1970 AMC 12 Problem 23

Attempt Problem 23 of the 1970 AMC 12 below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 1970 AMC 12 solutions, or check the answer key.

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23.

The number

10!

(

10

is written in base

10

), when written in the base

12

system, ends in exactly

k

zeros. The value of

k

is:

1

2

3

4

5

A trailing base-

12

zero contributes one factor of

12=2^2\cdot3

Compare the exponents of

2

and

3

10!

The prime exponents in

10!

are

\begin{gathered} v_2(10!)=5+2+1=8,\\ v_3(10!)=3+1=4. \end{gathered}

Each factor of

12=2^2\cdot3

uses two factors of

2

and one of

3.

Hence the number of trailing base-

12

zeros is

\min\left(\left\lfloor\frac82\right\rfloor,4\right)=4.

Therefore, the correct answer is D.

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