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

Revision as of 20:26, 23 June 2016

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

The product $(8)(888\dots8)$ , where the second factor has $k$ digits, is an integer whose digits have a sum of $1000$ . What is $k$ ?

$\textbf{(A)}\ 901\qquad\textbf{(B)}\ 911\qquad\textbf{(C)}\ 919\qquad\textbf{(D)}\ 991\qquad\textbf{(E)}\ 999$

No solution yet!

2014 AMC 10A (Problems • Answer Key • Resources)
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 (Problems • Answer Key • Resources)
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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 ==Solution==
-Note that for <math>k\ge{2}</math>, <math>8 \cdot \underbrace{888...8}_{k \text{ digits}}=7\underbrace{111...1}_{k-2 \text{ ones}}04</math>, which has a digit sum of <math>7+k-2+0+4=9+k</math>. Since we are given that said number has a digit sum of <math>1000</math>, we have <math>9+k=1000 \Rightarrow k=\boxed{\textbf{(D) }991}</math>.
+No solution yet!
 ==See Also==