# 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, 4 is even.)

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If you look at integer multiplication, only a finite number of digits of any two numbers number can influence any digit in the result. Enumerate all cases for those. –  Raphael Dec 11 '10 at 10:08

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.

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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: $$01, 03, 09, 27, 81, 43, 29, 87,\dots,89,67.$$
HINT $\rm\quad 3^4 = 1\ (mod\ 20)\ \Rightarrow\ 3^n\ =\ 20\ k + \{1,3,9,27\}$