Aptitude - Numbers - Discussion

Discussion :: Numbers - General Questions (Q.No.6)

6. 

How many of the following numbers are divisible by 132 ?
264, 396, 462, 792, 968, 2178, 5184, 6336

[A]. 4
[B]. 5
[C]. 6
[D]. 7

Answer: Option A

Explanation:

132 = 4 x 3 x 11

So, if the number divisible by all the three number 4, 3 and 11, then the number is divisible by 132 also.

264 11,3,4 (/)

396 11,3,4 (/)

462 11,3 (X)

792 11,3,4 (/)

968 11,4 (X)

2178 11,3 (X)

5184 3,4 (X)

6336 11,3,4 (/)

Therefore the following numbers are divisible by 132 : 264, 396, 792 and 6336.

Required number of number = 4.


Reshma said: (Aug 23, 2010)  
I DIN'T understand the logic, Could u please explain it ?

Gaurav said: (Oct 5, 2010)  
These three numbers are the LCM of the Given number i.e. 132.

Hiji said: (Nov 3, 2010)  
Why can't we take 2*2*3*11 ? Why it is only 4*3*11 ?

Hema said: (Nov 11, 2010)  
Why have we taken only 4, 3, 11 why can't we take 2?

Priya said: (Nov 24, 2010)  
All the numbers are even, so not necessary to take 2 as a to check. By taking the minimum nos to use, we split 132 as 4*3*11

Deepa said: (Feb 4, 2011)  
Thats is nothing but the 132 is the multiplication of 11*3*4 the no which are divisible by these 3 are the answers.

Emela said: (Jun 9, 2011)  
I need a shortcut method to solve this problem please.

Jahanzeb said: (Jun 15, 2011)  
Logic is not clearly explain the answer.

Please explain me in easiest way.

Reuben said: (Jul 24, 2011)  
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

Sunny said: (Aug 18, 2011)  
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

Amar said: (Aug 23, 2011)  
@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..!!

Santhosh said: (Nov 6, 2011)  
But it is lengthy process. Divisible each number by all these three elements. Is there any shortcut to solve such type of questions ?

Aruna said: (Dec 1, 2011)  
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.

Kirankumar said: (Apr 20, 2012)  
We take 4 instead of 2*2. Because all the numbers which are divisible by 4 also divisible by 2. And take the multiplication of same numbers.

Ramu said: (Aug 31, 2012)  
I DIN'T understand the logic, Could u please explain it ?

Senthi said: (Sep 8, 2012)  
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

Prasanta said: (Sep 18, 2012)  
Thank you senthi. I agree with your answer.

Amulak said: (Oct 20, 2012)  
Hey guys.

First we take coprime factors, that's why we take 4,3,and 11 and not 2.

coprime nos are those no whose HCF is 1.

Hope now it is clear.;)

Dr R Vasudevan said: (Jan 23, 2013)  
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.

Samaptra Das said: (Feb 21, 2013)  
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".

Sandhya said: (Sep 12, 2013)  
Why can't we take 2*6*11?

Gopalakrishnan said: (Jan 25, 2014)  
LCM mean least. We want to minimise the factors as much as possible. So in the above request we split the 6 by their multiples. (i.e) 3 and 2.

Mufariq said: (Feb 20, 2014)  
Dear all its so easy.

See to all which number is divide complete on 132:264, 396, 462, 792, 968, 2178, 5184, 6336 and don't leave no reminder.

That is,

264/132=2, 396/132=21, 462/132=21.4 and so on.

Siranjeevi said: (Jun 13, 2014)  
1+3+2=6.

Ex.
2+6+4=12.

12 is divided by 6.

Manigundan said: (Jul 3, 2014)  
Could you please explain this logic? I need a shortcut for this.

Aamin said: (Sep 12, 2014)  
@Siranjeevi. If you go by that method answer comes 7 whereas answer is 4.

Supriya said: (Feb 4, 2015)  
@Siranjeevi:

As per your method, after adding it gets 12 18 12 18 23 18 18 18.

So except 12 nothing is divide by 6.

Is there any other fast method?

Amit said: (Jul 1, 2015)  
This is easy though time consuming.

Chethanya said: (Jul 9, 2015)  
Why can't we do it in this method.

264/132 = 2.
396/132 = 3.
462/132 = 3.5.
792/132 = 6.
968/132 = 7.31.
2178/132 = 16.5.
5184/132 = 39.27.
6336/132 = 48.

BY seeing above 4 numbers are divisible by 132.

Yuvaraj said: (Sep 10, 2015)  
88 can be factorized to 11*4*2. If 9988 is divisible by 11, 4 & 2 but not by 88. Can't understand this logic. Can any one explain?

Riya Singha said: (Sep 11, 2015)  
Why we take only 11, 4, 3? Please answer.

Vara said: (Sep 12, 2015)  
Here answer is 4. But according to @Siranjeevi 6, can you clarify?

Gaurav said: (Sep 24, 2015)  
11, 3, 4 is H.C.F of 132 that's why, we take it.

Manish said: (Sep 26, 2015)  
=132/11=22.

=132/4=33.

=132/3=44.

Srinu said: (Nov 16, 2015)  
I will give one number and give clearly explanation step by step 132 find out factors?

How it will takes factors reply fast?

Jake said: (Jan 13, 2016)  
I just got confused with the logic, there are many answers to this not one. I agree with dividing each by 132 to get a whole number with no reminder. Its more satisfying.

Ravip said: (Mar 14, 2016)  
Can you please explain me what is the divisibility condition for 11.

Devarsh said: (Jun 16, 2016)  
It's simple and easy.

Total nos = 8,
132/8 = 16.5.

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

132 - 128 = 4.

Anonynmous said: (Jul 9, 2016)  
We can't take 2 as it will not sufficient for 132 but if we take 2 then we will be checking for 66.

V.Sethuramalingam said: (Jul 25, 2016)  
Smart thinking and superb explanation @Chethanya.

Koala said: (Aug 27, 2016)  
@Yuvaraj.

I don't think the given solution I applicable for every number.

Eg: 9988 is divisible by all the factors of 88 (4 x 2 x 11).

But then too it is not divisible by 88.

Anvesha said: (Sep 8, 2016)  
@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.

Adil said: (Sep 21, 2016)  
This is the wrong answer. In this logic, 462 can also divide by 4, 3, 11.

Shiv said: (Oct 22, 2016)  
I don't understand why only we take 4, 3 and 11.

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

Sushmitha said: (Jul 2, 2017)  
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.

Reshma said: (Jul 5, 2017)  
Can you give divisibility rule of more numbers?

Nishant said: (Jul 18, 2017)  
462 is also divisible by 11 and gives 42 so why we won't take it?

It's answer should be 5.

Merlin said: (Aug 9, 2018)  
What does it mean the required number of number is 4?

Supraja said: (Sep 18, 2018)  
Please explain the answer shortly.

Lalitha said: (Sep 24, 2018)  
@All.

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

Rushikesh said: (Feb 26, 2019)  
What is the divisibility property of 11?

Please, anyone, explain.

Rajat Tamoli said: (Jun 7, 2019)  
Factor found of 132 = 2*2*11*3.
So that's the reason for taking 4*11*3.

Raju said: (Aug 24, 2019)  
968 & 2178 is divisible by 11.

Kalai said: (Sep 24, 2019)  
@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.

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