2014 AMC 10A Problems/Problem 20

Revision as of 18:50, 7 February 2014 by Minimario (talk | contribs) (Solution)

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 $k$?

$\textbf{(A)}\ 901\qquad\textbf{(B)}\ 911\qquad\textbf{(C)}\ 919\qquad\textbf{(D)}}\ 991\qquad\textbf{(E)}\ 999$ (Error compiling LaTeX. Unknown error_msg)

Solution

Note that $8 \cdot \underbrace{888...8}_{k \text{ digits}}=7\underbrace{111...1}_{k-2 \text{ ones}}04$, which has a digit sum of $7+k-2+0+4=9+k$. Since we are given that said number has a digit sum of $1000$, we have $9+k=1000 \Rightarrow k=\boxed{\textbf{(D) }991}$

See Also

2014 AMC 10A (ProblemsAnswer KeyResources)
Preceded by
Problem 19
Followed by
Problem 21
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All AMC 10 Problems and Solutions

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