Question

Which of the following numbers are divisible by $9$?
${\text{(A) 81}}$
${\text{(B) 99}}$
${\text{(C) 108}}$
${\text{(D) All of the above}}$

Answer 1

Hint: Here we have to use this property that a number is divisible by $9$ if the sums of all the digits in the number are divisible by $9$. So we have to check the option one by one, then we will conclude the required answer.

Complete step-by-step answer:
First we add the digits in the given number contained in the option, if they add up to be $9$ or add up to be a number which is divisible by $9$ then we can conclude that the number is divisible by $9$.
Consider ${\text{81}}$:
The sum of digits in the number $81$ are:
$8 + 1 = 9$
Since $9$ is divisible by $9$ we can conclude that $81$ is divisible by $9$
Again, consider $99$
The sum of digits in the number $99$ are
$9 + 9 = 18$
Since $18$ is divisible by $9$ we can conclude that $99$ is divisible by $9$
Now consider $108$:
The sum of digits in the number $108$ are
$1 + 0 + 8 = 9$
Since $9$ is divisible by $9$ we can conclude that $108$ is divisible by $9$

Option D is the correct answer.

Note: Here we use this property that if the sum of digits of a number is divisible by $9$ then that number is divisible by $9$ only holds for numbers which are multiples of \[3\].
If this same property is used for another number then the answer won’t be correct.
For example, consider the divisibility of $111$ by $3$, using the same method we find that this property also works with the number $3$.
Also, consider the divisibility of $52$ by $7$, using the same method we get the sum of digits of $52$ to be $7$, still the number $52$ is not divisible by $7$ because $7$ is not a multiple of $3$ therefore this property won’t work with $7$