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

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

Revision as of 19:51, 17 December 2019

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

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 4 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 3 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 $9\lceil \frac{321}9 \rceil$.

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
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All AIME Problems and Solutions

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