Difference between revisions of "1951 AHSME Problems/Problem 18"
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== 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 = 21^2</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>21^2\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 any even, since <math>21^2</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 == |
Latest revision as of 11:56, 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 even number, 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 |
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