1951 AHSME Problems/Problem 18

Revision as of 12:29, 6 June 2024 by Doctor seventeen (talk | contribs) (Problem)
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)

Problem

The expression $21x^2+ax+21$ is to be factored into prime binomial factors and without a numerical monomial factor. This can be done if the value ascribed to $a$ 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 $21x^2+58x+21$. 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 number, 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
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50
All AHSME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png