Difference between revisions of "2019 AIME I Problems/Problem 1"

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==Solution 2==
 
==Solution 2==
 
We can see that <math>9=9</math>, <math>9+99=108</math>, <math>9+99+999=1107</math>, all the way to ten nines when we have <math>11111111100</math>. Then, when we add more nines, we repeat the same process, and quickly get that the sum of digits is <math>\boxed{342}</math> since we have to add <math>9\lfloor \log 321 \rfloor</math> to the sum of digits, which is <math>9\lceil \frac{321}9 \rceil</math>.
 
We can see that <math>9=9</math>, <math>9+99=108</math>, <math>9+99+999=1107</math>, all the way to ten nines when we have <math>11111111100</math>. Then, when we add more nines, we repeat the same process, and quickly get that the sum of digits is <math>\boxed{342}</math> since we have to add <math>9\lfloor \log 321 \rfloor</math> to the sum of digits, which is <math>9\lceil \frac{321}9 \rceil</math>.
 
  
 
==Solution 3 (Pattern)==
 
==Solution 3 (Pattern)==
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==Video Solution==
 
==Video Solution==
 
https://www.youtube.com/watch?v=JFHjpxoYLDk
 
https://www.youtube.com/watch?v=JFHjpxoYLDk
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==Video Solution 2==
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https://youtu.be/TSKcjht8Rfk
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~IceMatrix
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==Video Solution 3==
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https://www.youtube.com/watch?v=nbtIBP6Auig&t=460s
  
 
==See Also==
 
==See Also==

Revision as of 18:47, 31 May 2020

Problem 1

Consider the integer \[N = 9 + 99 + 999 + 9999 + \cdots + \underbrace{99\ldots 99}_\text{321 digits}.\]Find the sum of the digits of $N$.

Solution 1

Let's express the number in terms of $10^n$. We can obtain $(10-1)+(10^2-1)+(10^3-1)+\cdots+(10^{321}-1)$. By the commutative and associative property, we can group it into $(10+10^2+10^3+\cdots+10^{321})-321$. We know the former will yield $1111....10$, so we only have to figure out what the last few digits are. There are currently $321$ 1's. We know the last four digits are $1110$, and that the others will not be affected if we subtract $321$. If we do so, we get that $1110-321=789$. This method will remove three $1$'s, and add a $7$, $8$ and $9$. Therefore, the sum of the digits is $(321-3)+7+8+9=\boxed{342}$.

-eric2020


A similar and simpler way to consider the initial manipulations is to observe that adding $1$ to each term results in $(10+100+... 10^{320}+10^{321})$. There are $321$ terms, so it becomes $11...0$, where there are $322$ digits in $11...0$. Then, subtract the $321$ you initially added.

~ BJHHar

Solution 2

We can see that $9=9$, $9+99=108$, $9+99+999=1107$, all the way to ten nines when we have $11111111100$. Then, when we add more nines, we repeat the same process, and quickly get that the sum of digits is $\boxed{342}$ since we have to add $9\lfloor \log 321 \rfloor$ to the sum of digits, which is $9\lceil \frac{321}9 \rceil$.

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, $n$, the sum is $11...10-n$, where the first term is of length $n+1$. Here, that is $\boxed{342}$.

~BJHHar

Video Solution

https://www.youtube.com/watch?v=JFHjpxoYLDk

Video Solution 2

https://youtu.be/TSKcjht8Rfk

~IceMatrix

Video Solution 3

https://www.youtube.com/watch?v=nbtIBP6Auig&t=460s

See Also

2019 AIME I (ProblemsAnswer KeyResources)
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

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