2010 AIME I Problems/Problem 2

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

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

Page

Toolbox

Search

2010 AIME I Problems/Problem 2

Problem

Solution

See also