2015 AMC 10B Problem 23

Question

Let

n

be a positive integer greater than

4

such that the decimal representation of

n!

ends in

k

zeros and the decimal representation of

(2n)!

ends in

3k

zeros. Let

s

denote the sum of the four least possible values of

n.

What is the sum of the digits of

s?

7

8

9

10

11

Answer: B

Concepts:trailing zeros Legendre’s Formula factorial

Difficulty rating: 1790

Small hint:

Count factors of

5

in

n!

and

(2n)!

Big hint:

Test the first ranges where

k=1

and

k=2

Solution:

The number of trailing zeros is the number of factors of

5

. For

5\le n\le9

,

n!

has

k=1

zero. We need

(2n)!

to have

3

zeros, which happens when

15\le2n\le19

. Thus

n=8,9

.

For

10\le n\le14

,

n!

has

k=2

zeros. We need

(2n)!

to have

6

zeros, which happens when

25\le2n\le29

. Thus

n=13,14

.

These are the four least possible values, so

s=8+9+13+14=44

. The sum of the digits of

s

is

8

.

Thus, the correct answer is B.