Difference between revisions of "Euler's Totient Theorem Problem 1 Solution"

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==Solution==
 
==Solution==
This is a direct application of Euler's Totient Theorem. Since <math> \phi(100)=40 </math>, this reduces to <math> 7^1-3^1\equiv \boxed{04}\pmod{100} </math>. -BorealBear
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This is a direct application of [[Euler's Totient Theorem]]. Since <math> \phi(100)=40 </math>, this reduces to <math> 7^1-3^1\equiv \boxed{04}\pmod{100} </math>. -BorealBear

Latest revision as of 16:33, 21 March 2023

Problem

(BorealBear) Find the last two digits of $7^{81}-3^{81}$.

Solution

This is a direct application of Euler's Totient Theorem. Since $\phi(100)=40$, this reduces to $7^1-3^1\equiv \boxed{04}\pmod{100}$. -BorealBear