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Source: — Data Sufficiency |
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by sumgb » Mon Aug 08, 2011 2:38 pm
I would quickly write down different numbers satisfying statement 1 (variable n) and then come up with 2 numbers for variable t and check the remainder of the product nt in both cases and try to eliminate choices.
n can take values such as 2,5,8,11,14,17
t can take values such as 3,8,13,18, etc.

you can see you can get remainder as 6 (if n= 2, t =3) or 1 (n=17, t =18), no definite answer even after combining statements, mark E and move on...
not a solid approach, but it'll work (and not so time consuming <- my opinion)

hope this helps.
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by gmatboost » Mon Aug 08, 2011 9:50 pm
This is a hard question.

From the question prompt:
n can be written as 3k+2
t can be written as 5j+3
This is because we know they are a certain amount (2, 3) more than a multiple of some number (3, 5)

Statement 1:
n - 2 is divisible by 5 means that n is 2 more than a multiple of 5
Recall that n is also 2 more than a multiple of 3 (from the prompt)
Putting those together, n is 2 more than a multiple of 3 and 5, which is also known as a multiple of 15
So, n is 2 more than a multiple of 15 and can be written as 15k + 2
Multiply this by t: (15k+2)(5j+3) = 75kj + 45k + 10j + 6
The first two terms are each evenly divisible by 15, but 10j is not necessarily, and we do not know anything more about j. Insufficient.

Statement 2:
t is divisible by 3
Recall that t is also 3 more than a multiple of 5 (from the prompt)
This is perhaps the hardest step of all:
Putting those together, t is 3 more than a multiple of 15 (it cannot be 8 more or 13 more, because then it would satisfy being 3 more than a multiple of 5, but would not satisfy being a multiple of 3)

So, t can be written as 15j + 3
Multiply this by n: (3k+2)(15j+3) = 45kj + 30j + 9k + 6
The first two terms are again each evenly divisible by 15, but 9k is not necessarily, and we do not know anything more about k. Insufficient.

Combined:
n = 15k+2
t = 15j+3
nt = (15k + 2)(15j + 3) = 225jk + 30j + 45k + 6

Now, the first three terms are each evenly divisible by 15. All we have left is +6, so the remainder when nt is divided by 15 is always 6. Sufficient.
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by sumgb » Tue Aug 09, 2011 3:40 am
oh yeah, the remainder is always 6, my bad :-( dont know what I was thinking
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by navami » Tue Aug 09, 2011 4:11 am
Precisely ..
From Question we get . n = a*3 + 2
t = b*5 + 3
n*t = 15 * a * b + 6 + 10b + 9a

6 + 10b + 9a / 15 = ?

1) statement 1 says n - 2 is multiple of 5 . and n = a*3 + 2 so n-2 = a*2 +2 -2 = a*3 is multiple of 5. or a is multiple of 5.

2) statement 2 says ( likewise) b is multiple of 3

so now we need to find if 6 + 10b + 9a / 15 = ?

10b and 9a would be divisible by 15 { From statement 1 + statement 2}
but what about 6 ??????
so we know that (6 + 10b + 9a / 15 = NO)
This time no looking back!!!
Navami
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