Difference between revisions of "Euler's Totient Theorem Problem 2 Solution"
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We now just need to find the last two digits of <math>3^27</math>. Using the [[Chinese Remainder Theorem]], we find that the last two digits are <math>3\pmod{4}</math> and <math>12\pmod{25}</math>. We guess and check to get <math>\boxed{87}</math>. | We now just need to find the last two digits of <math>3^27</math>. Using the [[Chinese Remainder Theorem]], we find that the last two digits are <math>3\pmod{4}</math> and <math>12\pmod{25}</math>. We guess and check to get <math>\boxed{87}</math>. | ||
~BorealBear | ~BorealBear | ||
| + | |||
| + | Link back to [[Euler's Totient Theorem]]. | ||
Revision as of 16:42, 21 March 2023
Problem
(BorealBear) Find the last two digits of 
.
Solution
Finding the last two digits is equivalent to finding 
. We can start by expanding the uppermost exponent: 
. Then, since 
, the exponent is equal to 
. We see that 
, so it simplifies to 
.
We now just need to find the last two digits of 
. Using the Chinese Remainder Theorem, we find that the last two digits are 
and 
. We guess and check to get 
.
~BorealBear
Link back to Euler's Totient Theorem.
