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

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We can factor <math> 21x^2 + ax + 21</math> as <math>(7x+3)(3x+7)</math>, which expands to <math>21x^2+42x+21</math>. So the answer is <math> \textbf{(D)}\ \text{some even number}</math>
 
We can factor <math> 21x^2 + ax + 21</math> as <math>(7x+3)(3x+7)</math>, which expands to <math>21x^2+42x+21</math>. So the answer is <math> \textbf{(D)}\ \text{some even number}</math>
 
== Solution 2==
 
== Solution 2==
−
Factoring <math>21x^2+ax+21</math> by grouping, we need to find some <math>b,c</math> such that <math>b\cdot c = 21</math>, and that <math>b+c=a</math>.
+
Factoring <math>21x^2+ax+21</math> by grouping, we need to find some <math>b,c</math> such that <math>b\cdot c = 21^2</math>, and that <math>b+c=a</math>.
−
Since <math>21\equiv 1\;(mod\;2)</math>, <math>b\land c \equiv 1\;(mod\;2)</math>, and <math>b+c \equiv 0\;(mod\;2)</math>. So <math>a</math> must be even. <math>a</math> cannot be <math>any</math> even number, since <math>21</math> only has 4 odd factors, so the answer is <math> \textbf{(D)}\ \text{some even number}</math>
+
Since <math>21^2\equiv 1\;(mod\;2)</math>, <math>b\land c \equiv 1\;(mod\;2)</math>, and <math>b+c \equiv 0\;(mod\;2)</math>. So <math>a</math> must be even. <math>a</math> cannot be <math>any</math> even number, since <math>21^2</math> only has 4 odd factors, so the answer is <math> \textbf{(D)}\ \text{some even number}</math>
  
 
== See Also ==
 
== See Also ==

Revision as of 18:13, 5 June 2024

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 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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