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divisor, remainder et.al.

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divisor, remainder et.al. [#permalink] New post 16 Feb 2006, 18:29
Can any one explain how to find the ans. to these kind of questions:
I do not have the OA or OE

When a number N is divided by d1, the remainder is r1. When this number is divided by d2, the remainder is r2. What is the remainder when this number is divided by d1d2.
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Re: divisor, remainder et.al. [#permalink] New post 17 Feb 2006, 07:38
believe2 wrote:
Can any one explain how to find the ans. to these kind of questions:
I do not have the OA or OE

When a number N is divided by d1, the remainder is r1. When this number is divided by d2, the remainder is r2. What is the remainder when this number is divided by d1d2.


i digged someting here: http://www.gmatclub.com/phpbb/viewtopic.php?t=17675

go to the attachments, banerjee managed to have something, though i couldnot proved his method/formula.

i believe, honghu should have better idea. :P
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Re: divisor, remainder et.al. [#permalink] New post 17 Feb 2006, 08:46
Professor wrote:
believe2 wrote:
Can any one explain how to find the ans. to these kind of questions:
I do not have the OA or OE

When a number N is divided by d1, the remainder is r1. When this number is divided by d2, the remainder is r2. What is the remainder when this number is divided by d1d2.


i digged someting here: http://www.gmatclub.com/phpbb/viewtopic.php?t=17675

go to the attachments, banerjee managed to have something, though i couldnot proved his method/formula.

i believe, honghu should have better idea. :P


Thanks Professor, for researching and finding the link.

....but this example has got me even more confused.
Here is the text:
When a number is successively divided by two divisors d1 and d2 and two remainders r1 and r2 are obtained, the remainder that will be obtained by the product of d1 and d2 is given by the relation d1r2 + r1.

Where d1 and d2 are in ascending order respectively and r1 and r2 are their respective remainders when they divide the number.

In this case, the d1 = 8 and d2 = 11. And r1 = 3 and r2 = 7. Therefore, d1r2 + r1 = 8*7 + 3 = 59.

........in this example, if you try to do the reverse starting with 59 dividing it by either 8 or 11 whichever way, I don't see how you can get an r2=7

....I think I need some serious therapy now!!!
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 [#permalink] New post 17 Feb 2006, 12:16
Hmmm, I'm sorry to blow the trust but I don't have a formula. :oops: I'll work some more to see if I can get anything. But I agree it doesn't seem that the formula d1r2+r1 is correct.

Let's do some examples

10 mod 3 is 1
10 mod 2 is 0
10 mod 6 is 4
but 3*0+1 is not 4.

Let's see another

16 mod 5 is 1
16 mod 3 is 1
16 mod 15 is 1
5*1+1 is definitely not 1.

Not working. Or have I not understood it correctly?
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 [#permalink] New post 17 Feb 2006, 20:02
I'm not able to proof or derive that formula or even get it to work! :shock:
  [#permalink] 17 Feb 2006, 20:02
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divisor, remainder et.al.

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