For a number to be divisible by 3 (multiple of 3), its digits must add to a multiple of 3.
I count
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
sets,
sets have 3 different digits,
Each of those
sets can be arranged in
different arrangements/permutations,
making
different three-digit numbers whose digits are all odd.
Among those
sets listed above,
there are also
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
three-digit numbers whose digits are all odd.
The remaining
of the
sets listed above contain only two different digits, one of them repeated.
Form each of those
sets, we can make
different three-digit numbers,
because there are 3 positions to place the unrepeated digit,
and that gives us another
three-digit numbers whose digits are all odd.
That makes a total of
three-digit numbers divisible by 3, whose digits are all odd.