Difference between revisions of "2014 AMC 10A Problems/Problem 20"
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==Problem== | ==Problem== | ||
| − | The product <math>(8)(888\dots8)</math>, where the second factor has <math>k</math> digits, is an integer whose digits have a sum of <math>1000</math>. What is <math>k</math>? | + | <!-- don't remove the following tag, for PoTW on the Wiki front page--><onlyinclude>The product <math>(8)(888\dots8)</math>, where the second factor has <math>k</math> digits, is an integer whose digits have a sum of <math>1000</math>. What is <math>k</math>?<!-- don't remove the following tag, for PoTW on the Wiki front page--></onlyinclude> |
<math>\textbf{(A)}\ 901\qquad\textbf{(B)}\ 911\qquad\textbf{(C)}\ 919\qquad\textbf{(D)}\ 991\qquad\textbf{(E)}\ 999</math> | <math>\textbf{(A)}\ 901\qquad\textbf{(B)}\ 911\qquad\textbf{(C)}\ 919\qquad\textbf{(D)}\ 991\qquad\textbf{(E)}\ 999</math> | ||
Revision as of 16:16, 18 November 2015
- 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 
, where the second factor has 
digits, is an integer whose digits have a sum of 
. What is ?
Solution
Note that for 
, 
, which has a digit sum of 
. Since we are given that said number has a digit sum of , we have 
See Also
| 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 | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
