Aptitude - Numbers - Discussion

Factor Found of 132=4*11*3.

Divisibility Rule
For 4 -Divided last two-digit with 4.
For 3 -Sum all the number
For 11-Add all the even place digit and all odd place digit . If the difference is 0,11 the number will divisible.

Example 1. 264
From 4 = Last two digit is 64
64/4 Divisible.
From 3 = 2+6+4
12 Divisible .
From 11 = (2+4)-(6)
0 divisible.
Here all are divisible.

Example 2. 462.
From 4 = Last two digit is 62.
62/4 Not Divisible.
From 3 = 4+6+2.
12 Divisible.
From 11 = (4+2) -(6).
0 divisible.

This number is Not Divisible.

But if we consider 132=11*6*2.

Then the numbers divisible by 11, 6, 2 are 396, 462, 792, 2178, and 6336 tell me then why the answer is not 5 instead of 4 anyone please tell me.

@All.

Here is my explanation.

At first, we are having the number 132, which can be written as 4 * 3 * 11.

It can be obtained by 'L division' method. For each number, we are first checking, whether it can be divided by 4, if yes proceeding to the next number 3, if yes, and then to 11. If a number got divided by 4, 3, or 11 and it might be divided by 136.

And in between, if any of the condition becomes fails, there is no point in checking the number divisible with another number. So, we ignore the number.

In order to make the calculation faster, we are taking the minimum product representation of the divisor.

@Rushikesh.

Divisibility rule for 11: Difference between the sum of digits at odd places and sum of digits at even places is either 0 or a number that is divisible by 11.

It's simple and easy.

Total nos = 8,
132/8 = 16.5.

As given, we can do 16 * 8 = 128.

132 - 128 = 4.

@All.

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

We can tell this as prime factorization. So, the prime factors of 132 is 4, 11, 3.

I think the answer is 7.

Only 968 & 5184 are not divisible by 132.

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

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

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