Question

If the seven digit number $8a2b3c5$ is divisible by $99$ find the number of possible values of $a+b+c$.

Answer 1

$8a2b3c5$ is divisible by $99$

$\therefore$ it should be divisible by $9$ and $11$

for $9\Rightarrow 8+a+2+b+3+c+5$

$=18+a+b+c$

the sum should be divisible by $9$

$\therefore a+b+c=0,9,18,27$ (max value of 3 digits sum = 27)

For $11$ sum of odd sum of even = should be $0$ (or) divisible by $11$

$\therefore (8+2+3+5)-(a+b+c)=0$ or multiple by $11$

$18-(a+b+c)=0$ (or) multiple of 11.

we have $4$ possibilities for $a+b+c$ from (i)

$\therefore 18-(a+b+c)=0$ (or) multiple of $11$

$a+b+c\ne 0,9,27$

$18-18=0$

$\therefore$ the only possible value of $a+b+c$ is $18$

$\therefore a+b+c=18$