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AIME 2010 I · 第 2 题

AIME 2010 I — Problem 2

专题
Contest Math
难度
L4
来源
AIME

题目详情

Problem

Find the remainder when 9×99×999×⋯×99⋯9⏟999 9’s9 \times 99 \times 999 \times \cdots \times \underbrace{99\cdots9}_{\text{999 9's}} is divided by 10001000.

解析

Solution

Note that 999≡9999≡⋯≡99⋯9⏟999 9’s≡−1(mod1000)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=997999 - 3 + 1 = 997 integers, so all those integers multiplied out are congruent to −1(mod1000)- 1\pmod{1000}. Thus, the entire expression is congruent to −1×9×99=−891≡109(mod1000)- 1\times9\times99 = - 891\equiv\boxed{109}\pmod{1000}.

Solution 2

The expression also equals (10−1)(100−1)…(10999−1)(10-1)(100-1)\dots({10^{999}}-1). To find its modular 1,000, remove all terms from 1,000 and after. Then the expression becomes (10−1)(100−1)(−1)(mod1000)≡−891(mod1000)≡109(mod1000)(10-1)(100-1)(-1) \pmod{1000} \equiv -891 \pmod{1000} \equiv \boxed{109}\pmod{1000}

By maxamc

Video Solution by OmegaLearn

https://youtu.be/orrw4VydBTk?t=140

~ pi_is_3.14

Video Solution

https://www.youtube.com/watch?v=-GD-wvY1ADE&t=78s

Video Solution by WhyMath

https://youtu.be/EMTcFZB9KvA

~savannahsolver