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

Revision as of 20:22, 4 July 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 $999\equiv - 1\pmod{1000}$ , $9999\equiv - 1\pmod{1000}$ , $\dots$ , $\underbrace{99\cdots9}_{\text{999 9's}}\equiv - 1\pmod{1000}$ (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 $( - 1)(9)(99) = - 891\equiv\boxed{109}\pmod{1000}$ .

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

Revision as of 20:22, 4 July 2013

Problem

Solution

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