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

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==Problem 1==
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==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>.
  
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==Solution==
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==Solution 1==
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Let's express the number 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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Let's express the number 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 four digits are <math>1110</math>, 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 three <math>1</math>'s, and add a <math>7</math>, <math>8</math> and <math>9</math>. Therefore, the sum of the digits is <math>(321-3)+7+8+9=\boxed{342}</math>.
  
 
-eric2020
 
-eric2020
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-another Eric in 2020
  
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A similar, but separate way to consider the initial calculations 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...0-321</math>, where there are 322 digits in <math>11...0</math>. ~ BJHHar
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A similar and simpler way to consider the initial manipulations is to observe that adding <math>1</math> to each term results in <math>(10+100+... 10^{320}+10^{321})</math>. There are <math>321</math> terms, so it becomes <math>11...0</math>, where there are <math>322</math> digits in <math>11...0</math>. Then, subtract the <math>321</math> you initially added.
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~ BJHHar
  
 
==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>.
  
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==Video Solution==
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==Solution 3 (Pattern)==
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Observe how adding results in the last term but with a <math>1</math> concatenated in front and also a <math>1</math> subtracted (<math>09</math>, <math>108</math>, <math>1107</math>, <math>11106</math>). 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>.
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~BJHHar
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==Solution 4 (Official MAA)==
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Write <cmath>\begin{align*}N &=(10-1)+(10^2-1)+\cdots+(10^{321}-1)\\
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&=10+10^2+10^3+10^4+10^5+10^6+\cdots 10^{321}-321 \\
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&=1110-321+10^4+10^5+10^6+\cdots+10^{321}\\
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&=789+10^4+10^5+10^6+\cdots+10^{321}\\
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\end{align*}</cmath>
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The sum of the digits of <math>N</math> is therefore equal to <math>7+8+9+(321-3)=342</math>.
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==Video Solution #1(Using Smart Manipulation)==
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https://youtu.be/JQdad7APQG8?t=22
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==Video Solution 2==
 
https://www.youtube.com/watch?v=JFHjpxoYLDk
 
https://www.youtube.com/watch?v=JFHjpxoYLDk
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 +
==Video Solution 3==
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https://youtu.be/TSKcjht8Rfk
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~IceMatrix
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==Video Solution 4==
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https://youtu.be/9X18wCiYw9M
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~Shreyas S
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==Video Solution 5==
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https://youtu.be/hzbzEAo9ezA
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~savannahsolver
  
 
==See Also==
 
==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}}
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[[Category:Intermediate Number Theory Problems]]
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 06:57, 9 February 2023

Problem

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 -another Eric in 2020


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

Solution 4 (Official MAA)

Write \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*} The sum of the digits of $N$ is therefore equal to $7+8+9+(321-3)=342$.

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

https://youtu.be/TSKcjht8Rfk

~IceMatrix

Video Solution 4

https://youtu.be/9X18wCiYw9M

~Shreyas S

Video Solution 5

https://youtu.be/hzbzEAo9ezA

~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

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