1. The numbers $62,63,64,65,66,67,68,69$ and 70 are divided by, in some order, the numbers $1,2,3,4,5,6,7,8$ and 9 , resulting in nine integers. The sum of these nine integers is $S$. What are the possible values of $S$ ?
SOLUTION
The only multiple of nine in the first list is 63 and so this must be divided by 9 , yielding a quotient of 7 .

The only multiple of eight in the first list is 64 and so this must be divided by 8 , yielding a quotient of 8 .
The only remaining multiple of seven is 70 and so this must be divided by 7 , yielding a quotient of 10 .

Similarly we must have 66 divided by 6,65 divided by 5 and 68 divided by 4 , yielding quotients of 11,13 , and 17 respectively.
The only remaining multiple of three is 69 and so this must be divided by 3 , the only remaining multiple of two is 62 and so this must be divided by 2, yielding quotients of 23 and 31 respectively.
Finally we must have 67 divided by 1 , yielding a quotient of 67 .
This means that the value of $S$ is unique and $S=7+8+10+11+13+17+23+31+67=187$.