Lets play with groups U(n)

Number Theory Level 3

Find the last two digits of 2 3 123 23^{123} .


The answer is 67.

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Amar Mavi by Amar Mavi , 25, India

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2 solutions

Otto Bretscher
Feb 2, 2016

Since ϕ ( 100 ) = 40 \phi(100)=40 , we have 2 3 123 ≡ 2 3 3 ≡ 67 ( m o d 100 ) 23^{123}\equiv 23^3 \equiv \boxed{67} \pmod{100}

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Sir can u explain what is carmichael lambda i thing it is euler(100)=20

Amar Mavi - 5 years, 4 months ago

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We can do it in terms of the Euler phi as well... I will change my solution since people may be more familiar with phi.

Otto Bretscher - 5 years, 4 months ago
Deepak Kumar
Feb 4, 2016

Binomial can also be used

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Please post a solution using binomial expansion.

Harsh Khatri - 5 years, 4 months ago

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