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>[(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 111...11110, with 321 ones and one zero as digits. Subtract 321 to obtain 111...10790, with 318 ones, 2 zeros, a 7 and a 9 as digits. This sums to 334.
  
 
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{{AIME box|year=2019|n=I|before=First Problem|num-a=2}}
 
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Revision as of 22:27, 14 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

The above sequence is equivalent to $(10^1-1)+(10^2-1)+(10^3-1)+\cdots+(10^321-1)$. We can take out the ones so that this becomes $[(10^1)+(10^2)+(10^3)+\cdots+(10^321)]-321$. Now, notice that $(10^1)+(10^2)+(10^3)+\cdots+(10^321)$ is equal to 111...11110, with 321 ones and one zero as digits. Subtract 321 to obtain 111...10790, with 318 ones, 2 zeros, a 7 and a 9 as digits. This sums to 334.

2019 AIME I (ProblemsAnswer KeyResources)
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First Problem
Followed by
Problem 2
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