2010 AMC 10B Problems/Problem 25
Problem
Let , and let be a polynomial with integer coefficients such that
, and
.
What is the smallest possible value of ?
Solution
We observe that because , if we define a new polynomial such that , has roots when ; namely, when .
Thus since has roots when , we can factor the product out of to obtain a new polynomial such that .
Then, plugging in values of we get
Thus, the least value of must be the . Solving, we receive , so our answer is .
Critique
The above solution is incomplete. What is really proven is that 315 is a factor of , if such an exists. That only rules out answer A.
To prove that the answer is correct, one could exhibit a polynomium that satisfies the requirements with . Here's one: .
You get that from the matrix and and computing which comes out as the all-integer coefficients above.
Critique of the Critique
Once you find that is a factor of 315, you can instantly find that the answer is 315 because the problem asks for the smallest value of .
See also
2010 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 24 |
Followed by Last question | |
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 | ||
All AMC 10 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.