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:

$1$

$2$

$3$

$4$

$5$

Answer: D

Concepts:number base trailing zeros prime factorization factorial

Difficulty rating: 1830

Small Hint:

A trailing base- $12$ zero contributes one factor of $12=2^2\cdot3$

Big Hint:

Compare the exponents of $2$ and $3$ in $10!$

Solution:

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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