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Trick to find divisibility by 7,13,17,19,23,29.

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Trick to find divisibility by 7,13,17,19,23,29. [#permalink] New post 06 Aug 2012, 20:45
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A trick to find out the divisibility by 7,13,17,19,23,29 using check multipliers.

The trick:
step 1:Multiply the units digit by the check multiplier then add this to the original number after removing its tens digit.
step2:Check if this new number is divisible by our number if not then you can again do step 1.This can be done until the number is reduced to a number which can be easily said to be divisible by our number.

Finding check multiplier:
If we want to find the check multiplier for 13: Multiply 13 to get the units digit as 9. In this case 13*3 = 39.So the check multiplier is +1 the tens digit i.e 3+1 = 4.
The trick is to get the units digit as 9. This can be used even for 7 where 7*7=49 so the check multiplier is 5.

Checking with examples:
Is 169 divisible by 13?
First find the check multiplier for 13 = 13*3 = 39 = 3+1 = 4.
9*4 = 36 (multiply the units digit with the check multiplier. )
36+16 = 52 (Add 36 to 16.16 is obtained by removing the units digit from 169.)
Now we know 52 is a multiple of 13. So 169 is divisible by 13. If you are not sure if 52 is divisible by 13 then reduce 52 again. 5+2*4 =13. Now 13 is divisible by13! :-D

This same trick can be used for numbers like 7,17,19,23,29. For these numbers also try to get the units digit as 9 and hence the check multiplier.

Please let me know how you like this trick. I got this from a vedic maths book.
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Re: Trick to find divisibility by 7,13,17,19,23,29. [#permalink] New post 08 Aug 2012, 03:32
Thanks, thats good.
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Re: Trick to find divisibility by 7,13,17,19,23,29. [#permalink] New post 09 Aug 2012, 09:46
That's pretty interesting. Does the book also say how exactly this works? What are the properties of the numbers that allow this system to be applied?
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Re: Trick to find divisibility by 7,13,17,19,23,29. [#permalink] New post 10 Aug 2012, 23:36
I don't quite understand your explanation. Could you please define the check multiplier and elaborate on the process of using it?
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Re: Trick to find divisibility by 7,13,17,19,23,29. [#permalink] New post 28 Aug 2012, 17:21
I want to make a comment that as far as the GMAT is concerned, these divisibility rules are completely irrelevant. Now if you would like to have fun with divisibility, then that is a different story. Almost all of the divisions on the GMAT are very basic, and the goal of the test writers is not to test a student's ability to crunch numbers rapidly.

Cheers,
Dabral
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Re: Trick to find divisibility by 7,13,17,19,23,29. [#permalink] New post 15 Oct 2012, 20:12
dabral wrote:
I want to make a comment that as far as the GMAT is concerned, these divisibility rules are completely irrelevant. Now if you would like to have fun with divisibility, then that is a different story. Almost all of the divisions on the GMAT are very basic, and the goal of the test writers is not to test a student's ability to crunch numbers rapidly.

Cheers,
Dabral

I agree to this..
These divisibility tests may be useful in India's CAT exam but not for GMAT
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Re: Trick to find divisibility by 7,13,17,19,23,29. [#permalink] New post 02 Jul 2013, 08:33
while finding check multiplier, especially in case of 17, i think it is not tens digit+1 but apart from last digit remaining number+1
eg : 17 X 7 = 119 so check multiplier is 11+1=12
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Re: Trick to find divisibility by 7,13,17,19,23,29. [#permalink] New post 02 Jul 2013, 09:53
Expert's post
rahulsn84 wrote:
A trick to find out the divisibility by 7,13,17,19,23,29 using check multipliers.

The trick:
step 1:Multiply the units digit by the check multiplier then add this to the original number after removing its tens digit.
step2:Check if this new number is divisible by our number if not then you can again do step 1.This can be done until the number is reduced to a number which can be easily said to be divisible by our number.

Finding check multiplier:
If we want to find the check multiplier for 13: Multiply 13 to get the units digit as 9. In this case 13*3 = 39.So the check multiplier is +1 the tens digit i.e 3+1 = 4.
The trick is to get the units digit as 9. This can be used even for 7 where 7*7=49 so the check multiplier is 5.

Checking with examples:
Is 169 divisible by 13?
First find the check multiplier for 13 = 13*3 = 39 = 3+1 = 4.
9*4 = 36 (multiply the units digit with the check multiplier. )
36+16 = 52 (Add 36 to 16.16 is obtained by removing the units digit from 169.)
Now we know 52 is a multiple of 13. So 169 is divisible by 13. If you are not sure if 52 is divisible by 13 then reduce 52 again. 5+2*4 =13. Now 13 is divisible by13! :-D

This same trick can be used for numbers like 7,17,19,23,29. For these numbers also try to get the units digit as 9 and hence the check multiplier.

Please let me know how you like this trick. I got this from a vedic maths book.


Divisibility rules relevant for the GMAT are given here: math-number-theory-88376.html
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Re: Trick to find divisibility by 7,13,17,19,23,29. [#permalink] New post 29 May 2014, 03:14
My friend this trick isn't working for 17.

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Re: Trick to find divisibility by 7,13,17,19,23,29. [#permalink] New post 29 May 2014, 04:20
DonQuixote wrote:
That's pretty interesting. Does the book also say how exactly this works? What are the properties of the numbers that allow this system to be applied?


Separate the last digit from the rest of the number.
Let x be the units digit, and let y be the rest of the number.

So for any number N, N = 10y+x.

Using rahulsn84's method, we say that N should be divisible by prime p if y+Mx is divisible by p, where M is some 'check multiplier'.

10y + x = y + Mx
10y + x = 10y + 10Mx
x = 10Mx
0 = 10Mx - x
0 = x(10M - 1)

Now the question is, is x(10M - 1) divisible by p? We have defined the check multiplier such that 10M - 1 is divisible by p. So yes, p|(10M - 1).

Check multiplier for 7 is 5. 10*5 - 1 = 49, which is 7*7.
Check multiplier for 13 is 4. 10*4 - 1 = 39, which is 13*3.
Check multiplier for 17 is 12. 10*12 - 1 = 119, which is 17*7.
Check multiplier for 19 is 2. 10*2 - 1 = 19, which is 19*1.
Check multiplier for 23 is 7. 23*4 - 1 = 69, which is 23*3.
etc.

This works for all primes.

You can simplify (or complicate, depending on how you see things) this method by also realizing that you don't things to end in 9, but ending it with 1 works as well.

For example, for p = 17, instead of using a check multiplier of 12, we can use a check multiplier of 5. 17*3 = 51 = 10*5 + 1. In this case, instead of adding Mx, we just need to subtract Mx. That is, 10y + x = y - Mx. Doing the algebra, we get x(10M + 1) = 0.

But yes, as so many others pointed out, this will not be on the GMAT.
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Re: Trick to find divisibility by 7,13,17,19,23,29.   [#permalink] 29 May 2014, 04:20
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