Difference between revisions of "2019 AIME I Problems/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>. | ||
+ | |||
+ | ==Solution 1== | ||
+ | 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 | ||
+ | -another Eric in 2020 | ||
+ | |||
+ | |||
+ | |||
+ | 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. | ||
+ | |||
+ | ~ BJHHar | ||
+ | |||
+ | ==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>. | ||
+ | |||
+ | ==Solution 3 (Pattern)== | ||
+ | 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>. | ||
+ | |||
+ | ~BJHHar | ||
+ | |||
+ | ==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 | ||
+ | |||
+ | ==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== | ||
{{AIME box|year=2019|n=I|before=First Problem|num-a=2}} | {{AIME box|year=2019|n=I|before=First Problem|num-a=2}} | ||
+ | |||
+ | [[Category:Intermediate Number Theory Problems]] | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 07:57, 9 February 2023
Contents
Problem
Consider the integer 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.