Exponential Digits

We can solve this (and other questions like it) using a combination of the Chinese Remainder Theorem and Euler's Theorem.

First of all, Euler: if $a$ and $n$ are coprime, then $a^{\varphi (n)} \equiv 1 \pmod{n}$ , where $\varphi (n)$ is the Euler Totient function.

In this case, we have $\varphi (100) = 40$ ; so by the theorem, we have

$3^{385} \equiv (3^{40})^9 \cdot 3^{25} \equiv 3^{25} \pmod{100}$

It's absolutely possible to finish here and just work out the powers needed, but sometimes it's useful to use the Chinese Remainder Theorem as well.

We have $100=4 \cdot 25$ .

$3^{25} \equiv (-1)^{25} \equiv 3 \pmod{4}$

and, using Euler again (and that $\varphi (25) = 20$ ),

$3^{25} \equiv 3^5 = 243 \equiv 18 \pmod{25}$

The Chinese Remainder Theorem tells us that there is a unique solution to both of these congruences modulo $100$ . The quickest way to solve these is to combine them; the second one tells us that $3^{385} = 18+25k$ for some $k$ ; looking at this modulo $4$ we have $3^{385} \equiv 2+k \equiv 3$ so that $k \equiv 1 \pmod4$ ; this gives the answer $18+25=43$ with tens digit $\boxed4$ .

3^385 mod 100 = 43 therefore we have a 4 in the tens digit

In order to approach this problem, you will need to find a pattern. Since the base of the exponent is 3, an easy counting number, you can just write all the possibilities down. *

There is a repeated cycle for the power of threes. The pattern starts with 03, and progresses with:
09, 27, 81, 43, 29, 87, 61, 83, 49, 47, 41, 23, 69, 07, 21, 63, 89, 67, 01, 03 ,
containing the tens and ones digits. Note: The italicized number is the beginning of the same sequence.

As you can see, the repeated length of the numbers are 20.

Since the problem states that you need to solve $3^{385}$ , you can just divide 385 by 20 because the sequence repeats. The sequence repeats 19 times which results in a remainder of 5 .

The remainder means how many numbers you will be counting on from the beginning of the list:
03, 09, 27, 81, 43 , 29, 87, 61, 83, 49, 47, 41, 23, 69, 07, 21, 63, 89, 67, 01

In this case, 43 is the $5^{th}$ number in the list.

The tens digit of 43 is $\boxed{4}$ .

* I will have another solution posted soon.