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When a positive integer n is divided by 3, the remainder is

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Director
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When a positive integer n is divided by 3, the remainder is [#permalink] New post 14 Jul 2007, 13:07
When a 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.
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 [#permalink] New post 14 Jul 2007, 14:17
Statement 1) n>17
but it could be 32 47....
Statement 2)
t is divisible by 3 it could be 3, 6, 9

hence E
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Re: GMATPrep Remainder [#permalink] New post 14 Jul 2007, 15:17
asaf wrote:
When a 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.


C for me.

1: if n-2 is divisible by 5, then n = 15k+2 and n could be 2, 17, 32, 47 and so on.
2: if t is divisible by 3, then t = 15k + 3 and t could be 3, 18, 33, 48 and so on.

1+2: nt = (15k+2) (15k+3) = 225k^2 + 60k + 6. so the reminder is always 6. suff...
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Re: GMATPrep Remainder [#permalink] New post 14 Jul 2007, 15:53
asaf wrote:
When a 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.


Got C
n/3 = int +2/3 => n = 3*int + 2
t/5 = int + 3/5 => t = 5*int + 3
n*t = (3*int + 2) * (5*int +3)

(1) n-2 is divisible by 5; we know that n-2 = 3*int; therefore, 3*int is divisible by 5. So we have: n = 3*5*int + 2. Plug this in n*t, we still have unknown values.

(2) more of the same

Together, we have remainder of 6.
Re: GMATPrep Remainder   [#permalink] 14 Jul 2007, 15:53
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When a positive integer n is divided by 3, the remainder is

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