Find the value of $ a + b + c $ for which $ 708,a6b,8c9 $ is a multiple of $ 99 $.

Question

Can anyone help me with this? I know that $ 99 = 11 \times 3 \times 3 $, but I don’t know what to do next.

Suppose that $ a $, $ b $ and $ c $ are distinct digits for which the number $ 708,a6b,8c9 $ is a multiple of $ 99 $. Find $ a + b + c $.

Answer 1

Hint:

Since it is a multiple of $9$, we must have

$$7+0+8+a+6+b+8+c+9 \equiv 0 \mod 9$$

Since it is a multiple of $11$, we must have

$$7-0+8-a+6-b+8-c+9 \equiv 0 \mod 11$$

Also, $a+b+c \leq 27$