Difference between revisions of "2010 AIME I Problems/Problem 2"
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== Problem == | == Problem == | ||
Find the [[remainder]] when <math>9 \times 99 \times 999 \times \cdots \times \underbrace{99\cdots9}_{\text{999 9's}}</math> is divided by <math>1000</math>. | Find the [[remainder]] when <math>9 \times 99 \times 999 \times \cdots \times \underbrace{99\cdots9}_{\text{999 9's}}</math> is divided by <math>1000</math>. | ||
Revision as of 14:45, 9 August 2018
Contents
Problem
Problem
Find the remainder when 
is divided by 
.
Solution
Note that 
(see modular arithmetic). That is a total of 
integers, so all those integers multiplied out are congruent to 
. Thus, the entire expression is congruent to 
.
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
