Each of eight, consecutive, three digit numbers is divisble by its last digit. What is the sum of digits of the smallest number? 
Each of eight, consecutive, three digit numbers is divisble by its last digit. What is the sum of digits of the smallest number?

My approach:
Each number we can denote like that:
$\overline{abc},\ \overline{abc}+1,\ ...,\ \overline{abc}+7\quad$
,where$\ \overline{abc} \;$is the smallest.
Sum of these numbers: $\ 8(a+b+c)+28$
Also each number has the same second from the last digit, because there is no number divisble by $0$. That means sequance starts with number ending with digit $1\,$ or $2$.
This question is from this year's Kangaroo competition.
 A: As you have observed, you cannot have any number ending with a $0$, so your first two digits are fixed.
Division tests for numbers ending with $1,2,5,6$ add no information.
From the division test for $3$, you know that the first two digits must sum to a multiple of $3$.
From the division test for $7$, you know that the number obtained by truncating the last digit and subtracting away twice the last digit from it will give a multiple of $7$. This means that the first two digits have to also be a multiple of $7$.
From these two conditions, the first two digits must be a multiple of $21$, which means they can be $21,42,63,84$.
From the division test for $4$, you know the last two digits must form a number divisible by $4$.
The only one that fits this condition is the first two digits $84$.
So the smallest number is $841$, digit sum $13$.
For rigour, you need to exclude the possibility of the sequence starting with first digit $2$. If this were the case, the final number would end with $9$, and the division test for $9$ demands the sum of the first two digits should also be a multiple of $9$. But this would mean that $63$ is the only possibility, which would conflict with the rule for the last digit $4$ ($34$ is not a multiple of $4$). So we have verified that the above is the only sequence of numbers that meets the criteria.
A: Note that if the number is $\overline{abc} = 100a + 10b + c$, for it to be divisible by $c$ means $c \mid 100a + 10b$. As such, since $100a + 10b$ is the same for all $8$ numbers, to have it be divisible by their $8$ last digits means the $\operatorname{lcm}$ of those digits must divide $100a + 10b$.
Next, as you stated, the first ending digit must be either $1$, so it goes to $8$, or $2$, in which case it goes to $9$. For the first case, you have
$$\operatorname{lcm}(1,2,3,4,5,6,7,8) = 8(7)(5)(3) = 840 \tag{1}\label{eq1A}$$
$$\operatorname{lcm}(2,3,4,5,6,7,8,9) = 9(8)(7)(5) = 2\text{,}520 \tag{2}\label{eq2A}$$
However, $100a + 10b \lt 991$, so \eqref{eq1A} must be true, giving that the smallest number is $841$, with a digit sum of $8 + 4 + 1 = 13$.
