Question

If the number $15!,$ that is, $15\cdot14\cdot13\cdots1,$ ends with $k$ zeros when given to the base $12$ and ends with $h$ zeros when given to the base $10,$ then $k+h$ equals:

$5$

$6$

$7$

$8$

$9$

Answer: D

Concepts:factorial trailing zeros prime factorization number base

Difficulty rating: 2000

Small Hint:

Count the powers of $2,$ $3,$ and $5$ in $15!$

Big Hint:

A base- $12$ zero uses $2^2\cdot3$ , while a base- $10$ zero uses $2\cdot5$

Solution:

The prime valuations are $\begin{gathered} v_2(15!)=11,\\ v_3(15!)=6,\\ v_5(15!)=3. \end{gathered}$ Hence $k=\min(\lfloor\frac{11}{2}\rfloor,6)=5$ in base $12,$ and $h=\min(11,3)=3$ in base $10.$ Thus $k+h=8.$

Therefore, the correct answer is D.

1965 AMC 12 Problem 33

Answer: D

Difficulty rating: 2000

Small Hint:

Big Hint:

Solution:

Problem 33 in Other Years