Difference between revisions of "Euler's Totient Theorem Problem 1 Solution"
Borealbear (talk | contribs) (Created page with "==Problem== (BorealBear) Find the last two digits of <math> 7^{81}-3^{81} </math>. ==Solution== This is a direct application of Euler's Totient Theorem. Since <math> \phi(10...") |
Megaboy6679 (talk | contribs) (→Solution) |
||
| (One intermediate revision by one other user not shown) | |||
| Line 3: | Line 3: | ||
==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>. | + | 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 
.
Solution
This is a direct application of Euler's Totient Theorem. Since 
, this reduces to 
. -BorealBear
