There are unique integers a_2, a_3, a_4, a_5, a_6, a_7 such that (5)/(7) = (a_2)/(2!) + (a_3)/(3!) + (a_4)/(4!) + (a_5)/(5!) + (a_6)/(6!) + (a_7)/(7!), where 0 ≤ a_i \lt i for i = 2, 3, …, 7. Find a_2 + a_3 + a_4 + a_5 + a_6 + a_7.

1999 AMC 12 Problem 25 — Solution

1999 AMC 12 Problem 25

Below is the professionally curated solution for Problem 25 of the 1999 AMC 12, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 1999 AMC 12 solutions, or check the answer key.

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Concepts:number base factorial modular arithmetic

Difficulty rating: 2030

25.

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 $0 \le a_i \lt i$ for $i = 2, 3, \ldots, 7.$ Find $a_2 + a_3 + a_4 + a_5 + a_6 + a_7.$

$8$

$9$

$10$

$11$

$12$

Solution:

Multiplying by $7! = 5040$ gives $3600 = 2520a_2 + 840a_3 + 210a_4 + 42a_5 + 7a_6 + a_7.$ Reducing modulo $7,$ $a_7 = 2.$

Then $\dfrac{3600 - 2}{7} = 514 = 360a_2 + 120a_3 + 30a_4 + 6a_5 + a_6.$ Reducing modulo $6$ gives $a_6 = 4,$ and continuing this way yields $a_5 = 0, a_4 = 1, a_3 = 1, a_2 = 1.$

The sum is $1 + 1 + 1 + 0 + 4 + 2 = 9.$

Thus, the correct answer is B.

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