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Difference between revisions of "1989 AHSME Problems/Problem 27"
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<math>\mathrm{(A)}\ 14 \text{ or } 15\ \mathrm{(B)}\ 15 \text{ or } 16\ \mathrm{(C)}\ 16 \text{ or } 17\ \mathrm{(D)}\ 17 \text{ or } 18\ \mathrm{(E)}\ 18 \text{ or } 19</math>
 
<math>\mathrm{(A)}\ 14 \text{ or } 15\ \mathrm{(B)}\ 15 \text{ or } 16\ \mathrm{(C)}\ 16 \text{ or } 17\ \mathrm{(D)}\ 17 \text{ or } 18\ \mathrm{(E)}\ 18 \text{ or } 19</math>
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==Solution==
 
==Solution==
 
This is equivalent to finding the powers of <math>k</math> with coefficient <math>28</math> in the expansion of <math>(k^2+k^4+k^6+k^8+...)^2(k+k^2+k^3+k^4+...)</math>.
 
This is equivalent to finding the powers of <math>k</math> with coefficient <math>28</math> in the expansion of <math>(k^2+k^4+k^6+k^8+...)^2(k+k^2+k^3+k^4+...)</math>.
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<math>=k^5(1+k)(\tbinom{2}{2}+\tbinom{3}{2}k^2+\tbinom{4}{2}k^4+\tbinom{5}{2}k^6+...)</math>, the last part having general term <math>\tbinom{j}{2}k^{2j-4}</math>
 
<math>=k^5(1+k)(\tbinom{2}{2}+\tbinom{3}{2}k^2+\tbinom{4}{2}k^4+\tbinom{5}{2}k^6+...)</math>, the last part having general term <math>\tbinom{j}{2}k^{2j-4}</math>
 
from which it is easy to see that, since <math>\tbinom{8}{2}=28</math>, the last part contains the term <math>28k^{12}</math> and the whole result <math>28k^{17}+28k^{18}</math>. So the answer is <math>\mathrm{(D)}</math>.
 
from which it is easy to see that, since <math>\tbinom{8}{2}=28</math>, the last part contains the term <math>28k^{12}</math> and the whole result <math>28k^{17}+28k^{18}</math>. So the answer is <math>\mathrm{(D)}</math>.
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== See also ==
 +
{{AHSME box|year=1989|num-b=26|num-a=28}} 
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[[Category: Intermediate Algebra Problems]]
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 07:09, 22 October 2014

Problem

Let $n$ be a positive integer. If the equation $2x+2y+z=n$ has 28 solutions in positive integers $x$, $y$, and $z$, then $n$ must be either

$\mathrm{(A)}\ 14 \text{ or } 15\ \mathrm{(B)}\ 15 \text{ or } 16\ \mathrm{(C)}\ 16 \text{ or } 17\ \mathrm{(D)}\ 17 \text{ or } 18\ \mathrm{(E)}\ 18 \text{ or } 19$

Solution

This is equivalent to finding the powers of $k$ with coefficient $28$ in the expansion of $(k^2+k^4+k^6+k^8+...)^2(k+k^2+k^3+k^4+...)$.

But this is $\left(\frac{k^2}{1-k^2}\right)^2\cdot\frac{k}{1-k}\ =\ k^5\cdot\left(\frac{1}{1-k}\right)^3\cdot\left(\frac{1}{1+k}\right)^2\ =\ k^5\cdot(1+k)\cdot\left(\frac{1}{1-k^2}\right)^3$

$=k^5(1+k)(\tbinom{2}{2}+\tbinom{3}{2}k^2+\tbinom{4}{2}k^4+\tbinom{5}{2}k^6+...)$, the last part having general term $\tbinom{j}{2}k^{2j-4}$ from which it is easy to see that, since $\tbinom{8}{2}=28$, the last part contains the term $28k^{12}$ and the whole result $28k^{17}+28k^{18}$. So the answer is $\mathrm{(D)}$.

See also

1989 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 26 		Followed by
Problem 28
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 • 26 • 27 • 28 • 29 • 30
All AHSME Problems and Solutions

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