# 1999 AHSME Problems/Problem 25

## Problem

There are unique integers $a_{2},a_{3},a_{4},a_{5},a_{6},a_{7}$ such that $$\frac {5}{7} = \frac {a_{2}}{2!} + \frac {a_{3}}{3!} + \frac {a_{4}}{4!} + \frac {a_{5}}{5!} + \frac {a_{6}}{6!} + \frac {a_{7}}{7!}$$

where $0\leq a_{i} < i$ for $i = 2,3,\ldots,7$. Find $a_{2} + a_{3} + a_{4} + a_{5} + a_{6} + a_{7}$. $\textrm{(A)} \ 8 \qquad \textrm{(B)} \ 9 \qquad \textrm{(C)} \ 10 \qquad \textrm{(D)} \ 11 \qquad \textrm{(E)} \ 12$

## Solution

Multiply out the $7!$ to get $$5 \cdot 6! = (3 \cdot 4 \cdots 7)a_2 + (4 \cdots 7)a_3 + (5 \cdot 6 \cdot 7)a_4 + 42a_5 + 7a_6 + a_7 .$$

By Wilson's Theorem (or by straightforward division), $a_7 + 7(a_6 + 6a_5 + \cdots) \equiv 5 \cdot 6! \equiv -5 \equiv 2 \pmod{7}$, so $a_7 = 2$. Then we move $a_7$ to the left and divide through by $7$ to obtain $$\frac{5 \cdot 6!-2}{7} = 514 = 360a_2 + 120a_3 + 30a_4 + 6a_5 + a_6.$$

We then repeat this procedure $\pmod{6}$, from which it follows that $a_6 \equiv 514 \equiv 4 \pmod{6}$, and so forth. Continuing, we find the unique solution to be $(a_2, a_3, a_4, a_5, a_6, a_7) = (1,1,1,0,4,2)$ (uniqueness is assured by the Division Theorem). The answer is $9 \Longrightarrow \boxed{\mathrm{(B)}}$.

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 