What is the average of all positive integers that have four digits when written in base 3, but two digits when written in base 6? Write your answer in base 10.

4 digits base 3

smallest 1000 = 3^3 = 27

biggest 2222 = 2*27+2*9+2*3+2 = 80

3 digits base 6

smallest 100 = 36

biggest 1000-1 = 6^3-1 = 4095

So the intersection of these is 36 to 80 there are 80-36+1 = 45 of them

sum of numbers from 36 to 80 = (45/2)(72+44) = 45/2*116= 45*58 = 2610

average = 2610/45 =58

A more straight forward approach would be to say that of there are 45 numbers the median on is in the 46/2 = 23rd position. In this case the median and the mean will be the same.

36+23-1 = 58 Just as expected.

So the answer is 58 base 10

It has just been pointed out to me that I have answered for 3 digits base 6 instead of 2 digits.

You will need to adjust what i have done to get the correct answer.