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

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==Solution==
 
==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>
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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>.
  
 
==See Also==
 
==See Also==

Revision as of 16:08, 26 January 2016

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

Note that for $k\ge{2}$, $8 \cdot \underbrace{888...8}_{k \text{ digits}}=7\underbrace{111...1}_{k-2 \text{ ones}}04$, 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
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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