Difference between revisions of "2019 AIME I Problems/Problem 1"
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| − | ==Problem | + | ==Problem== |
Consider the integer <cmath>N = 9 + 99 + 999 + 9999 + \cdots + \underbrace{99\ldots 99}_\text{321 digits}.</cmath>Find the sum of the digits of <math>N</math>. | Consider the integer <cmath>N = 9 + 99 + 999 + 9999 + \cdots + \underbrace{99\ldots 99}_\text{321 digits}.</cmath>Find the sum of the digits of <math>N</math>. | ||
| Line 7: | Line 7: | ||
-eric2020 | -eric2020 | ||
| − | - | + | -another Eric in 2020 |
| Line 23: | Line 23: | ||
~BJHHar | ~BJHHar | ||
| − | ==Video Solution== | + | ==Solution 4 (Official MAA)== |
| + | Write <cmath>\begin{align*}N &=(10-1)+(10^2-1)+\cdots+(10^{321}-1)\\ | ||
| + | &=10+10^2+10^3+10^4+10^5+10^6+\cdots 10^{321}-321 \\ | ||
| + | &=1110-321+10^4+10^5+10^6+\cdots+10^{321}\\ | ||
| + | &=789+10^4+10^5+10^6+\cdots+10^{321}\\ | ||
| + | \end{align*}</cmath> | ||
| + | The sum of the digits of <math>N</math> is therefore equal to <math>7+8+9+(321-3)=342</math>. | ||
| + | ==Video Solution #1(Using Smart Manipulation)== | ||
| + | https://youtu.be/JQdad7APQG8?t=22 | ||
| + | |||
| + | ==Video Solution 2== | ||
https://www.youtube.com/watch?v=JFHjpxoYLDk | https://www.youtube.com/watch?v=JFHjpxoYLDk | ||
| − | ==Video Solution | + | ==Video Solution 3== |
https://youtu.be/TSKcjht8Rfk | https://youtu.be/TSKcjht8Rfk | ||
~IceMatrix | ~IceMatrix | ||
| − | ==Video Solution | + | ==Video Solution 4== |
https://youtu.be/9X18wCiYw9M | https://youtu.be/9X18wCiYw9M | ||
~Shreyas S | ~Shreyas S | ||
| + | |||
| + | ==Video Solution 5== | ||
| + | https://youtu.be/hzbzEAo9ezA | ||
| + | |||
| + | ~savannahsolver | ||
==See Also== | ==See Also== | ||
Latest revision as of 06:57, 9 February 2023
Contents
Problem
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 1
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 four digits are 
, and that the others will not be affected if we subtract . If we do so, we get that 
. This method will remove three 
's, and add a 
, 
and 
. Therefore, the sum of the digits is 
.
-eric2020 -another Eric in 2020
A similar and simpler way to consider the initial manipulations is to observe that adding to each term results in 
. There are terms, so it becomes 
, where there are 
digits in . Then, subtract the 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 concatenated in front and also a subtracted (
, 
, 
, 
). Then for any index of terms, 
, the sum is 
, where the first term is of length 
. Here, that is .
~BJHHar
Solution 4 (Official MAA)
Write 
The sum of the digits of is therefore equal to 
.
Video Solution #1(Using Smart Manipulation)
https://youtu.be/JQdad7APQG8?t=22
Video Solution 2
https://www.youtube.com/watch?v=JFHjpxoYLDk
Video Solution 3
~IceMatrix
Video Solution 4
~Shreyas S
Video Solution 5
~savannahsolver
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. 
