+0

# Base numbers

0
94
2

What is the average of all positive integers that have four digits when written in base 3, but three digits when written in base 5? Write your answer in base 10.

May 21, 2021

#1
+1

OK, young person! Here is my attempt at this:

Here are ALL 3-digit numbers in base 5, which are 2-digit numbers in base 8:

100, 101, 102, 103, 104, 110, 111, 112, 113, 114, 120, 121, 122, 123, 124, 130, 131, 132, 133, 134, 140, 141, 142, 143, 144. If you count them you should get 25 such numbers. And their average is: [100 + 144] /2 = 122 in base 5, which is the exact median of the above list. Converting [100 + 144] / 2 =122  from base 5 to base 10, you get: [25 + 49] / 2 =37 in base 10.

May 21, 2021
#2
+1

What is the average of all positive integers that have four digits when written in base 3, but three digits when written in base 5?

4 digits base 3            3^3 to  3^4-1 (base 10)   which is   27 to 80 (base 10)

3 digits base 5            5^2  to  5^3-1 (base 10) which is   25 to 124 (base 10)

The intersection is            27 to  80  (base 10)

80-27+1= 54   numbers here

sum =  54/2 ( 27+80)

$$average = \frac{54(107)}{2*54}=53.5$$

May 22, 2021
edited by Melody  May 22, 2021