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

$1$

$3$

$5$

$7$

$9$

Answer: E

Concepts:prime valuations trailing digits modular arithmetic

Difficulty rating: 2630

Small Hint:

Write $a_n=n(n+1)\cdots(n+9)$ and compare its powers of $2$ and $5$

Big Hint:

An odd rightmost nonzero digit first becomes possible when the ten-term block contains $5^7$

Solution:

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.

1998 AMC 12 Problem 30

Answer: E

Difficulty rating: 2630

Small Hint:

Big Hint:

Solution:

Problem 30 in Other Years