Data Sufficiency

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.

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.

I think the answer is C. What does the OA say?

answer is C.

stardust: any explaination?

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

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

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.

(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.

(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).

Well, you can edit your posts by clicking on the edit button

Regards

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

(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.