1999 AMC 12 Problem 25

Attempt Problem 25 of the 1999 AMC 12 below, then check your answer against the professionally curated solution 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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25.

There are unique integers

a_2,

a_3,

a_4,

a_5,

a_6,

a_7

such that

\begin{aligned} &\frac{5}{7} = \frac{a_2}{2!} + \frac{a_3}{3!} + \frac{a_4}{4!} \\ &\quad {}+ \frac{a_5}{5!} + \frac{a_6}{6!} + \frac{a_7}{7!}, \end{aligned}

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

Answer: B

Concepts:number base factorial modular arithmetic

Difficulty rating: 2030

Small hint:

Multiply both sides by

7!

Big hint:

Reduce modulo

7,

then

6,

then

5, \ldots

to peel off one digit at a time

Solution:

Multiplying by

7! = 5040

gives

\begin{aligned} &3600 = 2520a_2 + 840a_3 \\ &\quad {}+ 210a_4 + 42a_5 \\ &\quad {}+ 7a_6 + a_7. \end{aligned}

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.

Problem 25 in other years