1982 AHSME Problems/Problem 8

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We know that $n \choose {2}$ $-$ $n \choose 1$ $=$ ${n} \choose 3$ $-$ ${n} \choose 2,$ because they form an arithmetic sequence, and expanding, we have by the definitions in the problem: \[\frac{n(n-1)(n-2)(n-3)...}{2(n-2)(n-3)...}-n=\frac{n(n-1)(n-2)(n-3)(n-4)...}{6((n-3)(n-4)...)}-\frac{n(n-1)(n-2)(n-3)...}{2(n-2)(n-3)...}.\]

Canceling out the $(n-2)!$ and the $(n-3)!$ from each side of the equals sign, we have $\frac{n(n-1)}{2}-n = \frac {n(n-1)(n-2)}{6}-\frac{n(n-1)}{2}.$ Getting rid of the fractions by cross multiplication, and getting n on one side, we have $n^3 - 9n^2 + 14n = 0,$ 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 $x+y=-9$ and $xy=14.$ By guess and check, our integers are -7 and -2, so $n(n-7)(n-2)=0!!!$ According to the problem, $n>3,$ so we have n=7 or 2, thus $\boxed{\left(B\right)n=7}$

~ab2024

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