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Difference between revisions of "2010 AIME I Problems/Problem 2"
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== Solution ==
 
== Solution ==
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Note that <math>999\equiv - 1\pmod{1000}</math>, <math>9999\equiv - 1\pmod{1000}</math>, <math>\dots</math>, <math>\underbrace{99\cdots9}_{\text{999 9's}}\equiv - 1\pmod{1000}</math> (see [[modular arithmetic]]). That is a total of <math>999 - 3 + 1 = 997</math> integers, so all those integers multiplied out are congruent to <math>- 1\pmod{1000}</math>. Thus, the entire expression is congruent to <math>( - 1)(9)(99) = - 891\equiv\boxed{109}\pmod{1000}</math>.
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Note that <math>999\equiv 9999\equiv\dots \equiv\underbrace{99\cdots9}_{\text{999 9's}}\equiv -1 \pmod{1000}</math> (see [[modular arithmetic]]). That is a total of <math>999 - 3 + 1 = 997</math> integers, so all those integers multiplied out are congruent to <math>- 1\pmod{1000}</math>. Thus, the entire expression is congruent to <math>- 1\times9\times99 = - 891\equiv\boxed{109}\pmod{1000}</math>.
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== Solution 2 ==
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The expression also equals <math>(10-1)(100-1)\dots({10^{999}}-1)</math>. To find its modular 1,000, remove all terms from 1,000 and after. Then the expression becomes <math>(10-1)(100-1)(-1) \pmod{1000} \equiv -891 \pmod{1000} \equiv \boxed{109}\pmod{1000}</math>
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By maxamc
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== Video Solution by OmegaLearn ==
 +
https://youtu.be/orrw4VydBTk?t=140
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 +
~ pi_is_3.14
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== Video Solution ==
 +
https://www.youtube.com/watch?v=-GD-wvY1ADE&t=78s
 +
 
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==Video Solution by WhyMath==
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https://youtu.be/EMTcFZB9KvA
 +
 
 +
~savannahsolver
  
 
== See Also ==
 
== See Also ==

Latest revision as of 17:46, 30 April 2023

Problem

Find the remainder when $9 \times 99 \times 999 \times \cdots \times \underbrace{99\cdots9}_{\text{999 9's}}$ is divided by $1000$.

Solution

Note that $999\equiv 9999\equiv\dots \equiv\underbrace{99\cdots9}_{\text{999 9's}}\equiv -1 \pmod{1000}$ (see modular arithmetic). That is a total of $999 - 3 + 1 = 997$ integers, so all those integers multiplied out are congruent to $- 1\pmod{1000}$. Thus, the entire expression is congruent to $- 1\times9\times99 = - 891\equiv\boxed{109}\pmod{1000}$.

Solution 2

The expression also equals $(10-1)(100-1)\dots({10^{999}}-1)$. To find its modular 1,000, remove all terms from 1,000 and after. Then the expression becomes $(10-1)(100-1)(-1) \pmod{1000} \equiv -891 \pmod{1000} \equiv \boxed{109}\pmod{1000}$

By maxamc

Video Solution by OmegaLearn

https://youtu.be/orrw4VydBTk?t=140

~ pi_is_3.14

Video Solution

https://www.youtube.com/watch?v=-GD-wvY1ADE&t=78s

Video Solution by WhyMath

https://youtu.be/EMTcFZB9KvA

~savannahsolver

See Also

2010 AIME I (Problems • Answer Key • Resources)
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All AIME Problems and Solutions

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