1961 AHSME Problems/Problem 35

Problem

The number $695$ is to be written with a factorial base of numeration, that is, $695=a_1+a_2\times2!+a_3\times3!+ \ldots a_n \times n!$ where $a_1, a_2, a_3 ... a_n$ are integers such that $0 \le a_k \le k,$ and $n!$ means $n(n-1)(n-2)...2 \times 1$. Find $a_4$

$\textbf{(A)}\ 0\qquad \textbf{(B)}\ 1\qquad \textbf{(C)}\ 2\qquad \textbf{(D)}\ 3\qquad \textbf{(E)}\ 4$

Solution

This problem can be approached similarly to other base number problems.

Since $120 < 695 < 720$, divide $695$ by $120$. The quotient is $5$ and the remainder is $95$, so rewrite the number as \[695 = 5 \cdot 24 + 95\] Similarly, dividing $95$ by $24$ results in a quotient of $3$ and a remainder of $23$, so the number can be rewritten as \[695 = 5 \cdot 24 + 3 \cdot 24 + 23\] Repeat the steps to get \[695 = 5 \cdot 24 + 3 \cdot 24 + 3 \cdot 6 + 2 \cdot 2 + 1\] The answer is $\boxed{\textbf{(D)}}$. One can also stop at the second step by noting $23 < 24$.

See Also

1961 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 34
Followed by
Problem 36
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