Difference between revisions of "2005 AIME II Problems/Problem 13"
(→Solution) |
(→Solution 2) |
||
| Line 11: | Line 11: | ||
==Solution 2== | ==Solution 2== | ||
| − | We know that <math>P(n)-(n+3)=0 | + | We know that <math>P(n)-(n+3)=0</math> so P(n) has two distinct solutions so <math>P(x)</math> is at least quadratic. Let us first try this problem out as if <math>P(x)</math> is a quadratic polynomial. Thus <math>P(x)=ax^2+bx+c</math> where <math>a,b,c</math> are all integers. |
== See also == | == See also == | ||
Revision as of 22:31, 7 August 2018
Contents
Problem
Let 
be a polynomial with integer coefficients that satisfies 
and 
Given that 
has two distinct integer solutions 
and 
find the product 
Solution
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 
.
Solution 2
We know that 
so P(n) has two distinct solutions so is at least quadratic. Let us first try this problem out as if is a quadratic polynomial. Thus 
where 
are all integers.
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. 
