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Remainders

by tom4lax » Tue Sep 01, 2009 5:58 pm
When positive integer N is divided by 3, the remainder is 2; and when positive integer T is divided by 5, the remainder is 3. What is the remainder when the product NT is divided by 15?

I) n-2 is divisible by 5
II) t is divisible by 3

OA to follow shortly
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by mehravikas » Tue Sep 01, 2009 9:24 pm
Is it C?

From 2 we know that T/3 is an integer. so we need to find out the remainder of N/5.

Statement 1 tells us that n-2 is divisible by 5.

So the remainder will be 2.
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by tom4lax » Wed Sep 02, 2009 4:40 am
C is the correct answer. The answer i have here is a bit different though, but what i was looking for was another way of looking at it, so thank you. Intuitively I guess I understand whats happening, but I would like a more thorough understanding.

I) n = 3k + 2. From information we have n = 5m + 2, so we can obtain n=15p+2 (this is the step I dont fully follow, if the remainders are the same we can multiple the divisors?)

II) t = 5s + 3. From information we have t is divisible by 3, (t=3k) or t = 15q + 3. (the remainders arent the same, whats going on here?)

Combined we have nt = (15p + 2)(15q + 2). Multiply out and you ger remainder of 6.

I think I am just missing a simple rule here.

Thoughts?
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by tienvunguyen » Wed Sep 02, 2009 8:03 am
Here is how I think the explanation should be:

1)
n=3k+2 so n-2=3k (n-2 divisible by 3)
From statement 1: n-2=5m (n-2 divisible by 5)
Therefore n-2=15n (n-2 should be divisible by 3*5=15)
Rewriting n-2=15n so n=15n+2

2)
t=5s+3 so t-3=5s (t-3 divisible by 5)
t=3k so t-3=3q (if t is divisible by 3, t-3 is also divisible by 3) or to put it mathematically, t-3=3k-3=3*(k-1)=3*q
Therefore t-3=15v (t-3 should be divisible by 3*5=15)
Rewriting t-3=15v so t=15v+3

The answer is C. Because nt=(15n+2)(15v+3)=(15n)(15v)+2*(15v)+3*(15n)+6
So the remainder is 6. (It is much simpler when I think about it than when I put it in words)
Last edited by tienvunguyen on Thu Feb 04, 2010 2:40 pm, edited 1 time in total.
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by tom4lax » Wed Sep 02, 2009 10:17 am
great response, thanks so much for the help
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by mehravikas » Wed Sep 02, 2009 12:10 pm
Hi @tienvunguyen,

How can you say that if n-2 is divisible by 5 then n - 2 should be divisible by 15

n = 22
n - 2 is divisible by 5
n - 2 is not divisible by 15
tienvunguyen wrote:Here is how I think the explanation should be:

1)
n=3k+2 so n-2=3k (n-2 divisible by 3)
n=5m+2 so n-2=5m (n-2 divisible by 5)
Therefore n-2=15n (n-2 should be divisible by 3*5=15)
Rewriting n-2=15n so n=15n+2
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by fightthegmat » Mon Sep 07, 2009 7:44 pm
tom4lax wrote:C is the correct answer. The answer i have here is a bit different though, but what i was looking for was another way of looking at it, so thank you. Intuitively I guess I understand whats happening, but I would like a more thorough understanding.

I) n = 3k + 2. From information we have n = 5m + 2, so we can obtain n=15p+2 (this is the step I dont fully follow, if the remainders are the same we can multiple the divisors?)

II) t = 5s + 3. From information we have t is divisible by 3, (t=3k) or t = 15q + 3. (the remainders arent the same, whats going on here?)

Combined we have nt = (15p + 2)(15q + 2). Multiply out and you ger remainder of 6.

I think I am just missing a simple rule here.

Thoughts?
You said ,the answer you have is different though. Is the OA : D
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by boy141 » Mon Sep 07, 2009 9:43 pm
wait wait,

can someone really break it down? I am still lost on the n=3k+2. how did we get there?
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by tom4lax » Wed Sep 09, 2009 5:37 pm
Sorry, meant to say that the explanation that I have is different. OA is C.
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by heshamelaziry » Wed Sep 09, 2009 7:52 pm
boy141 wrote:wait wait,

can someone really break it down? I am still lost on the n=3k+2. how did we get there?
This is a rule that I understan and the only part I understand from this F problem.

Say you do long division for 11/2, on top you will have 5 in the bottom yu will have 1. so 51/2(5.5). So 11 equals (2 * quotient) + 1.

Hope this helps
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by heshamelaziry » Wed Sep 09, 2009 7:56 pm
tienvunguyen wrote:Here is how I think the explanation should be:

1)
n=3k+2 so n-2=3k (n-2 divisible by 3)
n=5m+2 so n-2=5m (n-2 divisible by 5)
Therefore n-2=15n (n-2 should be divisible by 3*5=15)
Rewriting n-2=15n so n=15n+2

2)
t=5s+3 so t-3=5s (t-3 divisible by 5)
t=3k so t-3=3q (if t is divisible by 3, t-3 is also divisible by 3) or to put it mathematically, t-3=3k-3=3*(k-1)=3*q
Therefore t-3=15v (t-3 should be divisible by 3*5=15)
Rewriting t-3=15v so t=15v+3

The answer is C. Because nt=(15n+2)(15v+3)=(15n)(15v)+2*(15v)+3*(15n)+6
So the remainder is 6. (It is much simpler when I think about it than when I put it in words)
How did you get this: n=5m+2 so n-2=5m (n-2 divisible by 5) from statement 1 ?

Thanks
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by tienvunguyen » Thu Feb 04, 2010 2:36 pm
Haven't check my profile for a while. Sorry for the late response.

@mehravikas:
I meant because n - 2 is divisible by 3 AND n - 2 is divisible by 5 -> n-2 is divisible by 15

@heshamelaziry:
Statement 1 says: n-2 is divisible by 5. So n-2=5m (m is integer). I did not have to say n=5m+2 at the beginning, which is irrelevant and further complicated my explanation. I've edited my post.
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by sanju09 » Sat Feb 06, 2010 4:45 am
tom4lax wrote:When positive integer N is divided by 3, the remainder is 2; and when positive integer T is divided by 5, the remainder is 3. What is the remainder when the product NT is divided by 15?

I) n-2 is divisible by 5
II) t is divisible by 3

OA to follow shortly
Let's take N = 3 p + 2 and T = 5 q + 3, where p and q are whole numbers, then the product N T = 15 p q + 9 p + 10 q + 6, which when divided by 15 would leave the same remainder as it would be when 9 p + 10 q + 6 is divided by 15, because 15 p q already has 15 as its factor and hence this part of N T won't leave any remainder (or it would leave 0 remainder) when divided by 15. So, let's only concentrate on the remainder when 9 p + 10 q + 6 is divided by 15.

(1) If N - 2 is divisible by 5, then 3 p or p has 5 as its factor and 9 p has 15 as its factor. It means that 9 p + 10 q + 6 when divided by 15 would leave the same remainder as it would be when 10 q + 6 is divided by 15. We don't know q. Insufficient

(2) If T is divisible by 3, 5 q or q has 3 as its factor and 10 q has 15 as its factor. It means that 9 p + 10 q + 6 when divided by 15 would leave the same remainder as it would be when 9 p + 6 is divided by 15. We don't know p. Insufficient

Taken together

When both 9 p and 10 q are separately divisible by 15, the remainder when 9 p + 10 q + 6 is divided by 15 would be 6.

Sufficient

[spoiler]C[/spoiler]
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by tanviet » Tue Jan 25, 2011 12:21 am
I wish to share most simple way

from statement 1

n=3k+2, n-2=3k, because n-2 is divisible my 5 we have 3k is divisible by 5, WE HAVE K IS DIVISIBLE BY 5, K=5X
SO n=3k+2=15x+2


t= 5y+3

from statement 1 we have nt=(5y+3)(15x+2)+ 15x(5y+3) +10y+6 INSUFFICIENT

I think you know how to go ahead

IMO C

gmatprep is very nice and should be studied carefully.
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by GMATGuruNY » Tue Jan 25, 2011 4:23 am
tom4lax wrote:When positive integer N is divided by 3, the remainder is 2; and when positive integer T is divided by 5, the remainder is 3. What is the remainder when the product NT is divided by 15?

I) n-2 is divisible by 5
II) t is divisible by 3

OA to follow shortly
I find it quicker and easier to plug in values.

N is 2 more than a multiple of 3 = 2, 5, 8, 11, 14, 17...
T is 3 more than a multiple of 5 = 3, 8, 13, 18, 23, 28...

Statement 1: N is 2 more than a multiple of 5 = 2, 7, 12, 17, 22...
Numbers common to both lists for N = 2, 17...
Our original list for T = 3, 8, 13, 18, 23, 28...
Possible remainders when NT is divided by 15:
(2*3)/15 = 0 R6
(2*8)/15 = 1 R1
Since R=6 and R=1, insufficient.

Statement 2: T is a multiple of 3 = 3, 6, 9, 12, 15, 18...
Numbers common to both lists for T = 3, 18...
Our original list for N = 2, 5, 8, 11, 14, 17...
Possible remainders when NT is divided by 15:
(2*3)/15 = 0 R6
(5*3)/15 = 1 R0
Since R=6 and R=0, insufficient.

Statements 1 and 2:
Our final list for N = 2, 17...
Our final list for T = 3, 18...
Possible remainders when NT is divided by 15:
(2*3)/15 = 0 R6
(2*18)/15 = 2 R6
(17*3)/15 = 3 R6
(17*18)/15 = 20 R6

Since in each case R=6, sufficient.

The correct answer is C.
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