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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==
  
 
{{AIME box|year=2019|n=I|before=First Problem|num-a=2}}
 
{{AIME box|year=2019|n=I|before=First Problem|num-a=2}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 01:20, 15 March 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 terms 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 $(321-3)+7+8+9=\boxed{342}$.

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

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