Difference between revisions of "2004 AIME I Problems/Problem 7"
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=== Solution 3 (Bash)=== | === Solution 3 (Bash)=== | ||
− | Consider the set <math>[-1, 2,-3,4,-5,6,-7,8,-9,10,-11,12,-13,14,-15]</math> | + | Consider the set <math>[-1, 2,-3,4,-5,6,-7,8,-9,10,-11,12,-13,14,-15]</math>. Denote by <math>S</math> all size 2 subsets of this set. Replace each element of <math>S</math> by the product of the elements. Now, the quantity we seek is the sum of each element. Since consecutive elements add to <math>1</math> or <math>-1</math>, we can simplify this to <math>|-1\cdot(-7)+2\cdot(-9)-3\cdot(-6)+4\cdot(-10)-5\cdot(-5)+\ldots+12\cdot(-14)-13\cdot(-1)+14\cdot(-15)|=|-588|=\boxed{588}</math>. |
== See also == | == See also == |
Revision as of 14:32, 24 August 2017
Problem
Let be the coefficient of in the expansion of the product Find
Solution
Solution 1
Let our polynomial be .
It is clear that the coefficient of in is , so , where is some polynomial divisible by .
Then and so , where is some polynomial divisible by .
However, we also know .
Equating coefficients, we have , so and .
Solution 2
Let be the set of integers . The coefficient of in the expansion is equal to the sum of the product of each pair of distinct terms, or . Also, we know that where the left-hand sum can be computed from:
and the right-hand sum comes from the formula for the sum of the first perfect squares. Therefore, .
Solution 3 (Bash)
Consider the set . Denote by all size 2 subsets of this set. Replace each element of by the product of the elements. Now, the quantity we seek is the sum of each element. Since consecutive elements add to or , we can simplify this to .
See also
2004 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 6 |
Followed by Problem 8 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
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