Difference between revisions of "2005 AIME II Problems/Problem 13"
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As above, we define <math>Q(x)=P(x)-x+7</math>, noting that it has roots at <math>17</math> and <math>24</math>. Hence <math>P(x)-x+7=A(x-17)(x-24)</math>. In particular, this means that | As above, we define <math>Q(x)=P(x)-x+7</math>, noting that it has roots at <math>17</math> and <math>24</math>. Hence <math>P(x)-x+7=A(x-17)(x-24)</math>. In particular, this means that | ||
Revision as of 20:23, 13 October 2016
Problem
Let 
be a polynomial with integer coefficients that satisfies 
and 
Given that 
has two distinct integer solutions 
and 
find the product 
Solution
As above, we define 
, noting that it has roots at 
and 
. Hence 
. In particular, this means that
. Therefore, 
satisfy 
, where 
, 
, and 
are integers. This cannot occur if 
or 
because the product 
will either be too large or not be a divisor of 
. We find that 
and 
are the only values that allow to be a factor of . Hence the answer is 
.
See also
| 2005 AIME II (Problems • Answer Key • Resources) | ||
| Preceded by Problem 12 |
Followed by Problem 14 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
| All AIME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
