Difference between revisions of "2010 AMC 10B Problems/Problem 25"
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Then, plugging in values of <math>2,4,6,8,</math> we get | Then, plugging in values of <math>2,4,6,8,</math> we get | ||
− | < | + | <cmath>P(2)-a=(2-1)(2-3)(2-5)(2-7)Q(2) = -15Q(2) = -2a</cmath> |
− | < | + | <cmath>P(4)-a=(4-1)(4-3)(4-5)(4-7)Q(4) = 9Q(4) = -2a</cmath> |
− | < | + | <cmath>P(6)-a=(6-1)(6-3)(6-5)(6-7)Q(6) = -15Q(6) = -2a</cmath> |
− | < | + | <cmath>P(8)-a=(8-1)(8-3)(8-5)(8-7)Q(8) = 105Q(8) = -2a</cmath> |
<math>-2a=-15Q(2)=9Q(4)=-15Q(6)=105Q(8).</math> | <math>-2a=-15Q(2)=9Q(4)=-15Q(6)=105Q(8).</math> |
Revision as of 16:38, 12 June 2011
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 |