For integers k and n such that 1≤ k< n, let nk=(n!)/(k!(n-k)!). Then ((n-2k-1)/(k+1)) nk is an integer

(B) for all even values of k and n, but not for all k and n

(C) for all odd values of k and n, but not for all k and n

(D) if k=1 or n-1, but not for all odd values of k and n

1976 AMC 12 Problem 23

Attempt Problem 23 of the 1976 AMC 12 below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 1976 AMC 12 solutions, or check the answer key.

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23.

For integers

k

and

n

such that

1\le k\lt n,

let

\binom nk=\frac{n!}{k!(n-k)!}.

Then

\left(\frac{n-2k-1}{k+1}\right)\binom nk

is an integer

for all

k

and

n

for all even values of

k

and

n,

but not for all

k

and

n

for all odd values of

k

and

n,

but not for all

k

and

n

k=1

n-1,

but not for all odd values of

k

and

n

n

is divisible by

k,

but not for all even values of

k

and

n

Answer: A

Concepts:combinations algebraic manipulation

Difficulty rating: 2080

Small hint:

Split

n-2k-1

into

(n-k)-(k+1)

Big hint:

Try rewriting the expression as a difference of two adjacent binomial coefficients

Solution:

Using

\binom n{k+1}=\frac{n-k}{k+1}\binom nk,

and writing

\frac{n-2k-1}{k+1}=\frac{n-k}{k+1}-1,

the given expression becomes

\binom n{k+1}-\binom nk.

This is an integer for every permitted

k

and

n.

Therefore, the correct answer is A.