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When positive integer n is divided by 25, remainder is 13,

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Senior Manager
Joined: 04 Jun 2007
Posts: 374
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When positive integer n is divided by 25, remainder is 13, [#permalink]  02 Nov 2007, 16:51
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When positive integer n is divided by 25, remainder is 13, what is n?
1) n<100
2) when n is divided by 20, remainder is 3

when i approached the question, i can see that either statement by itself is insufficient.

question remains, is it C or E? i know the answer, but i am not understanding how it is. so please provide explanations.
Director
Joined: 11 Jun 2007
Posts: 932
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Kudos [?]: 70 [0], given: 0

Re: divisibility DS [#permalink]  02 Nov 2007, 16:58
r019h wrote:
When positive integer n is divided by 25, remainder is 13, what is n?
1) n<100
2) when n is divided by 20, remainder is 3

when i approached the question, i can see that either statement by itself is insufficient.

question remains, is it C or E? i know the answer, but i am not understanding how it is. so please provide explanations.

from the stem we get n = 25k + 13
should be C

1) n<100 so possible values for n are
25(1) + 13 = 38
25(2) + 13 = 63
25(3) + 13 = 88
n could be 38, 63, and 88
not suff

2) 20k + 3
20(1) + 3 = 23
20(2) + 3 = 43
20(3) + 3 = 63
20(4) + 3 = 83
20(5) + 3 = 103
n could have endless possibilities
not suff

putting 1 and 2 together, only 63 is possible
so C
Senior Manager
Joined: 04 Jun 2007
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Re: divisibility DS [#permalink]  02 Nov 2007, 17:10
beckee529 wrote:
r019h wrote:
When positive integer n is divided by 25, remainder is 13, what is n?
1) n<100
2) when n is divided by 20, remainder is 3

when i approached the question, i can see that either statement by itself is insufficient.

question remains, is it C or E? i know the answer, but i am not understanding how it is. so please provide explanations.

from the stem we get n = 25k + 13
should be C

1) n<100 so possible values for n are
25(1) + 13 = 38
25(2) + 13 = 63
25(3) + 13 = 88
n could be 38, 63, and 88
not suff

2) 20k + 3
20(1) + 3 = 23
20(2) + 3 = 43
20(3) + 3 = 63
20(4) + 3 = 83
20(5) + 3 = 103
n could have endless possibilities
not suff

putting 1 and 2 together, only 63 is possible
so C

maybe I was reading the OA wrong, coz I got C and the OA was E....I need to double check, it's probably my mistake in reading the OA.

i got C too, this is what I did
since i deduced n could be 38,63, 88 frm statement 1 i put that value of n into statement 2-

n=20k+3
38= 20k+3
35= 20k <-- not possible

63=20k+3
60= 20k <-- possible

88= 20k+ 3
85= 20k <-- not possible
VP
Joined: 28 Mar 2006
Posts: 1384
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Re: divisibility DS [#permalink]  02 Nov 2007, 18:01
r019h wrote:
When positive integer n is divided by 25, remainder is 13, what is n?
1) n<100
2) when n is divided by 20, remainder is 3

when i approached the question, i can see that either statement by itself is insufficient.

question remains, is it C or E? i know the answer, but i am not understanding how it is. so please provide explanations.

n=25k + 13

where k can be 1,2 or 3 (from 1)

when k=1 n = 38 and remainder = 18
when k=2 n=63 and the remainder =3
when k=3 n=88 and the remainder = 8

Should be C
SVP
Joined: 05 Jul 2006
Posts: 1519
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Kudos [?]: 115 [0], given: 39

When positive integer n is divided by 25, remainder is 13, what is n?
1) n<100
2) when n is divided by 20, remainder is 3

N= 25X+13

from 1

25x+13<1000 thus 25x<87 ie: x< 87/25 , x< 3 12/25 ie x could be 1 ,2,3

not suff

from 2

n= 20y+3 ie: 25x+13 = 20y+3 ie: 20y-25x=10 ie: 4y-5x=0 ie 4y = 5x

x has to be even and y has to be a multiple of 5....insuff

both together x = 2 .suff

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When positive integer n is divided by 25, remainder is 13,

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