Difference between revisions of "1988 AHSME Problems/Problem 15"
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==Solution== | ==Solution== | ||
| − | + | Using polynomial division, we find that the remainder is <math>(2a+b)x+(a+b+1)</math>, so for the condition to hold, we need this remainder to be <math>0</math>. This gives <math>2a+b=0</math> and <math>a+b+1=0</math>, so <math>b=-2a</math> and <math>a-2a+1=0 \implies a=1 \implies b=-2</math>, which is <math>\boxed{\text{A}}.</math> | |
Latest revision as of 17:23, 26 February 2018
Problem
If 
and 
are integers such that 
is a factor of 
, then is
Solution
Using polynomial division, we find that the remainder is 
, so for the condition to hold, we need this remainder to be 
. This gives 
and 
, so 
and 
, which is 
See also
| 1988 AHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 14 |
Followed by Problem 16 | |
| 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. 
