Difference between revisions of "1986 AHSME Problems/Problem 24"
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==Solution== | ==Solution== | ||
| − | p(x) must be a factor of 3(x^4+6x^2+25)-(3x^4+4x^2+28x+5)=14x^2-28x+70=14(x^2-2x+5). | + | <math>p(x)</math> must be a factor of <math>3(x^4+6x^2+25)-(3x^4+4x^2+28x+5)=14x^2-28x+70=14(x^2-2x+5)</math>. |
| + | Therefore <math>p(x)=x^2 -2x+5</math> and <math>p(1)=4</math>. | ||
| − | + | The answer is <math>\fbox{(D) 4}</math>. | |
== See also == | == See also == | ||
Latest revision as of 17:53, 1 April 2018
Problem
Let 
, where 
and 
are integers.
If 
is a factor of both 
and 
, what is 
?
Solution
must be a factor of 
.
Therefore 
and 
.
The answer is 
.
See also
| 1986 AHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 23 |
Followed by Problem 25 | |
| 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 | ||
| All AHSME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
