AMC 12 solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

18.

The expression

21x^2+ax+21

is to be factored into two linear prime binomial factors with integer coefficients. This can be done if

a

is:

Any odd number

Some odd number

Any even number

Some even number

Zero

Write the factors as

(Ax+B)(Cx+D)

Because

AC=BD=21,

all four integer factors are odd

Suppose

\begin{aligned} &21x^2+ax+21\\ &\quad=(Ax+B)(Cx+D). \end{aligned}

Then

AC=BD=21,

A,B,C,D

are odd. Therefore

a=AD+BC

is even. Some even values work; for example,

\begin{aligned} &(3x+7)(7x+3)\\ &\quad=21x^2+58x+21. \end{aligned}

Not every even value works.

Thus, the correct answer is D.

Problem 18 in other years