Question 978706
I count {{{highlight(41)}}} out of the {{{5*5*5=125}}} three-digit numbers whose digits are all odd.
I counted it through a shortcut, but it is still a messy, cumbersome way to count it.
Maybe the kids at the artofproblemsolving forum would have an elegant, streamlined way to count them.


HOW I COUNT THEM:
For a number to be divisible by 3 (multiple of 3), its digits must add to a multiple of 3.
I count {{{13}}} sets of 3 digits with all digits are odd, and they add to
a multiple of 3 (3, 6, 9, 12, 15, 18, 21, 24, or 27).
Those sets are:
{1,1,1} , {1,1,7} ,
{1,3,5} , {1,5,9} ,
{1,7,7} ,
{3,3,3} , {3,3,9} ,
{3,5,7} ,
{3,9,9} ,
{5,5,5} ,
{5,7,9} ,
{7,7,7} , and
{9,9,9} .


Among those {{{13}}} sets, {{{4}}} sets have 3 different digits,
Each of those {{{4}}} sets can be arranged in {{{3*2=6}}} different arrangements/permutations,
making {{{4*6=24}}} different three-digit numbers whose digits are all odd.


Among those {{{13}}} sets listed above,
there are also {{{5}}} sets made of just one single repeated digit,
and each one of those sets can be arranged just one way,
to form just one three-digit number,
so from them we can get another {{{5}}} three-digit numbers whose digits are all odd.


The remaining {{{4}}} of the {{{13}}} sets listed above contain only two different digits, one of them repeated.
Form each of those {{{4}}} sets, we can make {{{3}}} different three-digit numbers,
because there are 3 positions to place the unrepeated digit,
and that gives us another {{{4*3=12}}} three-digit numbers whose digits are all odd.


That makes a total of {{{24+5+12=highlight(41)}}} three-digit numbers divisible by 3, whose digits are all odd.