1961 AMC 12 Problem 35

Question

The number

695

is to be written with a factorial base of numeration, that is,

\begin{aligned} 695={}&a_1+a_2\cdot2!+a_3\cdot3!\\ &+\cdots+a_n\cdot n!, \end{aligned}

where

a_1,

a_2,

a_3,

\ldots,

a_n

are integers such that

0\le a_k\le k,

and

n!

means

n(n-1)(n-2)\cdots2\cdot1.

Find

a_4.

0

1

2

3

4

Begin with the largest factorial not exceeding

695

After removing the

5!

contribution, divide the remainder by

4!

Since

5!=120,

695=5\cdot120+95.

4!=24,

and

95=3\cdot24+23.

Therefore the coefficient

a_4

is

3.

Thus, the correct answer is D.

Problem 35 in other years