1998 AMC 12 Problem 30

Question

For each positive integer

n,

let

a_n=\frac{(n+9)!}{(n-1)!}.

Let

k

denote the smallest positive integer for which the rightmost nonzero digit of

a_k

is odd. The rightmost nonzero digit of

a_k

is

1

3

5

7

9

Write

a_n=n(n+1)\cdots(n+9)

and compare its powers of

2

and

5

An odd rightmost nonzero digit first becomes possible when the ten-term block contains

5^7

The five even terms in any ten consecutive integers contribute at least

2^8.

Thus the rightmost nonzero digit can be odd only when the block contains at least eight factors of

5,

first possible when it contains

5^7=78125.

For

n=5^7-9,

the block has

v_2=9

and

v_5=8,

so the digit remains even. For

n=5^7-8=78117,

both valuations are

8.

Cancelling

2^8 5^8

and multiplying the remaining odd unit digits gives

\begin{aligned} &7\cdot9\cdot1\cdot3\cdot9\cdot3\\ &\qquad\cdot1\cdot1\cdot3 \equiv9\pmod{10}. \end{aligned}

Hence the first odd rightmost nonzero digit is

9,

and E is correct.

Problem 30 in other years