1999 AMC 12 Problema 25

Question

Existen enteros únicos

a_2,

a_3,

a_4,

a_5,

a_6,

a_7

tales que

\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}

donde

0 \le a_i \lt i

para

i = 2,

3,

\ldots,

7.

Halla

a_2 + a_3 + a_4 + a_5 + a_6 + a_7.

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

Respuesta: B

Conceptos:base numérica factorial aritmética modular

Nivel de dificultad: 2030

Pista pequeña:

Multiplica ambos lados por

7!

Multiply both sides by

7!

Pista grande:

Reduce módulo

7,

luego

6,

luego

5, \ldots

para ir extrayendo una cifra a la vez.

Reduce modulo

7,

then

6,

then

5, \ldots

to peel off one digit at a time

Solución:

Multiplicando por

7! = 5040

se obtiene

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

Reduciendo módulo

7,

a_7 = 2.

Luego

\dfrac{3600 - 2}{7} = 514

= 360a_2 + 120a_3

+ 30a_4 + 6a_5

+ a_6.

Reduciendo módulo

6

se obtiene

a_6 = 4,

y continuando de esta manera se llega a

a_5 = 0, a_4 = 1, a_3 = 1, a_2 = 1.

La suma es

1 + 1 + 1 + 0 + 4 + 2 = 9.

Por lo tanto, la respuesta correcta es B.

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.