2014 AMC 10A Problems/Problem 20

Revision as of 18:27, 9 March 2020 by Alexwin0806 (talk | contribs) (Solution)
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 $\fbox{(D)}$

See Also

2014 AMC 10A (ProblemsAnswer KeyResources)
Preceded by
Problem 19
Followed by
Problem 21
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 10 Problems and Solutions
2014 AMC 12A (ProblemsAnswer KeyResources)
Preceded by
Problem 15
Followed by
Problem 17
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 12 Problems and Solutions

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