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In CAT ,in Quantitative section,we can save a lot of time by eliminating the answer options,and divisibility checks comes in handy in this process.especially if you know the checks for prime numbers like 11,13,17...
So lets check out this quantitative methods...

Divisibility Checks for 11

We know that rem (10^1/11)=-1
Therefore the rem(10^2/11)=+1

So for a no. in the form ‘abcdef’, the divisibility can be checked by taking the diff. b/w
Sum of the odd numbered digits and even numbered digits.Check whether that difference is divisible by 11.If that is divisible then the large no. is divisible by 11.

Eg. Take the no. 1331
D
Odd numbered dig. Are 1,3 from right, sum =4
Even numbered dig. are 3,1 from right , sum =4
Difference of the sum is 0. So its divisible by 11

Divisibility Checks for 13

We know that rem (10^3/13)=-1
Therefore the rem(10^6/13)=+1

So for a no. in the form ‘abcdef’, the divisibility can be checked by taking the diff. b/w
abc and def. Check whether that difference is divisible by 13.If that is divisible then the large no. is divisible by 13.

Eg. Check the divisibility of 214175?
Split the no. in to 2 as 214 and 175.
214-175= 39
Since the difference 39 is divisible by 13, the no. 214175 is also divisible by 13.

Divisibility Checks for 17

We know that rem (10^9/17)=-1
Therefore the rem(10^18/17)=+1

So for checking the divisibility of 17,we need to take blocks of 9 digits and do the same procedure as we do for that of 7 or 13.

The knowledge of divisibility of various numbers will help you to eliminate some answer options without doing a single calculation. This types of small time savings can make a big difference in the whole exam.(As I said earlier, reaching answer quickly is the important thing…..)

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Comments (2)

On November 20, 2013 at 11:58 PM , Unknown said...

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On December 25, 2014 at 3:41 AM , ravisharmas said...

Very informative post..
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