1. A four-digit number, $n$, is written as ' $A B C D$ ' where $A, B, C$ and $D$ are all different odd digits. It is divisible by each of $A, B, C$ and $D$. Find all the possible numbers for $n$.
SolUTION
' $A B C D$ ' is a number composed of four out of five digits $1,3,5,7,9$. Consider two cases:
Case 1.9 is not one of the digits $A, B, C, D$.
Number ' $A B C D$ ' is a multiple of 3 , so $A+B+C+D$ is also a multiple of 3 . But $A+B+C+D=$ $1+3+5+7=16$, which is not a multiple of 3 . This is a contradiction, so there are no solutions in this case.

Case 2. 9 is one of the digits $A, B, C, D$.
Number ' $A B C D$ ' is a multiple of 9 , so $A+B+C+D$ is also a multiple of 9 .
Let $x$ be the digit not used in ' $A B C D$ '. Then $A+B+C+D=1+3+5+7+9-x=25-x$ and this must be a multiple of 9 . Hence $x=7$.
This means $1,3,5,9$ are the digits used, so ' $A B C D$ ' is a multiple of 5 . Hence $D=5$. We can rearrange $1,3,9$ as $A, B, C$ in six different ways.
All the possible numbers for $n$ are 1395, 1935, 3195, 3915, 9135, 9315. All of them are divisible by 1,3 and 9 (as their sum of digits is always 18), and also by 5 (as they all end in a 5).