Question

Which of the following numbers is divisible by 12?
$
  {\text{A}}{\text{. }}123452 \\
  {\text{B}}{\text{. }}79968 \\
  {\text{C}}{\text{. }}78968 \\
  {\text{D}}{\text{. }}79998 \\
 $

Answer 1

Hint: Here we go through by applying the divisibility rule of 12 to check which number is divisible by 12. And we know that if the number is divisible by 4 and 3 then the numbers are divisible by 12 and we know the divisibility rule of 3 and 4.

Complete step-by-step answer:

Here in the question we have to find the number which is divisible by 12 so we have to first know the divisibility rule of 12 i.e. Divisibility rule of 12 : Number divisible by both 3 and 4.
We know that Divisibility rule of 3 is when a number is divisible by 3 if the sum of the digits is divisible by 3. And the Divisibility rule of 4 is when a number is divisible by 4, if the last two digits are a multiple of 4.
By the above definition we will check the numbers which number is divisible by 12 or not.
Option (A) 123452
First we check the divisibility rule by 3 by adding their digits i.e. 1+2+3+4+5+2=17 and we know that 17 is not divisible by 3 so option A cannot be the answer.
Option (C) 78968
First we check the divisibility rule by 3 by adding their digits i.e. 7+8+9+6+8=28 and we know that 28 is not divisible by 3 so option C cannot be the answer.
Option (D) 79998
First we check the divisibility rule by 3 by adding their digits i.e. 7+9+9+9+8=42 and we know that 42 is divisible by 3 but the last two digits are not divisible by 4 so option D cannot be the answer.
Option (B) 79968
First we check the divisibility rule by 3 by adding their digits i.e. 7+9+9+6+8=39 and we know that 39 is divisible by 3 and last two digit is also divisible by 4 so option D is divisible by 12.

Hence option B is the correct answer.

Note: Whenever we face such a type of question the key concept for solving the question is to go through the divisibility rule of that number. So always keep in mind the divisibility rule of the numbers to find out the results easily.