Finding the last two digits of $3^{400}$

Question

I'm having some trouble with the following exercise:

Find the last two digits of the decimal representation of $3^{400}$.

This exercise Is under the Euler's theorem and Fermat's little theorem section of the worksheet so I think that we are supposed to use those to solve this. My problem is that this is the first time I'm seeing an exercise of this kind and I don't know how to proceed. Can someone give me a tip on how to do this or show me how it's done?

The last two digits are the same thing as the integer reduced mod 100. — Randall, Jan 19 '21 at 21:10

peter.petrov · Accepted Answer · 2021-01-19T22:12:18.783

1

It's a simple application of the totient theorem.

Applying it we get that

$$3^{\phi(100)}\equiv 1\pmod {100} $$ i.e.

$$3^{40}\equiv 1\pmod {100} $$

That's because $GCD(3, 100) = 1$ so we can apply the theorem.

Going on from there is pretty much easy.

Just raise the last congruence to the power of $10$ and you get

$$3^{400}\equiv 1\pmod {100} $$

So the last two digits are $0$ and $1$.

edited Jan 19 '21 at 22:12

answered Jan 19 '21 at 21:13

peter.petrov

12,568

2

What you've applied here is the totient theorem, not Fermat's little theorem. – Qiaochu Yuan Jan 19 '21 at 21:39
@QiaochuYuan Yeah, sorry. I know it as Euler's theorem actually. – peter.petrov Jan 19 '21 at 22:12

J. W. Tanner · Answer 2 · 2021-01-19T21:38:40.627

1

Alternatively, you could use binomial expansion:

$3^{400}=(10-1)^{200}\equiv-200\times10+1\equiv1\bmod100$

(in fact, $\bmod 1000$).

edited Jan 19 '21 at 21:38

answered Jan 19 '21 at 21:32

J. W. Tanner

60,406

Did it occur to you that this might have been asked before? Twice, in fact (ignoring the countless near duplicate that can be solved with the same tools). – Jyrki Lahtonen Jan 19 '21 at 21:49

Finding the last two digits of $3^{400}$

2 Answers2