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Difference between revisions of "1951 AHSME Problems/Problem 18"

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== Problem ==
 
== Problem ==
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The expression <math> 21x^2 +ax +21</math> is to be factored into two linear prime binomial factors with integer coefficients. This can be done if <math> a</math> is:
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The expression <math>21x^2+ax+21</math> is to be factored into prime binomial factors and without a numerical monomial factor. This can be done if the value ascribed to <math>a</math> is:
  
 
<math> \textbf{(A)}\ \text{any odd number} \qquad\textbf{(B)}\ \text{some odd number} \qquad\textbf{(C)}\ \text{any even number}</math>
 
<math> \textbf{(A)}\ \text{any odd number} \qquad\textbf{(B)}\ \text{some odd number} \qquad\textbf{(C)}\ \text{any even number}</math>

Latest revision as of 13:29, 6 June 2024

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 (Problems • Answer Key • Resources)
Preceded by
Problem 17 		Followed by
Problem 19
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All AHSME Problems and Solutions

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