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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Question : By interchanging the digits of a two-digit number, we get a number which is four times the original number minus 24. If the digit at the unit's place of the original number exceeds its digit at ten's place by 7, then the original number is:
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