2010 AIME I Problem 2

Attempt Problem 2 of the 2010 AIME I below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2010 AIME I solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

2.

Find the remainder when 999999999999 9’s9 \cdot 99 \cdot 999 \cdot \cdots \cdot \underbrace{99\ldots9}_{\text{999 9's}} is divided by 1000.1000.

Answer: 109
Concepts:modular arithmeticpattern recognition
Difficulty rating: 1950
Solution:

Work modulo 1000.1000. Every factor from the third one on ends in at least three 99s, so each is 1(mod1000).\equiv -1 \pmod{1000}. There are 999999 factors in all, hence 997997 of them are 1.\equiv -1.

The product is therefore 999(1)997891109(mod1000), \begin{aligned} &\equiv 9 \cdot 99 \cdot (-1)^{997} \\ &\equiv -891 \equiv 109 \pmod{1000}, \end{aligned} so the remainder is 109.109.

← Problem 1#1
Full Exam

Problem 2 in Other Years