'POWER OF 3'

If you are unfamiliar with modular arithmetic or congruence, then:

$a \equiv b \pmod{n}$ is read as " $a$ is congruent to $b$ modulo $n.$ " It means that $(a-b)$ is divisible by $n$ or $a$ and $b$ both leave equal remainder when divided by $n$ .

A number $n$ is said to have an even ten's digit if there exists integer $k$ such that $n\equiv k \mod20$ and $k<10.$

Also observe that $3^4 \equiv1 \mod20$ , multiplying both sides by $3^n$ we get $3^{n+4}\equiv 3^n \mod20$ , thus there are only 4 possible distinct remainders when $3^n$ is divided by $20$ .

So observing from the start( $n=0$ ),

$3^{0} \equiv 1\mod{20}$

$3^{1} \equiv 3\mod{20}$

$3^{2} \equiv 9\mod{20}$

$3^{3} \equiv 7\mod{20}$

All remainders are less than 10, which means that he ten's digit of any power of 3 is always even.

The last digit of 3 power any positive integer follows a certain order : $3, 9, 7, 1$

Now, $3^3 = 27, 3^4 = 81, 3^5 = 243, etc$ . So we come to know that the ten's digit of the first 3 powers of 3 is even.

Use induction on the exponent:

The last digit is either 1, 3, 7 or 9.

If 1 or 3, multiplying a power of 3 will not change the evenness of the second last digit.

If 7 or 9, multiplying will result in a carry over of 2, which again does not change the evenness of the second last digit.

We know that if $3^m \mod{20} \equiv n$ where $0>n>10$ , then it has an even tens digit because the remainder will not effect the tens digit if it is less than 10, and if it is in mod 20, then any number times 20 has an even tens digit

note that 3 and 20 are relatively prime, so using euler's extention of FLT, (note that $\phi(20)=8$ ) so $3^{8k+m} \mod{20} \equiv 3^m \mod{20}$

so we only need to go through m=0,1,2,3,4,5,6, and 7, and show that the remainders are all less than 10, then all the other powers are also proven

$3^{0} \mod{20} \equiv 1$

$3^{1} \mod{20} \equiv 3$

$3^{2} \mod{20} \equiv 9$

$3^{3} \mod{20} \equiv 7$

$3^{4} \mod{20} \equiv 1$

$3^{5} \mod{20} \equiv 3$

$3^{6} \mod{20} \equiv 9$

$3^{7} \mod{20} \equiv 7$

thus every remainder $\mod{20}$ for a power of $3^n$ is 1,3,7, or 9 so the tens digit must be even because any number mod 20 where the remainder is less than 10 will have an even tens digit