Power of 3

Relevant wiki: Modular Arithmetic - Problem Solving - Basic

We will prove by induction that the tens digit of $3^n$ is always even.

$3^0 \equiv 1 \bmod \ 20$

and

$\begin{aligned} & 3^n \bmod 20 \in \{1, 3, 9, 7\} \\ \implies & 3^{n + 1} \bmod 20 \in \{3, 9, 7, 1\} \end{aligned}$

Thus, the tens digit modulo $2$ is always even.

Relevant wiki: Induction - Introduction

Assume that there exists a power of 3 whose tens digits is odd. Let the first such power of 3 be $3^n$ . Then $3^{n-1}$ has an even tens digit. Consequently, multiplying this by 3 can only bring the tens digit to an odd number if the ones digit * 3 carries a 1 to the tens place. Thus the ones digit would have to be 4, 5, or 6. But the ones digit can only be 1, 3, 7, or 9, contradiction.

Relevant wiki: Induction - Introduction

We use induction 3^n has that property and the first steps are already above so if it is true for every number 3^k until 3^n (k, n integer): first we know that every power of 3 ends with one of those: 1,3,7,9 and we know that 3*2s is still divisible by 2 then the only problem for the 2nd digit evenness is the redemander of the first multiplication and that one is based on the numbers above 0,2 witch both are divisible by 2, witch means if 3^n has the proprety then 3^(n+1) has it too. Q.E.D.

We use the fact that $3^4 \equiv 1 \pmod{20}$ .

(1) For every $n=4k$ ,

$3^{4k}\equiv 1 \pmod{20}$

(2) For every $n=4k+1$ ,

$3^{4k+1}\equiv 3 \pmod{20}$

(3) For every $n=4k+2$ ,

$3^{4k+2}\equiv 9 \pmod{20}$

(4) For every $n=4k+3$ ,

$3^{4k+3}\equiv 27\equiv 7\pmod{20}$

Therefore for all $n$ , the tens digit of $3^n$ is even.

The one's digit of $3^n$ is only odd, 1,3,7,9. For values of $3^n$ ending in 1 or 3 there is no carry over to the 10's and for values of $3^n$ ending in 7 or 9 there only carry over to the 10's is 2. Starting with $3^2$ the result is 27, the 10's digit being even, causing all successive multiples be and odd number, 3, times an even number resulting in only even numbers for the 10's place,

the powers of 3 that have been given always have an even 10s unit. Multiplying the 10s therefore, by whatever number will always make it even. you can see that the units cycle through 1, 3, 9, and 27. None of these affect the 10s except the 27 which will add 2, therefore not affecting if it is odd or even. I hope this wasn't confusing

the last digit of 3's powers are: 1, 3, 7, 9

step1:- the carry, when these digits are multiplied by 3, is: 0, 2

step2:- when these carries are multiplied by 3 and the carry from step1 is added: always even.

There are two patterns we could notes: the first right number belongs to $\{7, 1, 3, 9\}$ and the second right number is always in the form $2\times n$ ord $2\times n+2$ with $n$ a positive integer, because moltiplicating each unit numbers $\{7, 1, 3, 9\}$ by 3 we have respectively the set $\{21, 3, 9, 27\}$ . Why? Because we have two cases: we should add nothing to the even decim number (wich is in the form $3\times 2\times n$ ) or add $2$ to it: in both the situations the second last digit remains even.

each power of 3 is obtained by the preceeding one multiplying by 3. The second digit of the third power (27) is even (2), while the first digit of any power of 3 ends in 1 - 3 - 9 or 7 When multiplying by 3, the first digit either doesn't give any charge (1 and 3) or gives 2 as a charge (9 or 7), while the second digit starting as a 2 on the third power and thus starting as an even digit, when multiplying by 3 gives in any case an even either because "even by 3" (first digit 1 - 3: no charge) or because "even by 3 plus 2" (first digit 7 - 9 : charge 2 to the second digit) [Sorry for the bad English]

3 x 6561 = 19683 8. 4 x 6561 = 26244 4. Keep doing the maths and itll add up to be an even number 100% of the time.

Note the units digit cycles through $7, 1, 3, 9$ and repeats (powers of $3\pmod {10}$ ). Then notice that the carry into the $10$ 's column is always either $0$ or $2$ and therefore always even. Finally, the $10$ 's column starts with $2$ . To get to the $10$ 's column of the next number, multiply by $3$ and add the carry. $\text{ODD}\times\text{EVEN}+\text{EVEN}=\text{EVEN}$ . By induction it'll always be even for $n>2$ .