Question

$5901AB04$ is an eight-digit number divisible by $792$ then $A+B$
$\left( A \right)\text{ 6}$
$\left( B \right)\text{ 7}$
$\left( C \right)\text{ 8}$
$\left( D \right)\text{ 9}$

Answer 1

Hint: In this question we have been given with an eight-digit number which is given as $5901AB04$ and the number is divisible by $792$. We will use the divisibility rule of $8$ and $9$ to get the required cases to solve for the value. We will also use the divisibility rule of the number $11$ to get the final case and then from all the cases, we will find the required solution.

Complete step-by-step answer:
We have been given with eight-digit number $5901AB04$ is divisible by $792$,

We find prime factors of 792 and get:
\[\Rightarrow 792=2\times 2\times 2\times 3\times 3\times 11\]
On multiplying the factors, we get:
\[\Rightarrow 792=8\times 9\times 11\]
Consider case $1$:
From divisibility rule of $8$ we know if last three digits of any number is divisible by $8$ then whole number is also divisible by $8$.
Here, Last three digits are $B04$, Here Value of could be $1,3,5,7$ or $9$ since $104$, $304$, $504$, $704$ and$904$ are divisible by $8$.
Consider case $2$:
From divisibility rule of $9$ we know if sum of all digits is divisible by $9$ then whole number is also divisible therefore, we get:
$\Rightarrow 5+9+0+1+A+B+0+4$
On simplifying, we get:
$\Rightarrow 19+A+B$
Now the numbers divisible by $9$ are $9,18,27,36,45$
So, $A+B$ has to be $8$ or $17$ since their sum with $19$ gives a multiple of $9$.
Therefore, we can say that the values of $A$ and $B$ will be as $3,5,8,9$ since $3$ and $5$ give sum as $8$ and $8$ and $9$ give sum as $17$.
Consider case $3$:
We know from divisibility rule that if the alternating sum of the digits is divisible by $11$ then whole number is also divisible by $11$ therefore, we get:
$\Rightarrow 5-9-0-1-A-B-0-4\text{ }$
On simplifying, we get:
$\Rightarrow A-B-9$
now from the earlier case we know that $A+B=8$, on rearranging, we get $A=8-B$, on substituting the value, we get:
$\Rightarrow 8-B-B-9$
On simplifying, we get:
$\Rightarrow -2B-1$
Now the possible values of $B$ are $5$ or $9$ from the earlier case therefore at $B=5$, we have:
$\Rightarrow -2\left( 5 \right)-1=-10-1=-11$, which is a number divisible by $11$.
And at $B=9$, we have:
$\Rightarrow -2\left( 9 \right)-1=-18-1=-19$, which is a number not divisible by $11$.
Therefore, from case $1,2$ and $3$, we can say $A+B=8$, which is the required solution

So, the correct answer is “Option C”.

Note: It is to be remembered that the divisibility rules we used were of the factors of the given divisor. since $8,9$ and $11$ are factors of the divisor, they will also be the factors of the dividend. In these types of questions, the solution should be split into various cases and then the use case should be applied to get the final solution.