Aptitude - Numbers - Discussion

968 & 2178 is divisible by 11.

Factor found of 132 = 2*2*11*3.
So that's the reason for taking 4*11*3.

462 is also divisible by 11 and gives 42 so why we won't take it?

It's answer should be 5.

You can't use the 11*6*2 because the rule is that the pairings have to be coprimes with the exception of prime numbers.
11, 3, 4 don't have any factors with each other.

Dividing each of the numbers with 132 directly is a lengthy and difficult process. So we take coprime factors, i.e 4, 3, and 11
(That's nothing but 132 is the multiplication of 11*3*4 , so the no's which are divisible by these 3 are also divisible by 132)
Divisibility rule for 11 : Difference between sum of digits at odd places and sum of digits at even places is either 0 or a number that is divisible by 11.
Divisibility rule for 3 : sum of digits is divisible by 3.
Divisibility rule for 4 : last 2 digits is divisible by 4.
---------------------------------------------------------------------------------
Round 1 : Just by seeing we can clearly tell that 264, 396, 462,792, 6336 are divisible by 11.
Round 2 : Now we have to check if above 5 numbers are also divisible by 3 and 4.
But 462 is not divisible by 4, because the last 2 digits ie 62 is not divisible by 4. Rest 4 of these numbers above are divisible by both 3 and 4.

So, the final answer is 4.

What does it mean the required number of number is 4?

Please explain the answer shortly.

Very Helpful. Thanks.

Can you give divisibility rule of more numbers?

What is the divisibility property of 11?

Please, anyone, explain.