2014 AMC 10A Problem 20

Question

The product

(8)(888\dots8),

where the second factor has

k

digits, is an integer whose digits have a sum of

1000.

What is

k?

901

911

919

991

999

Multiply a few examples to see the digit pattern

For

k\ge3

, the product has

k-2

digits equal to

1

The

k

-digit number made entirely of

8

s is

8\frac{10^k-1}{9}

, so the product is

\begin{aligned} 64\frac{10^k-1}{9} &=7\cdot10^k\\ &\quad+100\frac{10^{k-2}-1}{9}\\ &\quad+4. \end{aligned}

For

k\ge2

, this is the number whose digits are

7

, followed by

k-2

ones, then

0,4

.

This means that for any

k \geq 3,

the sum of the digits in the product is

7 + 4 + 0 + k - 2 = k + 9.

Finally, we get

k + 9 = 1000

k = 991.

Thus, D is the correct answer.

Problem 20 in other years