Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I am working through a modulo tutorial and have become stuck here: $$ 11^{32}(\operatorname{mod}13) = (11^{16})^2(\operatorname{mod}13)= 3^2(\operatorname{mod}13)= 9(\operatorname{mod}13) $$

My question is, how does $(11^{16})^2(\operatorname{mod}13)$ get reduced to $3^2(\operatorname{mod}13)$?

share|cite|improve this question
up vote 3 down vote accepted

Hint $\ $ By $\mu\!$ Fermat, $\rm\: mod\ 13\!:\ 11^{12}\equiv 1\:$ so $\rm\:11^{16}\equiv 11^{12}\cdot 11^4\equiv 1\cdot(-2)^4\equiv 3$

Generally, mod prime $\rm P\!:\ A\not\equiv0 \ \Rightarrow\ A^{P-1}\equiv 1\ \Rightarrow\ A^N \equiv\: A^{(N\ mod\ P-1)}$

Note also the use of least (balanced) residues $\rm\:11\equiv -2\pmod {13}\:$ to simplify calculations.

share|cite|improve this answer

Fermat's Little Theorem tells us that $11^{12} \equiv 1 \pmod {13}$, so $11^{16} \equiv 11^4 \equiv (-2)^4 \equiv 16 \equiv 3 \pmod {13}$

share|cite|improve this answer

Looks like everyone else is stumped by that step as well. Maybe wherever you got that problem had $11^{16}$ previously calculated. I would do the original problem similar to Bill.


share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.