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

Revision as of 14:17, 21 December 2013

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 $( - 1)(9)(99) = - 891\equiv\boxed{109}\pmod{1000}$ .

@@ Line 3: / Line 3: @@
 == Solution ==
-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 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>.
 == See Also ==

2010 AIME I (Problems • Answer Key • Resources)
Preceded by Problem 1	Followed by Problem 3
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Difference between revisions of "2010 AIME I Problems/Problem 2"

Revision as of 14:17, 21 December 2013

Problem

Solution

See Also