Difference between revisions of "2010 AIME I Problems/Problem 2"
Tempaccount (talk | contribs) (Remove extra problem section) |
(→Solution) |
||
Line 4: | Line 4: | ||
== Solution == | == Solution == | ||
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)(9)(99) = - 891\equiv\boxed{109}\pmod{1000}</math>. | 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)(9)(99) = - 891\equiv\boxed{109}\pmod{1000}</math>. | ||
+ | |||
+ | |||
+ | |||
+ | == Video Solution == | ||
+ | https://www.youtube.com/watch?v=-GD-wvY1ADE&t=78s | ||
== See Also == | == See Also == |
Revision as of 12:54, 28 October 2021
Contents
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
.
Video Solution
https://www.youtube.com/watch?v=-GD-wvY1ADE&t=78s
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.