24. Eight consecutive three-digit positive integers have the following property: each of them is divisible by its last digit. What is the sum of the digits of the smallest of these eight integers?
A 9
B 10
C 11
D 12
E 13
Solution
$\mathbf{E}$

Since we cannot divide by zero, the eight numbers must have the form $A B n$ where $n$ runs from 1 to 8 or from 2 to 9 . However if $A B n$ is divisible by $n$ for each $n$ from 2 to 9 , it is also divisible by $n$ for each $n$ from 1 to 8 and $A B 1$ is smaller than $A B 2$. Hence our solution will use the digits 1 to 8 . Note that $A B n$ is divisible by $n$ if and only if $A B 0$ is divisible by $n$. So we require the smallest number $A B 0$ which is divisible by $1,2,3,4,5,6,7,8$. The LCM of these 8 numbers is $8 \times 7 \times 5 \times 3=840$. The eight numbers are then $841,842,843,844,845,846,847,848$. The digit sum of 841 is 13 .