Difference between revisions of "2019 AIME I Problems/Problem 1"
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==Solution== | ==Solution== | ||
Let's express the terms in terms of <math>10^n</math>. We can obtain <math>(10-1)+(10^2-1)+(10^3-1)+\cdots+(10^{321}-1)</math>. By the commutative and associative property, we can group it into <math>(10+10^2+10^3+\cdots+10^{321})-321</math>. We know the former will yield <math>1111....10</math>, so we only have to figure out what the last few digits are. There are currently <math>321</math> 1's. We know the last 4 digits are 1110, and that the others will not be affected if we subtract <math>321</math>. If we do so, we get that <math>1110-321=789</math>. This method will remove 3 1's, and add a 7, 8 and 9. Therefore, the sum of the digits is <math>(321-3)+7+8+9=\boxed{342}</math>. | Let's express the terms in terms of <math>10^n</math>. We can obtain <math>(10-1)+(10^2-1)+(10^3-1)+\cdots+(10^{321}-1)</math>. By the commutative and associative property, we can group it into <math>(10+10^2+10^3+\cdots+10^{321})-321</math>. We know the former will yield <math>1111....10</math>, so we only have to figure out what the last few digits are. There are currently <math>321</math> 1's. We know the last 4 digits are 1110, and that the others will not be affected if we subtract <math>321</math>. If we do so, we get that <math>1110-321=789</math>. This method will remove 3 1's, and add a 7, 8 and 9. Therefore, the sum of the digits is <math>(321-3)+7+8+9=\boxed{342}</math>. | ||
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==See Also== | ==See Also== | ||
Revision as of 00:29, 15 March 2019
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 terms 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
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. 
