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DS-Divisibility question from GMATPREP

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DS-Divisibility question from GMATPREP [#permalink] New post 10 Nov 2006, 19:18
00:00
A
B
C
D
E

Difficulty:

(N/A)

Question Stats:

0% (00:00) correct 0% (00:00) wrong based on 0 sessions
Hi guys!

This one should be relatively simple for a lot of you. But was wondering as to what strategies people use when solving the DS remainder problems. Pluggin in numbers here takes ages and we all know that, this is one of the most commonly question type.

Here is the problem. Please explain. Thanks :-D

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 10 Nov 2006, 19:32
This is how I see, but I'm not sure if I'm right.

n=3x+2
t=5y+3

s1:3x is divisible by 5, so when n is divided by 15, remainder is 2. still insuff

s2:t should be divisible by 15

together: suff

Am I wrong?
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 [#permalink] New post 11 Nov 2006, 18:22
definitely C:

it is indeed difficult to plug in numbers directly into n and t
but lets change it to x and y as suggested above.

n = 3x+2 ; 3x= n-2
t = 5y+3 ; 5y = t-3

x and y are integers.

n*t = 15xy+9x+10y+6

s1: n-2 is divisible by 5, so 3x is divisible by 5, and because it is also divisible by 3, it means that 3x divisible by 15. so the remainder of n*t is connected to 10y+6, pick your y to be 0 or 1 to find it is insuff.

s2: t is divisible by 3, which means that t-3 is also divisible by 3, which means 5y divisible by 3 and hence by 15. still insuff (we left with 3x+6)

however together, we have 15xy divisible by 15, 9x divisible by 15 (from s1), 10y divisible by 15 (from s2). and a remainder 6.

suff.

can't think of simpler explanation though....but the idea, if you don't do lots of algebra, is that you do little algebra, and plug in numbers for x and y (which are not constrained), rather than plug in numbers for n and t (which are heavily constrained).

hope it helps.
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 [#permalink] New post 11 Nov 2006, 18:29
Thanks Hobbit!

By the way its nice to see that you are still active in the Math forum after getting a 780. Oh man if I had something even close to that I'll be on a long long vacation with lots of booze. :wink:

Good luck!
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 [#permalink] New post 12 Nov 2006, 11:15
uvs_mba wrote:
Nope the answer is C.

thanks



My bad....... :oops:

Given:
-------
n = 3p + 2 -------------(3)
t = 5q+ 3---------------(4)


From A n-2 is divisble by 5

So from (3) 3*p is divisble by 5
So p is factor of 5

But (A) alone NOT SUFF

Coming to (B) t is divible by 3

means from (4)
5q+ 3 is divisible by 3
which means q is a factor of 3
This is also NOT SUFF


Combining both and plug in values in
(3p + 2)(5q+ 3) where p is a factor of 5 and q is a factor of 3

Let p=5 and q=3
17*18 = some number which divided by 15 gives 6 reminder

Let p=10 and q=3 reminder is 6 when divided by 15....

So Answer is C
  [#permalink] 12 Nov 2006, 11:15
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