2010 AIME I Problems/Problem 2

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

2010 AIME I (Problems • Answer Key • Resources)
Preceded by Problem 1	Followed by Problem 3
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