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If n and r are positive integers and r is the remainder when

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If n and r are positive integers and r is the remainder when [#permalink] New post 30 Aug 2006, 10:08
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A
B
C
D
E

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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.
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 [#permalink] New post 30 Aug 2006, 10:26
Given n, r +ve integers

r = ? if 27 = n * k + r

S1: 20 = n * m + 7

or 13 = n * m

As 13 is prime, m =1, n = 13
or m=13, n = 1
As divisor is greater than remainder (only if the numbers are +ve),

n > 7 => n = 13. Sufficient.

S2: 27 = n * 2 + r
If n = 10, r = 7,
If n = 11, r = 5 .. and so on.
Not sufficient.

Answer: A
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 [#permalink] New post 30 Aug 2006, 10:28
will go with D

from st 1. we have n =13
so 27/13 gives r = 1
SUFF

from st2: same as above
SUFF

so D
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Re: tough remainder DS [#permalink] New post 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.

Is it A?

Will explain if correct.
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 [#permalink] New post 30 Aug 2006, 10:31
Unfortunately no OA was given :)
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 [#permalink] New post 30 Aug 2006, 10:42
apollo168 wrote:
Unfortunately no OA was given :)

OK then I will explain.
IMO its A.

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
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 [#permalink] New post 30 Aug 2006, 10:48
Thanks guys :)
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 [#permalink] New post 31 Aug 2006, 04:46
ps_dahiya wrote:
apollo168 wrote:
Unfortunately no OA was given :)

OK then I will explain.
IMO its A.

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.

(A)
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 [#permalink] New post 31 Aug 2006, 06:03
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.

Hence A.

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  [#permalink] 31 Aug 2006, 06:03
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