Difference between revisions of "2010 AMC 10B Problems/Problem 25"
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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 | 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>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> = -8 x^7+252 x^6-3248 x^5+22050 x^4-84392 x^3+179928 x^2-194592 x+80325 $ |
== See also == | == See also == |
Revision as of 13:13, 12 February 2017
Problem
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
= -8 x^7+252 x^6-3248 x^5+22050 x^4-84392 x^3+179928 x^2-194592 x+80325 $
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 |
See also The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.