2010 AIME I Problem 2

Question

Find the remainder when

9 \cdot 99 \cdot 999 \cdot \cdots \cdot \underbrace{99\ldots9}_{\text{999 nines}}

is divided by

1000.

Answer

Every factor from the third one on ends in at least three

9

s, so it is

\equiv -1 \pmod{1000}

Count how many factors are

\equiv -1 \pmod{1000}

and check whether that count is odd or even; only

9 \cdot 99

remains to compute

Work modulo

1000.

Every factor from the third one on ends in at least three

9

s, so each is

\equiv -1 \pmod{1000}.

There are

999

factors in all, hence

997

of them are

\equiv -1.

The product is therefore

\begin{aligned} &\equiv 9 \cdot 99 \cdot (-1)^{997} \\ &\equiv -891 \equiv 109 \pmod{1000}, \end{aligned}

so the remainder is

109.

Problem 2 in other years