# Difference between revisions of "2014 AMC 10A Problems/Problem 20"

The following problem is from both the 2014 AMC 12A #16 and 2014 AMC 10A #20, so both problems redirect to this page.

## Problem

The product $(8)(888\dots8)$, where the second factor has $k$ digits, is an integer whose digits have a sum of $1000$. What is $k$? $\textbf{(A)}\ 901\qquad\textbf{(B)}\ 911\qquad\textbf{(C)}\ 919\qquad\textbf{(D)}\ 991\qquad\textbf{(E)}\ 999$

## Solution

We can list the first few numbers in the form $8*(8....8)$

(Hard problem to do without the multiplication, but you can see the pattern early on) $8*8 = 64$ $8*88 = 704$ $8*888 = 7104$ $8*8888 = 71104$ $8*88888 = 711104$

By now it's clear that the numbers will be in the form $7$, $k-2$ $1$s, and $04$. We want to make the numbers sum to 1000, so $7+4+(k-2) = 1000$. Solving, we get $k = 991$, meaning the answer is $\fbox{(D)}$

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 