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Number theory Difficulty 3.6 AMC 10/12 Find the answer

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.

A number or a short expression. Spacing and $ signs are ignored.

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}.

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.