Aptitude - Numbers - Discussion

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.

@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.

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.

264, 396, 462, 792, 968, 2178, 5184, 6336
264, 396 and 792 are 2, 3, 6 multiples of 132 : leave them
7 multiple of 132 = 924 Hence omit 968
Similarly omit 462

2178 is not divisible by 4 ( last 2 digits)
5184 fails 11 divisibility test ((5+8=13); (1+4=5); 13-5 =8 is not a 11 multiple
6336 passes 3, 4, 11 divisibility tests
264, 396, 798 pass by observation divisibility by 132
6336 by detailed 3,4,11 divisibility pass

For 6336 alternate method is 50 times 132 is 6600
6600 -6336 = 264 - a multiple of 132: hence it passes.

If a number "A" is divisible by all factors of another number say "B", then "A" is divisible by "B"
thats the logic

example factors of 6= 2*3

any number that is divisible by both 2 and 3 will be divisible by 6. always
in this quetsion we have factorized 132 in to 4*3*11 so that we can check whether all numbers are divisible by them.. if yes then the number is divisible by 132.

ps: shortcut, u need to learn the divisibility rules for 11 and 3, google them

Hi,
its very easy to find whether a given no.is divisible by 3,4,11..
if sum of digits in the gn no. is divisible by 3..then the gn no.s divisible by 3.
if the last 2 digits('1's &'10' )is divisible by 4..then the whole no. is divisible by 4..
likewise,if sum of no.s in oddposition - sum of no.s in even position=0 then..the gn no is divisible by 11..
so..it is easy way to find the answer..
it s enough to know the simple rules of dividing process.

Looking at the series we can conclude it is series

132 --
264 (132+ 132)
396 (264+132)
462 (396+132 = 528)not equal
792 (660+132)
968( 792+132= 924) not equal
therefore 924+132 = 1056
1056+1056 = 2112
2178 (1056 + 1056 = 2112) not equal
5184 (2112 +2112 + 1056= 5280) not equal
6336 (6204+132 = 6336) equal

You have told that the number divisible by all the three number 4, 3 and 11, then the number is divisible by 132 also.

but,462 is divisible by 4,3 and 11,so according to your logic,so why 462 is not in the category that divisible by 132.WHY? please reply very soon

@Sunny

462 is not divisible by 4.

The logic is some what like this...If a number is divisible by p and q then its also divisible by p*q(product of two nos),to the condition that p and q are co-primes..extending the rule to one more level..!!

Look. 4=2*2. If it is already can divide by 4, then what is the need to check it with again 2 because as we know If a number "A" is divisible by all factors of another number say "B", then "A" is divisible by "B".