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> | + | 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 15:08, 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 
, 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. 
