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Difference between revisions of "1982 AHSME Problems/Problem 8"

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== Solution ==
 
== Solution ==
  
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We know that <math>n \choose {2}</math> <math>-</math> <math> n \choose 1</math> <math>=</math> <math> {n} \choose 3</math> <math>-</math> <math>{n} \choose 2,</math> because they form an arithmetic sequence, and expanding, we have by the definitions in the problem: <cmath>\frac{n(n-1)(n-2)(n-3)...}{2(n-2)(n-3)...}-n=\frac{n(n-1)(n-2)(n-3)(n-4)...}{6((n-3)(n-4)...)}-\frac{n(n-1)(n-2)(n-3)...}{2(n-2)(n-3)...}.</cmath>
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Since <math>\binom{n}{1}</math>, <math>\binom{n}{2}</math>, and <math>\binom{n}{3}</math> form an arithmetic progression, <math>\binom{n}{2} - \binom{n}{1} = \binom{n}{3} - \binom{n}{2}</math>.  Therefore, <cmath>\frac{n!}{2!\cdot(n-2)!}-\frac{n!}{1!\cdot(n-1)!}=\frac{n!}{3!\cdot(n-3)!}-\frac{n!}{2!\cdot(n-2)!}.</cmath>
  
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Canceling out the <math>(n-2)!</math> and the <math>(n-3)!</math> from each side of the equals sign, we have <math>\frac{n(n-1)}{2}-n = \frac {n(n-1)(n-2)}{6}-\frac{n(n-1)}{2}.</math> Getting rid of the fractions by cross multiplication, and getting n on one side, we have <math>n^3 - 9n^2 + 14n = 0,</math> and we can factor out the n, so n(n^2-9n+14)=0, and we are looking for two integers x and y such that <math>x+y=-9</math> and <math>xy=14.</math> By guess and check, our integers are -7 and -2, so <math>n(n-7)(n-2)=0!!!</math> According to the problem, <math>n>3,</math> so we have n=7 or 2, thus <math>\boxed{\left(B\right)n=7}</math>
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Simplifying these expressions yields <math>\frac{n(n-1)}{2}-n = \frac {n(n-1)(n-2)}{6}-\frac{n(n-1)}{2}</math>. Multiplying both sides by <math>6</math> and collecting all terms on one side yields <math>n^3 - 9n^2 + 14n = 0</math>, which factors to <math>n(n-7)(n-2)=0</math>. The solutions to this equation are <math>n \in \{0, 2, 7\}</math>, but <math>n > 3</math> so the only valid answer is <math>n = \boxed{(\mathbf{B})\ 7}</math>.
  
 
~ab2024
 
~ab2024

Revision as of 22:38, 29 June 2025

Problem

By definition, $r! = r(r - 1) \cdots 1$ and $\binom{j}{k} = \frac {j!}{k!(j - k)!}$, where $r,j,k$ are positive integers and $k < j$. If $\binom{n}{1}, \binom{n}{2}, \binom{n}{3}$ form an arithmetic progression with $n > 3$, then $n$ equals

$\textbf{(A)}\ 5\qquad \textbf{(B)}\ 7\qquad \textbf{(C)}\ 9\qquad \textbf{(D)}\ 11\qquad \textbf{(E)}\ 12$

Solution

Since $\binom{n}{1}$, $\binom{n}{2}$, and $\binom{n}{3}$ form an arithmetic progression, $\binom{n}{2} - \binom{n}{1} = \binom{n}{3} - \binom{n}{2}$. Therefore, \[\frac{n!}{2!\cdot(n-2)!}-\frac{n!}{1!\cdot(n-1)!}=\frac{n!}{3!\cdot(n-3)!}-\frac{n!}{2!\cdot(n-2)!}.\]

Simplifying these expressions yields $\frac{n(n-1)}{2}-n = \frac {n(n-1)(n-2)}{6}-\frac{n(n-1)}{2}$. Multiplying both sides by $6$ and collecting all terms on one side yields $n^3 - 9n^2 + 14n = 0$, which factors to $n(n-7)(n-2)=0$. The solutions to this equation are $n \in \{0, 2, 7\}$, but $n > 3$ so the only valid answer is $n = \boxed{(\mathbf{B})\ 7}$.

~ab2024

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