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# When a number is successively divided by two divisors d1 and

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Director
Joined: 17 Dec 2005
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When a number is successively divided by two divisors d1 and [#permalink]

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08 Jan 2006, 12:37
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

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.

I please you to give me a concrete example/question for this rule.

Thanks
Director
Joined: 24 Oct 2005
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08 Jan 2006, 12:44
D1 = 3.. D2 =4

NUMBER = 5

5 mod 3 = 2
5 mod 4 = 1

3*1 + 2 = 5
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Director
Joined: 17 Dec 2005
Posts: 547
Location: Germany
Followers: 1

Kudos [?]: 40 [0], given: 0

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08 Jan 2006, 12:53
Sorry my fault.

I think of an GMAT-like question.

I don't think that they say n=..., d♫=... and dÂ².......
08 Jan 2006, 12:53
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# When a number is successively divided by two divisors d1 and

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