Difference between revisions of "1951 AHSME Problems/Problem 18"
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== Solution == | == Solution == | ||
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== | ||
+ | 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>21^2\equiv 1 (mod2)</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 any even, 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 11:54, 19 March 2017
Contents
Problem
The expression is to be factored into two linear prime binomial factors with integer coefficients. This can be done if
is:
Solution
We can factor as
, which expands to
. So the answer is
Solution 2
Factoring by grouping, we need to find some
such that
, and that
.
Since
,
, and
. So
must be even.
cannot be any even, since
only has 4 odd factors, so the answer is
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.