Difference between revisions of "2019 AIME I Problems/Problem 1"
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| − | The above sequence is equivalent to <math>(10^1-1)+(10^2-1)+(10^3-1)+\cdots+(10^321-1)</math>. We can take out the ones so that this becomes <math> | + | The above sequence is equivalent to <math>(10^1-1)+(10^2-1)+(10^3-1)+\cdots+(10^(321)-1)</math>. We can take out the ones so that this becomes <math>(10^1)+(10^2)+(10^3)+\cdots+(10^321)-321</math>. Now, notice that <math>(10^1)+(10^2)+(10^3)+\cdots+(10^321)</math> is equal to <math>111...11110</math>, with 321 ones and one zero as digits. Subtract 321 to obtain <math>111...10790</math>, with 318 ones, 2 zeros, a 7 and a 9 as digits. This sums to <math>\boxed{334}</math>. |
{{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 21:28, 14 March 2019
Problem 1
Consider the integer ![\[N = 9 + 99 + 999 + 9999 + \cdots + \underbrace{99\ldots 99}_\text{321 digits}.\]](http://latex.artofproblemsolving.com/2/5/1/2511ced2d0525f01babe6c1df6bc04741aeae61c.png)
Find the sum of the digits of 
.
Solution
The above sequence is equivalent to 
. We can take out the ones so that this becomes 
. Now, notice that 
is equal to 
, with 321 ones and one zero as digits. Subtract 321 to obtain 
, with 318 ones, 2 zeros, a 7 and a 9 as digits. This sums to 
.
| 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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