Aptitude - Numbers - Discussion

What is the divisibility property of 11?

Please, anyone, explain.

@All.

Here, We can take 6, but again 6 having the factors of 2*3.

Please explain the answer shortly.

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

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

It's answer should be 5.

Can you give divisibility rule of more numbers?

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.
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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.

I don't understand why only we take 4, 3 and 11.

How I would be wrong if I take 2*11*3?

This is the wrong answer. In this logic, 462 can also divide by 4, 3, 11.

@Koala,

I guess you missed out on the fact that we in here are talking about co-primes while 4 and 2 don't happen to be co-prime factors. 88 = 8 * 11.

And 9988 is not divisible by 8.