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If n and r are positive integers and r is the remainder when [#permalink]
30 Aug 2006, 10:08

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If n and r are positive integers and r is the remainder when 27 is divided by n, r=?
1). When 20 is divided by n, remainder is 7.
2). When 27 is divided by n, quotient is 2, remainder is r.

Re: tough remainder DS [#permalink]
30 Aug 2006, 10:30

apollo168 wrote:

If n and r are positive integers and r is the remainder when 27 is divided by n, r=? 1). When 20 is divided by n, remainder is 7. 2). When 27 is divided by n, quotient is 2, remainder is r.

St1:
20 = nx + 7
i.e nx must be 13. So either n = 1 or n = 13. But n must be greater than 7 to get a remainder of 7. So n = 13.
Reaminder of 27/13 = 1: SUFF

St2:
27 = 2n + r
So r must be an odd integer. So
r =1 for n = 13
r = 3 for n = 12
r = 5 for n = 11
r = 7 for n = 10
: INSUFF _________________

St1: 20 = nx + 7 i.e nx must be 13. So either n = 1 or n = 13. But n must be greater than 7 to get a remainder of 7. So n = 13. Reaminder of 27/13 = 1: SUFF

St2: 27 = 2n + r So r must be an odd integer. So r =1 for n = 13 r = 3 for n = 12 r = 5 for n = 11 r = 7 for n = 10 : INSUFF

Statement 1 is sufficient as Dahiya has pointed out.

Statement 2, upon closer inspection, is just restating the question. No new information can be attained. What necromonger referred to as "twin trouble." Insuff.

I arrived upon A as well but without much of the hassle or even putting in numbers Should've prolly double checked!!!

Acc to the problem statement we have to find r if 27 = nk + r.

Now the first statement gives you 20 = nk + 7, so by substituting nk = 13 in the first statement we can solve for r. (since both should be true for all k). Therefore, SUFF.

In the second statement we get 27 = 2n + r, which is really a specific solution for the problem statement itself - like psdahiya says, no new information. INSUFF.

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