Difference between revisions of "2010 AMC 10B Problems/Problem 25"
m (→Solution) |
(→Solution) |
||
Line 24: | Line 24: | ||
Thus, the least value of <math>a</math> must be the <math>lcm(15,9,15,105)</math>. | Thus, the least value of <math>a</math> must be the <math>lcm(15,9,15,105)</math>. | ||
Solving, we receive <math>315</math>, so our answer is <math> \boxed{\textbf{(B)}\ 315} </math>. | Solving, we receive <math>315</math>, so our answer is <math> \boxed{\textbf{(B)}\ 315} </math>. | ||
+ | <math>\blacksquare</math> | ||
== See also == | == See also == |
Revision as of 16:07, 18 February 2013
Problem
Let , and let be a polynomial with integer coefficients such that
, and
.
What is the smallest possible value of ?
Solution
There must be some 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 .
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 |