Difference between revisions of "2016 AMC 10A Problems/Problem 20"
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All the desired terms are in the form <math>a^xb^yc^zd^w1^t</math>, where <math>x + y + z + w + t = N</math> (the <math>1^t</math> part is necessary to make stars and bars work better.) | All the desired terms are in the form <math>a^xb^yc^zd^w1^t</math>, where <math>x + y + z + w + t = N</math> (the <math>1^t</math> part is necessary to make stars and bars work better.) | ||
Since <math>x</math>, <math>y</math>, <math>z</math>, and <math>w</math> must be at least <math>1</math> (<math>t</math> can be 0), let <math>x' = x - 1</math>, <math>y' = y - 1</math>, <math>z' = z - 1</math>, and <math>w' = w - 1</math>, so <math>x' + y' + z' + w' + t = N - 4</math>. Now, we use stars and bars to see that there are <math>\binom{N}{4}</math> solutions to this equation. We have <math>\binom{14}{4} = 1001</math>, so <math>N = \boxed{14}</math>. | Since <math>x</math>, <math>y</math>, <math>z</math>, and <math>w</math> must be at least <math>1</math> (<math>t</math> can be 0), let <math>x' = x - 1</math>, <math>y' = y - 1</math>, <math>z' = z - 1</math>, and <math>w' = w - 1</math>, so <math>x' + y' + z' + w' + t = N - 4</math>. Now, we use stars and bars to see that there are <math>\binom{N}{4}</math> solutions to this equation. We have <math>\binom{14}{4} = 1001</math>, so <math>N = \boxed{14}</math>. | ||
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+ | ==Another solution== | ||
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+ | The number of terms that have all <math>a,b,c,d</math> raised to a positive power is <math>\binom{N-1}{3}+\binom{N-2}{3}+\cdots + \binom{4}{3}+\binom{3}{3}=\binom{N}{4}</math>. For <math>N=\boxed{14}</math>, <math>\binom{N}{4}=1001</math>. | ||
==See Also== | ==See Also== | ||
{{AMC10 box|year=2016|ab=A|num-b=19|num-a=21}} | {{AMC10 box|year=2016|ab=A|num-b=19|num-a=21}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 01:50, 2 February 2017
For some particular value of , when is expanded and like terms are combined, the resulting expression contains exactly terms that include all four variables and , each to some positive power. What is ?
Solution
All the desired terms are in the form , where (the part is necessary to make stars and bars work better.) Since , , , and must be at least ( can be 0), let , , , and , so . Now, we use stars and bars to see that there are solutions to this equation. We have , so .
Another solution
The number of terms that have all raised to a positive power is . For , .
See Also
2016 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 19 |
Followed by Problem 21 | |
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 | ||
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