Difference between revisions of "2019 AIME I Problems/Problem 1"
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Observe how adding results in the last term but with a 1 concatenated in front and also a 1 subtracted (09, 108, 1107, 11106). Then for any index of terms, <math>n</math>, the sum is <math>11...10-n</math>, where the first term is of length <math>n+1</math>. Here, that is <math>\boxed{342}</math>. | Observe how adding results in the last term but with a 1 concatenated in front and also a 1 subtracted (09, 108, 1107, 11106). Then for any index of terms, <math>n</math>, the sum is <math>11...10-n</math>, where the first term is of length <math>n+1</math>. Here, that is <math>\boxed{342}</math>. | ||
~BJHHar | ~BJHHar | ||
| + | |||
==Video Solution== | ==Video Solution== | ||
https://www.youtube.com/watch?v=JFHjpxoYLDk | https://www.youtube.com/watch?v=JFHjpxoYLDk | ||
Revision as of 21:00, 17 December 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 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
A similar and simpler way to consider the initial manipulations is to observe that adding 1 to each term results in 
. There are 321 terms, so it becomes 
, where there are 322 digits in . Then, subtract the 321 you initially added.
~ BJHHar
Solution 2
We can see that 
, 
, 
, all the way to ten nines when we have 
. Then, when we add more nines, we repeat the same process, and quickly get that the sum of digits is 
since we have to add 
to the sum of digits, which is 
.
Solution 3 (Pattern)
Observe how adding results in the last term but with a 1 concatenated in front and also a 1 subtracted (09, 108, 1107, 11106). Then for any index of terms, 
, the sum is 
, where the first term is of length 
. Here, that is .
~BJHHar
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. 
