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# When positive integer n is divided by 3, the remainder is 2;

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When positive integer n is divided by 3, the remainder is 2; [#permalink]

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02 Jun 2008, 10:24
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45% (medium)

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72% (02:05) correct 28% (03:48) wrong based on 18 sessions

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When positive integer n is divided by 3, the remainder is 2; and when positive integer t is divided by 5, the remainder is 3. What is the remainder when the product nt is divided by 15?

(1) n-2 is divisible by 5

(2) t is divisible by 3

OPEN DISCUSSION OF THIS QUESTION IS HERE: when-positive-integer-n-is-divided-by-3-the-remainder-is-86155.html
[Reveal] Spoiler: OA

Last edited by Bunuel on 30 Jul 2012, 03:28, edited 1 time in total.
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02 Jun 2008, 10:34
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Expert's post
C

Just peaking a few numbers:

n: positive integer n is divided by 3, the remainder is 2 : 2,5,8,11,14,17 (2 and 17 satisfy first condition)

t: positive integer t is divided by 5, the remainder is 3: 3,8,13,18 (3 and 18 satisfy second condition)

1. nt=2*3 and nt=2*8 --> r=6 and r=1 insufficient.
2. nt=2*3 and nt=5*3 --> r=6 and r=0 insufficient.
1&2) nt=2*3, nt=2*18, and nt=3*17 --> r=6, r=6, and r=6 sufficient.
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Kudos [?]: 3568 [0], given: 360

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02 Jun 2008, 10:51
10 Kudos from me if somebody will find FAST solution using formulas!
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02 Jun 2008, 11:05
Doh! messed up...didn't see n-2, just n...Walker is right!
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02 Jun 2008, 20:33
11
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walker wrote:
C

Just peaking a few numbers:

n: positive integer n is divided by 3, the remainder is 2 : 2,5,8,11,14,17 (2 and 17 satisfy first condition)

t: positive integer t is divided by 5, the remainder is 3: 3,8,13,18 (3 and 18 satisfy second condition)

1. nt=2*3 and nt=2*8 --> r=6 and r=1 insufficient.
2. nt=2*3 and nt=5*3 --> r=6 and r=0 insufficient.
1&2) nt=2*3, nt=2*18, and nt=3*17 --> r=6, r=6, and r=6 sufficient.

hi Walker may be this helps

n=3x+2-----------(1)
t=5y+3-----------(2)

Statement 1:
arranging (1)
n-2=3x this means 3x is divisible by 5 let 3x=3*5*a=15a
equation (1) is now n=15a+2
not suff

Statement 2:
(2)is divisible by 3 means 5y is divisible by 3
so let 5y=3*5*b=15b

eq (2) now is t=15b+3
not suff

combining
nt=15a+2*15b+3

all terms except 6 is divisible by 6 so remainder is 6
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Kudos [?]: 3568 [0], given: 360

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02 Jun 2008, 20:54
rohit929 wrote:
hi Walker may be this helps

n=3x+2-----------(1)
t=5y+3-----------(2)

Statement 1:
arranging (1)
n-2=3x this means 3x is divisible by 5 let 3x=3*5*a=15a
equation (1) is now n=15a+2
not suff

Statement 2:
(2)is divisible by 3 means 5y is divisible by 3
so let 5y=3*5*b=15b

eq (2) now is t=15b+3
not suff

combining
nt=15a+2*15b+3

all terms except 6 is divisible by 6 so remainder is 6

Should be nt=225ab+45a+30b+6=15[15ab+3a+2b]+6

Thanks for your reasoning. Your approach is interesting. It would be good to see your approach for other remainder-problems. As I promised +10.
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Re: DS number prop   [#permalink] 02 Jun 2008, 20:54
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