complex GmatPrep Prob..Prob with the OA

This topic has expert replies
User avatar
Senior | Next Rank: 100 Posts
Posts: 37
Joined: Tue Jan 18, 2011 2:50 am
Thanked: 2 times

complex GmatPrep Prob..Prob with the OA

by abidshariff » Sun Jun 19, 2011 7:37 pm
This is the approach I ve seen in one of the Manhattan Gmat forums. But I edited this to show u my point.. Can any of u guys solve my 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.But we do know that 10k+6 is certainly not divisible by 15. Since 10k+6 has either 6 or 1 as units digit, neither of which is an units digit in any of the numbers divisible by 15. So nt is not divisible by 15. We can answer the question.. SUFFUCIENT

(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.

[spoiler]So A.

But the OA is C[/spoiler]...Please help....................
Attachments
diffprob.jpg

Legendary Member
Posts: 1448
Joined: Tue May 17, 2011 9:55 am
Location: India
Thanked: 375 times
Followed by:53 members

by Frankenstein » Sun Jun 19, 2011 8:14 pm
abidshariff wrote:
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.But we do know that 10k+6 is certainly not divisible by 15. Since 10k+6 has either 6 or 1 as units digit, neither of which is an units digit in any of the numbers divisible by 15. So nt is not divisible by 15. We can answer the question.. SUFFUCIENT

(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.

[spoiler]So A.

But the OA is C[/spoiler]...Please help....................
Hi,
The fact that 10k+6 is not divisible by 15 doesn't help you find the remainder. The remainder can be 1,6 or 11.
Hence, Not sufficient
Cheers!

Things are not what they appear to be... nor are they otherwise

User avatar
Master | Next Rank: 500 Posts
Posts: 461
Joined: Tue May 10, 2011 9:09 am
Location: pune
Thanked: 36 times
Followed by:3 members

by amit2k9 » Sun Jun 19, 2011 10:35 pm
a n can be 2,5,8,11,14,17,20 and so on.
t can be 3,8,13,18,23,28 and so on.

n-2/5 = integer means n = 2 or 17. t can be anything.not sufficient.

b n cab be any value and t is 18. not sufficient.

a+b gives n= 17,32 and so on. t = 18.

thus gives 2 as remainder in both the conditions. C it is.
For Understanding Sustainability,Green Businesses and Social Entrepreneurship visit -https://aamthoughts.blocked/
(Featured Best Green Site Worldwide-https://bloggers.com/green/popular/page2)

User avatar
Senior | Next Rank: 100 Posts
Posts: 37
Joined: Tue Jan 18, 2011 2:50 am
Thanked: 2 times

by abidshariff » Mon Jun 20, 2011 7:34 am
Sorry guys..i completely messed it up..my problem with not seeing the question persists....with the gmat just a week away, I cant really allow this....God!!..Improve my concentration...