Difference between revisions of "2010 AMC 12B Problems/Problem 21"
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== Solution == | == Solution == | ||
− | + | We observe that because <math>P(1) = P(3) = P(5) = P(7) = a</math>, if we define a new polynomial <math>R(x)</math> such that <math>R(x) = P(x) - a</math>, <math>R(x)</math> has roots when <math>P(x) = a</math>; namely, when <math>x=1,3,5,7</math>. | |
+ | |||
+ | Thus since <math>R(x)</math> has roots when <math>x=1,3,5,7</math>, we can factor the product <math>(x-1)(x-3)(x-5)(x-7)</math> out of <math>R(x)</math> to obtain a new polynomial <math>Q(x)</math> such that <math>(x-1)(x-3)(x-5)(x-7)(Q(x)) = R(x) = P(x) - a</math>. | ||
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> | + | <math>-2a=-15Q(2)=9Q(4)=-15Q(6)=105Q(8).</math> |
− | + | Thus, the least value of <math>a</math> must be the <math>\text{lcm}(15,9,15,105)</math>. | |
− | <math> | + | Solving, we receive <math>315</math>, so our answer is <math> \boxed{\textbf{(B)}\ 315} </math>. |
− | <math> | + | To complete the solution, we can let <math>a = 315</math>, and then try to find <math>Q(x)</math>. We know from the above calculation that <math>Q(2)=42, Q(4)=-70, Q(6)=42</math>, and <math>Q(8)=-6</math>. Then we can let <math>Q(x) = T(x)(x-2)(x-6)+42</math>, getting <math>T(4)=28, T(8)=-4</math>. Let <math>T(x)=L(x)(x-8)-4</math>, then <math>L(4)=-8</math>. Therefore, it is possible to choose <math>T(x) = -8(x-8)-4 = -8x + 60</math>, so the goal is accomplished. As a reference, the polynomial we get is |
− | < | + | <cmath>P(x) = (x-1)(x-3)(x-5)(x-7)((-8x + 60)(x-2)(x-6)+42) + 315</cmath> |
− | + | <cmath> = -8 x^7+252 x^6-3248 x^5+22050 x^4-84392 x^3+179928 x^2-194592 x+80325 </cmath> | |
− | |||
== See also == | == See also == | ||
− | {{AMC12 box|year=2010|num-b=20|num-a=22 | + | {{AMC12 box|year=2010|ab=B|num-b=20|num-a=22}} |
+ | [[Category:Intermediate Algebra Problems]] | ||
+ | {{MAA Notice}} |
Revision as of 15:22, 17 June 2020
Problem 21
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 .
To complete the solution, we can let , and then try to find . We know from the above calculation that , and . Then we can let , getting . Let , then . Therefore, it is possible to choose , so the goal is accomplished. As a reference, the polynomial we get 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 |
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