GMAT Prep Question

This topic has expert replies
Junior | Next Rank: 30 Posts
Posts: 28
Joined: 13 Nov 2006
Thanked: 1 times

GMAT Prep Question

by TSonam » Wed Dec 12, 2007 10:19 am
1) When positive integer n is divided by 3, remainder is 2. When positive integer t is divided by 5, remainder is 3. What is the remainder when product nt is divided by 15.

a) n -2 is divisible by 5
b) t is divisible by 3

please provide explaination.

Master | Next Rank: 500 Posts
Posts: 315
Joined: 17 Aug 2006
Thanked: 23 times

by Suyog » Wed Dec 12, 2007 4:16 pm
Q. When positive integer n is divided by 3, remainder is 2. When positive integer t is divided by 5, remainder is 3. What is the remainder when product nt is divided by 15.

n can be 5 8 11 14 17 20 23 26 29 32 and so on
t can be 8 13 18 23 28 33 38 43 48 53 and so on

Each option alone is not suff since one speaks about n and other about t
Both together...

n -2 is divisible by 5 and t is divisible by 3
again, n can be 17 32 47 and so on
and t can be 18 48 and so on

again the remainder will be diff every time

Ans E.

Sorry....i didnt calculate properly.....the remainder is 6 everytime.....
Sorry guys... ;(

ans C.
Last edited by Suyog on Fri Dec 21, 2007 12:08 pm, edited 1 time in total.

Master | Next Rank: 500 Posts
Posts: 158
Joined: 03 Dec 2007
Thanked: 7 times

by StarDust845 » Fri Dec 14, 2007 12:38 pm
I think the answer is C. What does the OA say?

Junior | Next Rank: 30 Posts
Posts: 28
Joined: 13 Nov 2006
Thanked: 1 times

by TSonam » Tue Dec 18, 2007 6:11 am
answer is C.

stardust: any explaination?

I tried plugging in numbers but can't really come up with a concrete answer.

Master | Next Rank: 500 Posts
Posts: 460
Joined: 25 Mar 2007
Thanked: 27 times

by samirpandeyit62 » Tue Dec 18, 2007 9:54 pm
1) When positive integer n is divided by 3, remainder is 2. When positive integer t is divided by 5, remainder is 3. What is the remainder when product nt is divided by 15.

i.e n = 3a + 2 & t =5b+3

a) n -2 is divisible by 5

i.e n =5c + 2 also n =3a +2

so 5c +2 = 3a +2

or 5c = 3a

so here we can see that the values of a & c must be such that 5c= 3a

i.e c should be a multiple of 3 & a of 5

e.g c = 3 & a =5 i.e so n can be 17, 32 etc

& t = 5b+3 where b can be any value from 1,2,3........

so the reaminder will be diff each time INSUFF

b) t is divisible by 3

i.e t =3d also t =5b+3

so 3d = 5b +3 ------------------- (eqn 1)

also n = 3a+2 so nt =(3a+2)(3d),

we cannot determine the remainder here coz a can assume any value here INSUFF

COMBINE:

now we from 1 that n= 5c +2

& from 2, t =3d

so nt = (5c +2)(3d) = 15cd +6d

now when we dicvide this by 15 the factor 15cd will get divided evenly so reaminder will be produced by the divison 6d/15

now from eqn 1 we have

3d =5b+3 or

d =(5b + 3)/3

since d is an inetger so 5b+3 must be a multiple of 3

i.e it can be 18, 33......

so 6d = 18*6, 33*6......

in each case the remainder will be 3

C SUFF

in each case when we divide
Regards
Samir

Master | Next Rank: 500 Posts
Posts: 158
Joined: 03 Dec 2007
Thanked: 7 times

by StarDust845 » Wed Dec 19, 2007 9:02 am
This is how I solved it.

n = 3p + 2 for all p GE 0 (GE means greater than or equal to)
t = 5q + 3 for all q GE 0
nt = 15pq + 9p + 10q + 6 -- (1)

1) n - 2 is divisible by 3 implies p is divisible by 3 But we don't know aboyt q in (1) above. so INSUFF.
2) t is divisible by 3 implies q is divisible by 3 But we don't know aboyt p in (1) above so INSUFF.

Combining both we see that all the first 3 terms in (1) get divided by 15. Reminder is 6. Hence we know that nt is not divisible by 15. Hence C.

Master | Next Rank: 500 Posts
Posts: 158
Joined: 03 Dec 2007
Thanked: 7 times

by StarDust845 » Wed Dec 19, 2007 9:04 am
(Correcting typos in my earlier post)
This is how I solved it.

n = 3p + 2 for all p GE 0 (GE means greater than or equal to)
t = 5q + 3 for all q GE 0
nt = 15pq + 9p + 10q + 6 -- (1)

1) n - 2 is divisible by 5 implies p is divisible by 3 But we don't know about q in (1) above. so INSUFF.
2) t is divisible by 3 implies q is divisible by 3 But we don't know about p in (1) above so INSUFF.

Combining both we see that all the first 3 terms in (1) get divided by 15. Reminder is 6. Hence we know that nt is not divisible by 15. Hence C.

Master | Next Rank: 500 Posts
Posts: 158
Joined: 03 Dec 2007
Thanked: 7 times

by StarDust845 » Wed Dec 19, 2007 9:07 am
(Correcting typos AGAIN!!! in my earlier post)
This is how I solved it.

n = 3p + 2 for all p GE 0 (GE means greater than or equal to)
t = 5q + 3 for all q GE 0
nt = 15pq + 9p + 10q + 6 -- (1)

1) n - 2 is divisible by 5 implies p is divisible by 5 which implies second term is divisible by 15. But we don't know about q in (1) above. so INSUFF.
2) t is divisible by 3 implies q is divisible by 3 which implies 10q is divisible by 15 But we don't know about p in (1) above so INSUFF.

Combining both we see that all the first 3 terms in (1) get divided by 15. Reminder is 6. Hence we know that nt is not divisible by 15. Hence C.

Calista (Sorry, I should check more crefully for typos before posting).

User avatar
Legendary Member
Posts: 986
Joined: 20 Dec 2006
Location: India
Thanked: 51 times
Followed by:1 members

by gabriel » Wed Dec 19, 2007 10:40 am
Well, you can edit your posts by clicking on the edit button

Regards

Newbie | Next Rank: 10 Posts
Posts: 4
Joined: 05 May 2007
Thanked: 1 times

Re: GMAT Prep Question

by srawat_itpro » Fri Dec 21, 2007 12:11 am
TSonam wrote:1) When positive integer n is divided by 3, remainder is 2. When positive integer t is divided by 5, remainder is 3. What is the remainder when product nt is divided by 15.

a) n -2 is divisible by 5
b) t is divisible by 3

please provide explaination.
Hi

My two cents...

from Q:
n=3i + 2 [i>=0]
t=5j + 3 [j>=0]
so
nt = (3i+2)(5j+3)
=15ij + 9i + 10j + 6

Since the divisor is 15 then 15ij portion can be removed and we consider
Remainder = 9i + 10j + 6

Now----
1 & 2 can't be considered for seperately sufficient because each gives infor about either i or j sepearately which is not good enuf.

then consider two together
1. n - 2 is divisible by 5
=> 3i is divisible by 5
=> i=0,5,15,20,25....
We cant simply say for all i's the term 9i is divisible by 15.

2. t is divisble by 3
=> 5j + 3 divisible by 3
=> 5j divisible by 3
=> j=0,3,6,9,12.... etc
=> A simple viewing at remainder will show that for all j's 10j term is divisible by 15.

3) Now combine 1 & 2
t(n-2) is divisible by 3*15
tn - 2t is divisible by 15
tn remainder is inserted here
=> (9i + 10j + 6)-2(5j+3) is divisible ...
=> 9i is divisible by 15

so based on 3 and 2 we can say that both terms 9i and 10j in 9i + 10j + 6 are divisible by 15.

Which means 6 is the remainder always and (C) is the answer.

Pardon my being too simplistic at someplaces.

Rgds
-Sandy
Decide and do it !!!

User avatar
Senior | Next Rank: 100 Posts
Posts: 77
Joined: 09 Jul 2007
Location: US of A

by Auzbee » Tue Dec 25, 2007 2:26 pm
(Correcting typos AGAIN!!! in my earlier post)
This is how I solved it.

n = 3p + 2 for all p GE 0 (GE means greater than or equal to)
t = 5q + 3 for all q GE 0
nt = 15pq + 9p + 10q + 6 -- (1)

1) n - 2 is divisible by 5 implies p is divisible by 5 which implies second term is divisible by 15. But we don't know about q in (1) above. so INSUFF.
2) t is divisible by 3 implies q is divisible by 3 which implies 10q is divisible by 15 But we don't know about p in (1) above so INSUFF.

Combining both we see that all the first 3 terms in (1) get divided by 15. Reminder is 6. Hence we know that nt is not divisible by 15. Hence C.

Calista (Sorry, I should check more crefully for typos before posting).
I think StarDust845 solution for this problem is superb. Great thinking and superb logic SD845.