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Remainders!

by mj78ind » Thu Jul 08, 2010 2:09 am
n leaves remainder 2 when divided by 3.
t leaves remainder 3 when divided by 5.
what is the remainder when nt is divided by 15.

1. n - 2 is a multiple of 5
2. t is a multiple of 3

Do not know the OA ....but my computations say E

Disclaimer: found at another site liked it so posted here
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by Rich@VeritasPrep » Thu Jul 08, 2010 2:36 am
You can attack this one via plugging in numbers.

According to the prompt,

n could be 2, 5, 8, ...
t could be 3, 8, 13, ...

St (1) tells us...

n could be 2.

If t=3, then nt = 6, which has a remainder of 6 when divided by 15.
If t=8, then nt = 16, which has a remainder of 1 when divided by 15.

INSUFFICIENT

St (2) tells us...

t could be 3.

If n=2, then nt = 6, which has a remainder of 6 when divided by 15.
If n=5, then nt = 15, which has a remainder of 0 when divided by 15.

INSUFFICIENT

(1) and (2) together tell us...

n = 2, 17, 32,...
t = 3, 18, 33,...

Try various values of nt, and you'll notice that when you divide by 15, you always get a remainder of 6. SUFFICIENT
Last edited by Rich@VeritasPrep on Thu Jul 08, 2010 2:43 am, edited 1 time in total.
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by kvcpk » Thu Jul 08, 2010 2:40 am
n leaves remainder 2 when divided by 3.

So n-2 is divisible by 3

t leaves remainder 3 when divided by 5.

t-3 is divisible by 5

1. n - 2 is a multiple of 5

From both the statements, we come to know that n-2 is multiple of 15

so n can be 17,32,47...
t-3 is divisible by 5. So t can be 3,8,13,18..

Suppose n=17,t=3
then nt = 51 when div by 15, reminder is 6
Suppose n=17,t=8
then nt = 136 when div by 15, reminder is 1

Hence INSUFFICIENT.
but we have no idea about t yet. So INSUFFICEINT.

2. t is a multiple of 3

From both the statements we know that t-3 is multiple of 5 and t is multiple of 3.
if t is a multiple of 3, then t-3 is also a multiple of 3.
Hence, t-3 should be a multiple of 15.
for same reason as above, t can be 18,33,48...

n can be 2,5,8,11...

Suppose n=18,t=2
nt = 36 when div by 15 gives 6
suppose n=18,t=5
nt=90 when div by 15 gives 0
hence INSUFFICIENT

Combining both:
n-2 is multiple of 15 and t-3 is multiple of 15

n will be of the form.. 17,32,47.. = 15i+2
t will be of the form... 18,33,48... = 15j+3

(15j+2)(15j+3)
15(some value) + 6

Hence reminder is always 6.

pick C
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by jube » Thu Jul 08, 2010 4:47 am
n=3a +2
t=5b + 3

1. n-2 = 5j or n = 5j + 2 AND n=3a + 2. n can be 2, 17, 32, 62 ...
t= 5b + 3 = 3, 8, 13...
nt = 6 -- leaves remainder 6
nt = 16 -- leaves remainder 1
- insuff.


2.t=3k & t =5b + 3. t can be 3, 18, 33....
n=3a+2 = 2, 5, 8...
nt = 6 -- leaves remainder 6
nt = 15, leaves remainder 0
- insuff.

1&2:
n = 2, 17, 32...
t = 3, 18, 33....
nt = 6 -- leaves remainder 6
nt = 51 -- leaves remainder 6
nt = 36 -- leaves remainder 6 & so on...

hence, C
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by mj78ind » Thu Jul 08, 2010 5:42 am
So I set about solving this algebraically, and this is what I discovered which is pretty interesting:

let n = 3a + 2
t = 5b + 3
nt = (3a+2)*(5b+3)
or nt/15 = (15ab+9a+10b+6)/15,

=> ab + (9a+10b+6)/15 the part in bold will give the remainder



Stmt 1 n - 2 = 5x, thus n = 5x+2 = 3a + 2 or a = 5x/3
input in (9a+10b+6)/15, we get (15x+10b+6)/15 OR x + (10b+6)/15 thus remainder is still dependent on b, which can by anything. Hence INSUFFICIENT

Stmt 2 --- t = 3y = 5b + 3
input in (9a+10b+6)/15, we get (9a + 6y)/15 again dependent on a and y hence in INSUFFICIENT

Stmt 1 and 2 combined:
(15x + 6y)/15 which is x + 6y/15 , hence the remainder was dependent on y. And for the life of me I could not figure out how the remainder is 6 without picking numbers.#@!$
then it struck me the remainder is dependent on y and what is y dependent on?
Sure enough lets go to stmt 2, 3*(y-1) = b*5, y = b*5/3 + 1 and we know y has to be an integer, thus b has to be a multiple of 3. Thus, putting b = 3, 6, 9 etc.. we have y = 6, 11, 16 etc....... now going back to 6y/15 putting y in we see that the remainder is always 6! Voila!!

Took me half a day to realize this but hey it works !!

Hence C
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