Difference between revisions of "2010 AIME I Problems/Problem 2"
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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\times9\times99 = - 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\times9\times99 = - 891\equiv\boxed{109}\pmod{1000}</math>. | ||
| + | == Solution 2 == | ||
| + | The expression also equals <math>(10-1)(100-1)\dots({10^{999}}-1)</math>. To find its modular 1,000, remove all terms from 1,000 and after. Then the expression becomes <math>(10-1)(100-1)(-1) \pmod{1000} \equiv -891 \pmod{1000} \equiv \boxed{109}\pmod{1000}</math> | ||
| + | By maxamc | ||
== Video Solution by OmegaLearn == | == Video Solution by OmegaLearn == | ||
Latest revision as of 16:46, 30 April 2023
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 
.
Solution 2
The expression also equals 
. To find its modular 1,000, remove all terms from 1,000 and after. Then the expression becomes 
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
~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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
