1951 AHSME Problems/Problem 18

Problem

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:

$\textbf{(A)}\ \text{any odd number} \qquad\textbf{(B)}\ \text{some odd number} \qquad\textbf{(C)}\ \text{any even number}$ $\textbf{(D)}\ \text{some even number} \qquad\textbf{(E)}\ \text{zero}$

Solution

We can factor $21x^2 + ax + 21$ as $(7x+3)(3x+7)$ , which expands to $21x^2+42x+21$ . So the answer is $\textbf{(D)}\ \text{some even number}$

Solution 2

Factoring $21x^2+ax+21$ by grouping, we need to find some $b,c$ such that $b\cdot c = 21$ , and that $b+c=a$ . Since $21\equiv 1\;(mod\;2)$ , $b\land c \equiv 1\;(mod\;2)$ , and $b+c \equiv 0\;(mod\;2)$ . So $a$ must be even. $a$ cannot be $any$ even number, since $21$ only has 4 odd factors, so the answer is $\textbf{(D)}\ \text{some even number}$

1951 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 17	Followed by Problem 19
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1951 AHSME Problems/Problem 18

Contents

Problem

Solution

Solution 2

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