# 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 1(Modular Functions)

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 $9 \Longrightarrow \boxed{\mathrm{(B)}}$.

## Solution 2(Basic Algebra and Bashing)

We start by multiplying both sides by $7!$, and we get: $$3600=2520a_2+840a_3+210a_4+42a_5+7a_6+a_7$$ After doing some guess and check, we find that the answer is $\boxed{\textbf{(B)~9}}$.

~aopspandy

## Solution 3 (The Easiest and Most Intuitive)

Let's clear up the fractions: $$\frac{5}{7}=\frac{2520a_2+840a_3+210a_4+42a_5+7a_6+a_7}{7!}$$ $$3600=2520a_2+840a_3+210a_4+42a_5+7a_6+a_7$$ Notice that if we divide everything by $7$ then we would have:$$514+\frac{2}{7}=360a_2+120a_3+30a_4+6a_5+a_6+\frac{1}{7}a_7$$ Since $0 \le a_7<7$ and $a_7$ must be an integer, then we have $\frac{2}{7}=\frac{1}{7}a_7$, so $a_7=2$.

Similarly, if we divide everything by $6$, then we would have: $$85+\frac{4}{6}=60a_2+20a_3+5a_4+a_5+\frac{1}{6}a_6$$ Again, since $0 \le a_6<6$ and $a_6$ must be an integer, we have $\frac{4}{6}=\frac{1}{6}a_6$, so $a_6=4$.

The pattern repeats itself, so in the end we have $a_2=1$, $a_3=1$, $a_4=1$, $a_5=0$, $a_6=4$, $a_7=2$. So $a_2+a_3+a_4+a_5+a_6+a_7=\boxed{\textbf{(B)~9}}$

~BurpSuite, with a help from ostriches88

## Solution 4

By multiplying both sides by $7$ we get

$$5 = \frac72 a_2 + \frac76 a_3+ \frac{7}{24} a_4 + \frac{7}{120} a_5 + \frac{7}{720} a_6 + \frac{a_7}{720}$$

since $0 \le a_2 < 2$, if $a_2 = 0$ the rest of the right hand side will not add up to be $5$, so $a_2 = 1$

$$\frac32 = \frac76 a_3+ \frac{7}{24} a_4 + \frac{7}{120} a_5 + \frac{7}{720} a_6 + \frac{a_7}{720}$$

If $a_3 = 2$, $\frac76 a_3 = \frac73 > \frac32$, so $a_3 = 1$

$$\frac13 = \frac{7}{24} a_4 + \frac{7}{120} a_5 + \frac{7}{720} a_6 + \frac{a_7}{720}$$

If $a_4 = 2$, $\frac{7}{24} a_4 = \frac{7}{12} > \frac13$, so $a_4 = 1$

$$\frac{1}{24} = \frac{7}{120} a_5 + \frac{7}{720} a_6 + \frac{a_7}{720}$$

If $a_5 = 1$, $\frac{7}{120} a_5 = \frac{7}{120} > \frac{1}{24}$, so $a_5 = 0$

$$\frac{1}{24} = \frac{7}{720} a_6 + \frac{a_7}{720}$$

$30 = 7 a_6 + a_7$. Since $a_7 < 7$, $a_7 = 2$ and $a_6=4$

Therefore, $a_{2} + a_{3} + a_{4} + a_{5} + a_{6} + a_{7} = 1 + 1 + 1 + 0 + 4 + 2= \boxed{\textbf{(B) } 9}$.
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