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Difference between revisions of "1989 AHSME Problems/Problem 27"

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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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==Solution 2==
 +
Suppose <math>n</math> is even. Thus <math>z</math> is even, so let <math>z=2t</math>. Hence <math>x+y+t=\frac{n}{2}</math>. By stars and bars, this has
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<cmath>\binom{\frac{n}{2}-1}{3-1}</cmath>
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solutions. This implies that <math>\frac{n}{2}-1=8\to n=18</math>. If <math>z</math> is odd, let <math>z=2t-1</math>. Then <math>x+y+t=\frac{n+1}{2}</math>, in which <math>n=17</math> is a solution. The answer is then <math>\mathrm{(D)}</math>.
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~yofro
  
 
== See also ==
 
== See also ==

Latest revision as of 11:39, 5 July 2021

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 $\mathrm{(D)}$.


Solution 2

Suppose $n$ is even. Thus $z$ is even, so let $z=2t$. Hence $x+y+t=\frac{n}{2}$. By stars and bars, this has \[\binom{\frac{n}{2}-1}{3-1}\] solutions. This implies that $\frac{n}{2}-1=8\to n=18$. If $z$ is odd, let $z=2t-1$. Then $x+y+t=\frac{n+1}{2}$, in which $n=17$ is a solution. The answer is then $\mathrm{(D)}$.

~yofro

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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