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:

$1$

$60$

$15$

$120$

$30$

Answer: E

Concepts:divisibility modular arithmetic Fermat’s Little Theorem

Difficulty rating: 1790

Small Hint:

Show $n^5-n$ is always divisible by $2,$ $3,$ and $5$

Big Hint:

Use consecutive factors for $2$ and $3,$ and residues modulo $5$ for the remaining factor

Solution:

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.

1957 AMC 12 Problem 32

Answer: E

Difficulty rating: 1790

Small Hint:

Big Hint:

Solution:

Problem 32 in Other Years