Difference between revisions of "Euler's Totient Theorem Problem 2 Solution"
Borealbear (talk | contribs) (Created page with "==Problem== (BorealBear) Find the last two digits of <math> 3^{3^{3^{3}}} </math>. ==Solution== This problem is just asking for <math> 3^{3^{3^{3}}}\pmod{100} </math>. We ca...") |
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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]]. |
Latest revision as of 22:14, 15 August 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.