Difference between revisions of "2010 AMC 12B Problems/Problem 21"
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== Solution == | == Solution == | ||
| + | There must be some polynomial <math>Q(x)</math> such that <math>P(x)-a=(x-1)(x-3)(x-5)(x-7)Q(x)</math> | ||
| + | |||
| + | Then, plugging in values of <math>2,4,6,8,</math> we get | ||
| + | |||
| + | <math>P(2)-a=(2-1)(2-3)(2-5)(2-7)Q(2) = -15Q(2) = -2a</math> | ||
| + | <math>P(4)-a=(4-1)(4-3)(4-5)(4-7)Q(4) = 9Q(4) = -2a</math> | ||
| + | <math>P(6)-a=(6-1)(6-3)(6-5)(6-7)Q(6) = -15Q(6) = -2a</math> | ||
| + | <math>P(8)-a=(8-1)(8-3)(8-5)(8-7)Q(8) = 105Q(8) = -2a</math> | ||
| + | |||
| + | <math>-2a=-15Q(2)=9Q(4)=-15Q(6)=105Q(8).</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>. | ||
== See also == | == See also == | ||
{{AMC12 box|year=2010|num-b=20|num-a=22|ab=B}} | {{AMC12 box|year=2010|num-b=20|num-a=22|ab=B}} | ||
Revision as of 22:20, 7 February 2011
Problem 21
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 12B (Problems • Answer Key • Resources) | |
| Preceded by Problem 20 |
Followed by Problem 22 |
| 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 12 Problems and Solutions | |
