Difference between revisions of "Euler's Totient Theorem Problem 2 Solution"
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==Solution== | ==Solution== | ||
| − | + | Finding the last two digits is equivalent to finding <math>3^{3^{3^{3}}}\pmod{100}</math>. We can start by expanding the uppermost exponent: <math>3^{3^{27}}</math>. Then, since <math>\phi(100)=40</math>, the exponent is equal to <math> 3^{27}\pmod{40} </math>. We see that <math> 3^4=81\equiv1\pmod{40} </math>, so it simplifies to <math>3^3=27\pmod{40}</math>. | |
| − | We | + | 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 | ||
| + | |||
| + | Link back to [[Euler's Totient Theorem]]. | ||
Revision as of 17:42, 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.
