1976 AHSME Problems/Problem 23

Problem 23

For integers $k$ and $n$ such that $1\le k<n$ , let $C^n_k=\frac{n!}{k!(n-k)!}$ . Then $\left(\frac{n-2k-1}{k+1}\right)C^n_k$ is an integer

$\textbf{(A) }\text{for all }k\text{ and }n\qquad \\ \textbf{(B) }\text{for all even values of }k\text{ and }n,\text{ but not for all }k\text{ and }n\qquad \\ \textbf{(C) }\text{for all odd values of }k\text{ and }n,\text{ but not for all }k\text{ and }n\qquad \\ \textbf{(D) }\text{if }k=1\text{ or }n-1,\text{ but not for all odd values }k\text{ and }n\qquad \\ \textbf{(E) }\text{if }n\text{ is divisible by }k,\text{ but not for all even values }k\text{ and }n$

Solution

We know $C^n_k = \binom{n}{k}$ , so let's rewrite the expression as $\left(\frac{n-2k-1}{k+1}\right) \binom{n}{k}$ . Notice that $n-2k-1 = n-2(k+1)+1 = (n+1)-2(k+1).$

This allows us to rewrite the expression as $\left(\frac{(n+1)-2(k+1)}{k+1}\right) \binom{n}{k}.$

From here, we just have to do some algebra to get $\left(\frac{(n+1)-2(k+1)}{k+1}\right) \binom{n}{k} = \left( \frac{n+1}{k+1}-2 \right) \frac{n!}{k!(n-k)!}$ $= \frac{(n+1)!}{(k+1)!(n-k)!} - 2 \cdot \frac{n!}{k!(n-k)!}$ $= \binom{n+1}{k+1}-2\binom{n}{k},$ so $\left(\frac{n-2k-1}{k+1}\right)C^n_k$ is an integer $\boxed{\textbf{(A) }\text{for all }k\text{ and }n}$ . ~jiang147369

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1976 AHSME Problems/Problem 23

Problem 23

Solution