Modular arithmetic exercise

What are the last 3 digits of 99 9 123 999^{123} ?

(when it is expressed in base 10)


The answer is 999.

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1 solution

J Krol
Feb 12, 2015

Consider 99 9 123 m o d 1000 999^{123} mod 1000 . This is equal to ( 1 ) 123 m o d 1000 = 1 m o d 1000 = 999 (-1)^{123} mod 1000 = -1mod1000 =999 So the last three digits are 999.

Why did you solve it with (mod1000)?

Yassir Douslimi - 5 years, 2 months ago

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