Question 978706: Hi tutors, can you help me answer this question? Thanks
A three-digit number has all digits odd. How many such numbers are divisible by three?
A. 29
B. 36
C. 39
D. 40
E. 41
Answer by KMST(5328)   (Show Source):
You can put this solution on YOUR website! I count  out of the  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  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.
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