# 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 $8 \cdot (8....8)$

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

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 $\fbox{(D)}$

Another way to proceed is that we know the difference between the sum of the digits of each product and $k$ is always $9$, so we just do $1000-9=\boxed{\textbf{(D)991}}$.

## Solution 2(Educated Guesses if you have no time)

We first note that $125 \cdot 8 = 1000$ and so we assume there are $125$ 8s.

Then we note that it is asking for the second factor, so we subtract $1$(the original $8$ in the first factor).

Now we have $125-1=124.$ The second factor is obviously a multiple of $124$.

Listing the first few, we have $124, 248, 372, 496, 620, 744, 868, 992, 1116, 1240, ...$

We notice that the 4th answer choice is 1 less than a 992(a multiple of 124.)

Thus we make an educated guess that it is somehow less by 1, so we get $\fbox{(D)}$. ~mathboy282

We were just lucky; this method is NOT reliable. Please note that this probably will not work for other problems and is just a lucky coincidence.

## Video Solution

~savannahsolver

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