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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 <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>.
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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 <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 \[N = 9 + 99 + 999 + 9999 + \cdots + \underbrace{99\ldots 99}_\text{321 digits}.\]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 $\boxed{334}$.

2019 AIME I (Problems • Answer Key • Resources)
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