1969 AMC 12 Problem 23

Attempt Problem 23 of the 1969 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 1969 AMC 12 solutions, or check the answer key.

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23.

For any integer

n

greater than

1,

the number of prime numbers greater than

n!+1

and less than

n!+n

is:

Here

n!=1\cdot2\cdots(n-1)n.

0

1

\dfrac n2

for

n

even,

\dfrac{n+1}{2}

for

n

odd

n-1

n

Every integer strictly between the endpoints has the form

n!+k

with

2\le k\le n-1

Use the fact that

k

divides both

n!

and

n!+k

Every integer in the interval has the form

n!+k

for some

2\le k\le n-1.

Because

n!

is divisible by

k,

so is

n!+k,

and

n!+k\gt k.

Thus every such integer is composite. (For

n=2,

the interval is empty.) Hence there are no primes in the interval.

Therefore, the correct answer is A.

Problem 23 in other years