Difference between revisions of "2010 AIME I Problems/Problem 2"

Revision as of 07:06, 11 February 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 $- 1\times9\times99 = - 891\equiv\boxed{109}\pmod{1000}$ .

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

@@ Line 14: / Line 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)
Preceded by Problem 1	Followed by Problem 3
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

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