Fermat's little theorem

Question

Obtain residue class of $7^{9999}$ modulo 100 using the Little Fermat theorem.

But I have no idea how to proceed.

I would rather use Euler's theorem: http://en.wikipedia.org/wiki/Euler's_theorem — Beni Bogosel, Mar 26 '12 at 15:27
It can be done using just basic tools. But I believe that if we are going to use something related to Fermat's Theorem, we really want to use Euler's generaliation of Fermat's Theorem. — André Nicolas, Mar 26 '12 at 15:29
An hint on how to proceed with basic tools is to compute $7^4$ mod $100$ and then observe that... — Giovanni De Gaetano, Mar 26 '12 at 15:34
Have you looked at the first link on the right-hand side? http://math.stackexchange.com/questions/12846/using-fermats-little-theorem — dls, Mar 26 '12 at 15:34

score 7 · Answer 1 · answered Mar 26 '12 at 15:36

7

First observe that $$ 7^{4}=2401\equiv1\bmod100. $$ Now write $9999=4\cdot2499+3$ so that $$ 7^{9999}=(7^4)^{2499}\cdot7^3\equiv7^3=343=43\bmod 100. $$

answered Mar 26 '12 at 15:36

Andrea Mori

26,969

score 2 · Answer 2 · answered Mar 26 '12 at 15:35

2

Note that $7^4 = 2401 \equiv 1 \bmod 100$. Now divide $9999$ by $4$ with remainder.

answered Mar 26 '12 at 15:35

lhf

216,483

score 2 · Answer 3 · answered Mar 26 '12 at 15:38

2

You could notice that $7^8 \equiv 1 \mod 100$. This makes the problem a lot easier $$7^{9999}\equiv 7^{9 \cdot 1111} \equiv 7^{7} \equiv 43 \mod 100 $$

answered Mar 26 '12 at 15:38

Beni Bogosel

23,381

score 2 · Answer 4 · edited Mar 26 '12 at 16:53

2

Hint $\rm\ \ 4\:|\:7^{\:\!2}-1,\ 25\:|\:7^{\!\:2}+1\ \Rightarrow\ 100\:|\:7^{\:\!4}-1\:|\:7^{\:4\!\:N}-1\ \Rightarrow\ 100\:|\:7^{\:4\!\:N+3}-7^{\:\!3}$

Or: $\rm\ mod\ 4,25\!:\ \ 7^{\:\!4} \equiv 1\ \Rightarrow\ mod\ 100\!:\ \ 7^{\:\!4}\equiv 1\ \Rightarrow\ 7^{\:\!3 +4\!\:N}\equiv\: 7^{\:\!3} (7^{\:\!4})^N\equiv\: 7^{\:\!3}$

edited Mar 26 '12 at 16:53

answered Mar 26 '12 at 16:33

Bill Dubuque

272,048

Fermat's little theorem

4 Answers4