Difference between revisions of "2010 AMC 10B Problems/Problem 25"
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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 15:07, 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
![\[P(2)-a=(2-1)(2-3)(2-5)(2-7)Q(2) = -15Q(2) = -2a\]](http://latex.artofproblemsolving.com/9/a/f/9afd67021f0f85e87b25f156824b541f88c0306b.png)
![\[P(4)-a=(4-1)(4-3)(4-5)(4-7)Q(4) = 9Q(4) = -2a\]](http://latex.artofproblemsolving.com/4/1/c/41c5b8fac26a68834f50593977429f4698646e2e.png)
![\[P(6)-a=(6-1)(6-3)(6-5)(6-7)Q(6) = -15Q(6) = -2a\]](http://latex.artofproblemsolving.com/d/8/a/d8a09c7042d1dab945a5506a13a985074be1c9e4.png)
![\[P(8)-a=(8-1)(8-3)(8-5)(8-7)Q(8) = 105Q(8) = -2a\]](http://latex.artofproblemsolving.com/8/8/a/88a51f757dc2a65f7aa72e90e745ed6eb67f8ada.png)
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 | ||
