Difference between revisions of "2019 AIME I Problems/Problem 1"
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Revision as of 01:07, 17 March 2019
Contents
Problem 1
Consider the integer ![\[N = 9 + 99 + 999 + 9999 + \cdots + \underbrace{99\ldots 99}_\text{321 digits}.\]](http://latex.artofproblemsolving.com/2/5/1/2511ced2d0525f01babe6c1df6bc04741aeae61c.png)
Find the sum of the digits of 
.
Solution
Let's express the number in terms of 
. We can obtain 
. By the commutative and associative property, we can group it into 
. We know the former will yield 
, so we only have to figure out what the last few digits are. There are currently 
1's. We know the last 4 digits are 1110, and that the others will not be affected if we subtract . If we do so, we get that 
. This method will remove 3 1's, and add a 7, 8 and 9. Therefore, the sum of the digits is 
.
-eric2020
Video Solution
https://www.youtube.com/watch?v=JFHjpxoYLDk
See Also
| 2019 AIME I (Problems • Answer Key • Resources) | ||
| Preceded by First Problem |
Followed by Problem 2 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
| All AIME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
