One day on the board, Ms. Math wrote the digits 1 to 9. She then wrote a
certain number of fives and also a number of eights. The number of fives and the
number of eights were not necessarily the same. The mean (average) of all the
digits on the board is 6.4.
Determine the smallest number of digits that can be on the board
Since the average is 6.4 the total must be divisible by 32.
So we have to find number of fives (f) and number of eights (e)
satisfying the equation 5f+8e= 32k-45.
Trying to solve for k=2,3, ... etc and checking the integer solutions to fit the average TOTAL / (F+E) =6.4 we are ok when k=6:
5f+8e=147
f=(147-8e)/5
e=4 or 9 or 14, but only 14 provides the correct average.
ANSWER: 14 EIGHTS + 7 FIVES + (9 original digits) sum up to 112+35+45=192
192: (14+7+9)=192/30=6.4
Edited on April 30, 2018, 6:47 pm
my KISS solution
