Question : The six-digit number $\mathrm{N = 4a6b9c}$ is divisible by $99$, then the maximum sum of the digits of $\mathrm{N}$ is:
Option 1: 18
Option 2: 36
Option 3: 45
Option 4: 27
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Correct Answer: 27
Solution :
We have, the six-digit number $\mathrm{N = 4a6b9c}$ is divisible by $99$.
Divisibility rule for $9$,
A number is divisible by $9$ if the sum of digits is divisible by $9$.
As $\mathrm{N = 4a6b9c}$ is divisible by $9$.
The sum of the digits $=\mathrm{(4+a+6+b+9+c)}$
The sum of the digits $=\mathrm{(19+a+b+c)}$
$\mathrm{(19+a+b+c)}$ is divisible by $9$ _________(i)
Divisibility rule for $11$,
A number is divisible by $11$ if the difference between the sum of the digits at even places and the sum of the digits at odd places is divisible by $11$.
As $\mathrm{N = 4a6b9c}$ is divisible by $11$.
$\mathrm{(4+6+9)-(a+b+c)}$ is divisible by $11$
$\mathrm{19-(a+b+c)}$ is divisible by $11$ _________(ii)
From the equation (i) and (ii),
$\mathrm{(a+b+c)=8}$
The maximum sum of the digits $=19+(a+b+c)=19+8=27$
Hence, the correct answer is $27$.
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