It took me too much time to determine the answer. Is there a shortcut to this problem.

I don't think there is a shortcut. You just need to try different numbers.

Does someone have an insight into this problem?

Well, I personally would go for a numerical approach for this one, but here's a way to work it algebraically:

Let n = 3j + 2, where j is a positive integer
Let t = 5k + 3, where k is a positive integer

nt = (3j+2)(5k+3) = 15jk + 9j + 10k + 6

So the question is: what is the remainder after 15jk + 9j + 10k + 6 is divided by 15? Well, 15jk is clearly divisible by 15. If we show that 9j and 10k are divisible by 15 as well, then we can determine the remainder (6).

(1) The fact that n - 2 is divisible by 5 means that 3j is divisible by 5. So j can be written as 5x, where x is a positive integer. That means we can rewrite 9j (in the question) as 45x, which is divisible by 15. But we don't know whether 10k is divisible by 15. Insufficient.

(2) That t is divisible by 3 means that 5k + 3 is divisible by 3, and therefore 5k is divisible by 3. So k can be written as 3y, where y is a positive integer. That means we can rewrite 10k (in the question) as 30y, which is divisible by 15. But we don't know whether 9j is divisible by 15. Insufficient.

(1&2) nt = 15jk + 9j + 10k + 6 = 15jk + 45x + 30y + 6. The remainder must be 6. Sufficient.

Ray, great explanation above.
Can you also explain how you would use a "numerical approach" here?

Thanks

to all of the question posters:

forum rules dictate: DO NOT POST IMAGE FILES for text-only problems.
you should only use image files for problems that require them, usually because they contain diagrams.

rey was generous enough to provide an answer, even though he shouldn't have. to the most recent poster ("abovethehead"), please type the problem into the forum so that we can answer it for you. thanks.

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.

Is there a fast way to do this problem without using algebra?

Here's another approach. Hope it helps.

http://www.manhattangmat.com/forums/remainder-question-t7438.html?hilit=When%20positive%20integer%20n%20is%20divided%20by%203,

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Ben Ku
Instructor
ManhattanGMAT

I used the following approach for this one.

Statement 1: (n-2) is divisible by 5. But (n-2) also must be divisible by 3 since n has a remainder of 2 when n is divided by 3. Thus, (n-2) is divisible by 15 => when n is divided by 15, the remainder is 2.

Statement 2: t is divisible by 3 => (t-3) is aslo divisible by 3. But (t-3) is divisible by 5 since t has a remainder of 3 when t is divided by 5. So (t-3) is divisible by 15 and t has a remainder of 3 when t is divided by 15.

Combining Statement 1 and Statement 2: when nt is divided by 15, the remainder is 2X3=6.

HI
I faced a similar sum in GMAT Prep and tried solving it by above method, but i got it wrong ,later when i used substitution method i got it correct .Kindly let me know whether this method is to be specifically used or its generic

If P and N are positive integers and P>N , what is the remainder when P^2 – N^2 is divided by 15?
1) The remainder when P+N is divided by 5 is 1.
2) The remainder when P-N is divided by 3 is 1.

OA – E

hi -

if you're going to post a new problem, you must post it in a new thread (per the forum rules).

please also follow all the attendant rules about how to title your post, etc. (the forum rules are contained in the first post in each folder, right on top of page 1.)

thank you.

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.

Solution:
n = 3a + 2, t = 5b + 3
Statement 1:
n = 5c + 2 and n = 3a + 2 => n = 15d + 2
nt = (15d + 2)*(5b + 3) = 15d(5b + 3) + 10b + 6
Remainder = 10b + 6; it will have multiple values

Not Sufficient

Statement 2:
t = 3x and t = 5b + 3 => t = 15y + 3
nt = ( 3a + 2)*(15y + 3)
Same as above

Not sufficient

Statement 1 and 2
n = 5c + 2 and n = 3a + 2 => n = 15d + 2
t = 3x and t = 5b + 3 => t = 15y + 3
nt = (15d + 2) * (15y + 3) = 15d(15y + 3) + 2*15y + 2*3 = 15z + 6

Remainder = 6

Sufficient
Ans C

keep up the good work!

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Stacey Koprince
Instructor
Director of Online Community
ManhattanGMAT

I got D. Here's what I did
I assumed n = 5 and t = 8

I) n cannot be 5, after some time I got n = 17 i.e. 17 - 2 is divisible by 5 and 17%3=2. So (n*t)%15 = (17*8)%15 = 16

Another number say n = 32 also works out and again (32*8)%15 = 16. So on, so 16 is fixed remainder so I is suff.

II) Let t = 18 i.e. 18%3 = 0 and 18%5 = 3. So (n*t)%15 = (5*18)%15 = 0.

Another number say t = 33 also works out and again (5*33)%15 = 0. So on, so II is also suff.

Now what's wrong with my approach?

Thanks

Ashwin, for statement 1, you are right that n will be numbers such as 17, 32, 47, etc. However, t could be many numbers: 3, 8, 13, 15, etc., and you will get different results depending on what you use for t. Just using t=8 is not enough. Testing numbers is a great strategy, but you need to actively try to prove insufficiency for each statement by selecting different numbers (the numbers must still conform to the information given, of course).

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Jamie Nelson
ManhattanGMAT Instructor