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

1999 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 24	Followed by Problem 26
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1999 AHSME Problems/Problem 25

Contents

Problem

Solution 1(Modular Functions)

Solution 2(Basic Algebra and Bashing)

See also