1951 AHSME Problems/Problem 18

Revision as of 11:56, 19 March 2017 by Xiej (talk | contribs) (Solution 2)

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

See Also

1951 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 17
Followed by
Problem 19
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