In every power of 3 the tens digit is an even number

How to prove that in every power of $3$, with natural exponent, the tens digit
is an even number?

For example, $3 ^ 5 = 243$ and $4$ is even.

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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.

HINT $\rm\quad 3^4 = 1\ (mod\ 20)\ \Rightarrow\ 3^n\ =\ 20\ k + \{1,3,9,27\}$

Moron’s answer is very elegant.

Here is an easy but tedious way of dealing with many similar questions: For any $n$ and $k$, the last $k$ digits of the numbers in the sequence $n^1,n^2,n^3,\dots$ eventually repeat, so we only need to check a few cases.

For the question at hand, note that $3^2=9=10-1$, so $3^{20}=(10-1)^{10}$ has the form $100k+1$ for some $k$. This means that the last 2 digits of powers of 3 repeat each 20 numbers. So we just need to check the last two digits of $3^0,3^1,\dots,3^{19}$, and this can be done very quickly even by hand: $$01, 03, 09, 27, 81, 43, 29, 87,\dots,89,67.$$

Case 1: $n = 2k; k$ even.

$3^{2k} = 9^{k} = (10 – 1)^{k} = 10^{k} – {k \choose 2}^{k-1} + …. + {k \choose k-2}10^2 – {k \choose k – 1}10 + 1 = N*100 – k*10 + 1$. $k$ is even so the second digit is even.

Case 2: $n = 2k; k$ odd.

$3^{2k} = 9^{k} = (10 – 1)^{k} = 10^{k} – {k \choose 2}^{k-1} + …. – {k \choose k-2}10^2 + {k \choose k – 1}10 – 1 = N*100 + k*10 – 1$. $k$ is odd so $k*10 – 1 = (k – 1)*10 + 9$ so the second digit is even.

Case 3: $n = 2k + 1; k$ even.

$3^{2k} = 9^{k} = (10 – 1)^{k} = 10^{k} – {k \choose 2}^{k-1} + …. + {k \choose k-2}10^2 – {k \choose k – 1}10 + 1 = N*100 – k*10 + 1$.

$3^{2k+1} = $M*100 – 3k*10 + 3$. As $k$ is even the second digit is even.

Case 4: $n = 2k + 1; k$ odd.

$3^{2k} = 9^{k} = (10 – 1)^{k} = 10^{k} – {k \choose 2}^{k-1} + …. – {k \choose k-2}10^2 + {k \choose k – 1}10 – 1 = N*100 + k*10 – 1$.

$3^{2k + 1} = M*100 + 3k*10 – 7 = M*100 + (3k -1)*10 + 7$. As $3k -1$ is even, the second digit is even.

This also tells you what the first digit is: 1, 9, 3, or 7. Which gives a hint that might be a more efficient method.

Method 2:

$3^4 = 81 = 8*10 + 1$

So if $n = 4m + k; k = 0,1,2,3$

$3^n = (3^4)^m 3^k = (8*10 + 1)^m 3^k = (M*100 + 8*m*10 + 1)\{1,3,9,27\}$

$= N*100 + 8*m*\{1,3,9,7\}*10 \{+ 2*10\} + 1*\{1,3,9,7\}$

Or more elegantly as Bill Dubuque pointed out

$3^4 = 1 \mod (20) \implies 3^{4m + k} = (3^4)^m 3^k = 3^k \mod 20 \implies$

$ 3^n = \{1,3,9,7\} \mod 20$.

It’s actually interesting.

If you do a table of multiples of 1, 3, 7, 9 modulo 20, you will find a closed set, ie you can’t derive an 11, 13, 17 or 19 from these numbers.

What this means is that any number comprised entirely of primes that have an even tens-digit will itself have an even tens-digt. Such primes are 3, 7, 23, 29, 41, 43, 47, 61, 67, 83, 89, 101, 103, 107, 109 to a hundred.

If the tens-digit is odd, then it must be divisible by an odd number of primes of the form 11, 13, 17, 19 mod 20. Any even number of these would produce an even tens digit.

Considering $3^k \bmod 20$, we know from the Carmichael function that the values will cycle on a pattern of $\lambda(20)=4$ (or less). This is short enough to just evaluate:

$
3^1\equiv 3 \bmod 20 \\
3^2\equiv 9 \bmod 20 \\
3^3\equiv 7 \bmod 20 \\
3^4\equiv 1 \bmod 20
$

and the cycle repeats. As you can see, none of these values are greater than $10$ so the tens digit of any power of $3$ will be even (because these values will always be added to some multiple of $20$).

A slightly different approach:

Let $n$ be the exponent of the variable power of $3$. Therefore, the power of $3$ takes the form $3^n$.
Now, $n$ can either be even or odd. Hence two cases are formed:-

Case1: $n$ is even.

Since $n$ is even, let us say it is of the form $2k$.
Therefore, $3^n = 3^{2k} = (3^k)^2$. This means that is $n$ if even, $3^n$ is a square.
Notice that the even powers of $3$ result in a number with the last digit $\in \{1,9\}$
And $(2m + 1)^2 \equiv 1(mod 4)$
$\Rightarrow 3^n \equiv 1(mod4)$
$\Rightarrow 3^n -1 \equiv 0(mod4)$
$\Rightarrow 4 |(3^n-1)$.
We know that, if $4$ divides a number, it must divide $(10($$the$ $second$ $last$ $digit$$) + $$the$ $last$ $digit$$)$
Therefore, if $4 |(3^n-1)$,$ $ $ $ $ $ $ $ $ $ $ $ $4|(10($$the$ $second$ $last$ $digit$$) + $$the$ $last$ $digit$$)-1$
$\Rightarrow 4|(10($$the$ $second$ $last$ $digit$$) + 1-1)$ and $4|(10($$the$ $second$ $last$ $digit$$) + 9-1)$.
$\Rightarrow 4|(10($$the$ $second$ $last$ $digit$$) + 0)$ and $4|(10($$the$ $second$ $last$ $digit$$) + 8)$.
Since, $4| 0$ and $4| 8$, $4|(10($$the$ $second$ $last$ $digit$$)$
This is true only if $the$ $second$ $last$ $digit$ is even.

Case2: $n$ is odd.

Since $n$ is odd, let us say it is of the form $2k+1$.
Therefore, $3^n = 3^{2k+1} = (3^k)^2.(3)$. This means that if $n$ is odd, $3^n$ is a $($square $* 3)$.
Notice that the odd powers of $3$ result in a number with the last digit $\in \{3,7\}$
And $(2m + 1)^2 \equiv 1(mod 4)$
$\Rightarrow (2m + 1)^{2p+1} \equiv (2m + 1)(mod4)$
$\Rightarrow 3^n \equiv 3(mod4)$
$\Rightarrow 3^n -3 \equiv 0(mod4)$
$\Rightarrow 4 |(3^n-3)$.
We know that, if $4$ divides a number, it must divide $(10($$the$ $second$ $last$ $digit$$) + $$the$ $last$ $digit$$)$
Therefore, if $4 |(3^n-3)$,$ $ $ $ $ $ $ $ $ $ $ $ $4|(10($$the$ $second$ $last$ $digit$$) + $$the$ $last$ $digit$$)-3$
$\Rightarrow 4|(10($$the$ $second$ $last$ $digit$$) + 3-3)$ and $4|(10($$the$ $second$ $last$ $digit$$) + 7-3)$.
$\Rightarrow 4|(10($$the$ $second$ $last$ $digit$$) + 0)$ and $4|(10($$the$ $second$ $last$ $digit$$) + 4)$.
Since, $4| 0$ and $4| 4$, $4|(10($$the$ $second$ $last$ $digit$$)$
This is true only if $the$ $second$ $last$ $digit$ is even.

Therefore, the tens digit of any power of $3$ is always even.

By brute force:

The powers of $3$ modulo $100$ are periodically

$$01,03,09,27,81,43,29,87,61,83,49,47,41,23,69,07,21,63,89,67,01,\cdots$$


By induction:
$$3^0\bmod20=1\in\{1,3,9,7\}$$
and
$$3^n\bmod20\in\{1,3,9,7\}\implies 3^{n+1}\bmod20\in\{3,9,7,1\}.$$

So the tenth digit modulo $2$ is always $0$.