1957 AMC 12 Problem 32

Question

The largest of the following integers which divides each of the numbers of the sequence

1^5-1,

2^5-2,

3^5-3,

\ldots

n^5-n,

\ldots

is:

1

60

15

120

30

Show

n^5-n

is always divisible by

2,

3,

and

5

Use consecutive factors for

2

and

3,

and residues modulo

5

for the remaining factor

Factor

\begin{aligned} n^5-n &=n(n-1)(n+1)\\ &\quad\cdot(n^2+1). \end{aligned}

Among three consecutive integers, one is divisible by

3

and at least one is even, so the expression is divisible by

6.

Also

n^5\equiv n\pmod5

for every integer

n,

so it is divisible by

5.

Hence every term is divisible by

30.

Taking

n=2

gives

2^5-2=30,

so no larger listed integer can divide every term.

Thus, the correct answer is E.

Problem 32 in other years