Difference between revisions of "2010 AIME I Problems/Problem 2"
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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>. | 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>. | ||
| − | == See | + | == See Also == |
{{AIME box|year=2010|num-b=1|num-a=3|n=I}} | {{AIME box|year=2010|num-b=1|num-a=3|n=I}} | ||
[[Category:Introductory Number Theory Problems]] | [[Category:Introductory Number Theory Problems]] | ||
Revision as of 15:52, 12 April 2012
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 | ||
