# Find remainder with congruence theorem

• Feb 6th 2008, 01:09 PM
Find remainder with congruence theorem
This is a problem we ran over in class, but I don't fully understand it.

Find remainder when $\displaystyle 3^{1000000}$ is divided by 26.

Solution:

$\displaystyle 1000000 = (3)(333333)+1$, therefore $\displaystyle 3^{1000000} = 3^{(333333)(3)+1} = 3^{333333}(3)$

By using the congruence theorem, we know that if a congruence b, mod n, then a and b would have the same remainder upon being divided by n.

Then it goes that the remainder is 3, but how do you get that?
• Feb 6th 2008, 02:51 PM
ThePerfectHacker
Quote:

This is a problem we ran over in class, but I don't fully understand it.

Find remainder when $\displaystyle 3^{1000000}$ is divided by 26.

Solution:

$\displaystyle 1000000 = (3)(333333)+1$, therefore $\displaystyle 3^{1000000} = 3^{(333333)(3)+1} = 3^{333333}(3)$

By using the congruence theorem, we know that if a congruence b, mod n, then a and b would have the same remainder upon being divided by n.

Then it goes that the remainder is 3, but how do you get that?

Note that $\displaystyle 3^3 = 27$ this means that $\displaystyle 3^3 \equiv 27 \equiv 1(\bmod 26)$.
Thus, $\displaystyle (3^3)^{333333} \equiv 1^{333333} (\bmod 26)$.
Thus, $\displaystyle 3^{999999}\equiv 1(\bmod 26)$
Multiply both sides by $\displaystyle 3$ to get your answer.