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How to find a remainder?

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How to find a remainder? [#permalink] New post 19 Jun 2010, 18:13
is there any easy method especially when dealing with big numbers?
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Re: How to find a remainder? [#permalink] New post 20 Jun 2010, 21:23
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There are specific rules to remember divisibility by numbers up to 10.

1: All numbers
2: Even numbers
3: Sum of digits = 3
4: Divisible by 2 twice/Last two digits divisible by 4
5: Last digit = 5 or 0
6: Divisible by 3 and 2
7: No divisibility rule
8: Divisible by 2 thrice/last three digits divisible by 8
9: Sum of digits = 9
10: Last digit = 0

Now for bigger numbers, what you basically need to do is assume they are divisible and check for consistency with constrains given in the problem.

For example if a question asks if a number x is divisible by 12 and gives you additional constraints say:

x = 12a

And then check against restrictions. But I'd say that beyond this the method would be case specific. Hope this helps!
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Re: How to find a remainder? [#permalink] New post 20 Jun 2010, 21:28
Expert's post
sriharshagaruda wrote:
is there any easy method especially when dealing with big numbers?


You have an example perhaps?
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Re: How to find a remainder? [#permalink] New post 23 Jun 2010, 14:26
I had a similar question with a divisibility/remainder problem that I came across today from the MGMAT Number Properties question bank:

X is a positive integer. If X is divided by 11, the quotient is Y and the remainder is 3. If X is divided by 19, the remainder is 3. What is the remainder when Y is divided by 19?

The way to solve this problem is to write the two equations (X=11Y+3 and X=19Z+3) and set them equal to each other so (11Y+3=19Z+3 --> 11Y=19Z).

Then you know that 11Y is divisible by 19 because z is an integer, and specifically, Y is divisible by 19 since 11 is not. Thus, the remainder is 0.

Is it correct to say that the way to solve divisibility/remainder problems is to write 2 equations, set them equal to each other, and then make divisibility deductions?
Re: How to find a remainder?   [#permalink] 23 Jun 2010, 14:26
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