# 1976 AHSME Problems/Problem 23

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## Problem 23

For integers $k$ and $n$ such that $1\le k, 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