Question

If $34X68$ is to be divisible by $9$, then what is the value of the missing digit?
A.$5$
B.$6$
C.$9$
D.$8$

Answer 1

Hint: In order to find the missing digit which will make the whole number $34X68$ completely divisible by $9$, we should know the divisibility rules of $9$. According to the divisibility rule of $9$, the sum of all the digits of a number should be exactly divisible by $9$, then only the whole number will be exactly divisible by $9$.

Complete answer:
We are given with a number $34X68$, where a digit is missing which is represented by $X$.
Since, we know that according to the divisibility rule a number is completely divisible by $9$, if the sum of all the digits is divisible by $9$.
So, according to this statement, the sum of the digits of the numbers $34X68$ should be divisible by $9$, as $34X68$ given divisible by $9$.
Taking the sum of the digits:
$3 + 4 + X + 6 + 8$
$ \Rightarrow 7 + X + 14$
$ \Rightarrow 21 + X$ ……(1)
We need to choose a digit which makes the sum a multiple of $9$.
Let’s choose from the options given.
Starting from Option A, that is $5$:
Substituting $5$ in equation 1 and we get:
$
   \Rightarrow 21 + 5 \\
   = 26 \\
 $
Which is not a multiple of $9$, so it’s not possible.
Next for 2nd Option, that is $6$:
Substituting $6$ in equation 1 and we get:
$
   \Rightarrow 21 + 6 \\
   = 27 \\
 $
Which is a multiple of $9$ as we know that $9 \times 3 = 27$.
So, if we take $6$ in the missing digit, the number $34X68$ is to be divisible by $9$.
Hence, Option B is correct.

Note:
1.We can cross check our answers by substituting the digits one by one and dividing it by $9$, if the remainder gives zero then it is completely divisible by $9$, otherwise it is not divisible by $9$.
2.Similarly, for different numbers we have different divisibility rules: For example, a number to be divisible by $5$, the numbers should have $5$ or $0$ at its one’s place.