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gmatprep [#permalink]

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New post 29 Jan 2009, 04:20
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This topic is locked. If you want to discuss this question please re-post it in the respective forum.

I have asked this question earlier also but forgotten the explanation. pls tell a shortcut 2 solve this
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Re: gmatprep [#permalink]

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New post 29 Jan 2009, 05:41
From stmt1: n = 5a + 2
From stmt2: t = 3*5b + 3 = 15b + 3

Hence, nt = 75ab + 30b + 15a + 6

Thus, when nt is divided by 15, the remainder will be 6.

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Re: gmatprep [#permalink]

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New post 29 Jan 2009, 10:14
Can u pls explain how u got this? (highlighted in red)
scthakur wrote:
From stmt1: n = 5a + 2
From stmt2: t = 3*5b + 3 = 15b + 3
Hence, nt = 75ab + 30b + 15a + 6

Thus, when nt is divided by 15, the remainder will be 6.

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Re: gmatprep [#permalink]

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New post 29 Jan 2009, 11:54
ritula wrote:
Can u pls explain how u got this? (highlighted in red)
scthakur wrote:
From stmt1: n = 5a + 2
From stmt2: t = 3*5b + 3 = 15b + 3
Hence, nt = 75ab + 30b + 15a + 6

Thus, when nt is divided by 15, the remainder will be 6.


n=3k+2 -->(1)
t=5l+3 -->(2)

stat1:
n-2 = 5m
n=5m+2 -->(3)
from (1) and (2) it is clear
3k=5m

--> n= 15*a+2

stat2:
t=3x -->(4)

from (4) and (2)

3x=5l+3 ---> 5l must be multiple of 3
5l= 15b

t= 15b+3


nt = (15*a+2)*(15b+3) = ...(multiple of15)+6
remainder is 6
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Re: gmatprep [#permalink]

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New post 29 Jan 2009, 13:37
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IMO C.

The two solutions posted were saying the value of the remainder to be 6 but since it is a DS question we need not find the value. :wink: :wink:

Here is my approach.

Given in the question stem,

n - 2 is divisible by 3 since n is divisible by 3 leaving a remainder 2. -which means n-2 has atleast 3 as a factor ---fact 1
t - 3 is divisible by 5 since t is divisible by 5 leaving a remainder 3 - which mean t-3 has atleast 5 as a factor. ---fact 2

From stmt 1, it is said that n-2 is divisible by 5. Which means n-2 has atleast 5 as a factor. And from fact 1, we know that 3 is factor of n-2. So we can conclude that n-2 has atleast 3 and 5 as factors or in other words

n-2 = 15k where k is an interger. ==> n = 15k + 2 -----Eq 1

No additional information abt t. Hence insufficient.

From stmt 2, t is a divisible by 3, then t-3 is also divisible by 3. from fact 2, we know that t-3 has 5 also a factor. So we can conclude that t-3 has atleast 3 and 5 as factor or in other words,

t-3 = 15l where l is an integer ==> t = 15l + 3 -----Eq 2.

No additional information abt n is given. Hence insufficient

Now combine both the statements, we know that

n = 15k + 2
t = 15l + 3.

Rule to remember -

Arithmatic on remainders when divisiors are same states that when a and b are the remainders when x and y are divided by asome number n, then remainder left when product xy is divided by n is nothing but the product of the remainders a and b.


Here n and k are numbers divided by 15 leaving remainders 2 and 3. So we can easily apply the above rule to find the remainder when product nt is divided by 15.

So ans is C.

Last edited by mrsmarthi on 30 Jan 2009, 09:08, edited 1 time in total.

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Re: gmatprep [#permalink]

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New post 29 Jan 2009, 22:11
Superb! +1 frm me
pls edit ur post here(highlighted in red)
1 more thing can u pls tell sum source where these rules can be learnt bcos i wasnt aware that such rule exists.
mrsmarthi wrote:
IMO C.

The two solutions posted were saying the value of the remainder to be 6 but since it is a DS question we need not find the value. :wink: :wink:

Here is my approach.

Given in the question stem,

n - 2 is divisible by 3 since n is divisible by 3 leaving a remainder 2. -which means n-2 has atleast 3 as a factor ---fact 1
t - 3 is divisible by 5 since t is divisible by 5 leaving a remainder 3 - which mean t-3 has atleast 5 as a factor. ---fact 2

From stmt 1, it is said that n-2 is divisible by 5. Which means n-2 has atleast 5 as a factor. And from fact 1, we know that 3 is factor of n-2. So we can conclude that n-2 has atleast 3 and 5 as factors or in other words

n-2 = 15k where k is an interger. ==> n = 15k + 2 -----Eq 1

No additional information abt t. Hence insufficient.

From stmt 2, t is a divisible by 3, then t-3 is also divisible by 3. from fact 2, we know that t-3 has 5 also a factor. So we can conclude that t-3 has atleast 3 and 5 as factor or in other words,

t-3 = 15l where l is an integer ==> k = 15l + 3 -----Eq 2.

No additional information abt n is given. Hence insufficient

Now combine both the statements, we know that

n = 15k + 2
t = 15l + 3.

Rule to remember -

Arithmatic on remainders when divisiors are same states that when a and b are the remainders when x and y are divided by asome number n, then remainder left when product xy is divided by n is nothing but the product of the remainders a and b.


Here n and k are numbers divided by 15 leaving remainders 2 and 3. So we can easily apply the above rule to find the remainder when product nt is divided by 15.

So ans is C.

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Re: gmatprep [#permalink]

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New post 30 Jan 2009, 03:16
Great Explanation! Agreed with C.

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Re: gmatprep [#permalink]

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New post 30 Jan 2009, 09:11
Ritula,

When ever I find some useful shortcuts which simplifies the process, I make a note on that(In other words an entry into flash card). So I didn't remember the source of the rule that I mentioned. But probably it could be MGMAT - Number porperties Strategy guide.

BTW, thanks for pointing out the mistake. I have edited the post.

Last edited by mrsmarthi on 14 Mar 2009, 09:22, edited 1 time in total.

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Re: gmatprep [#permalink]

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New post 30 Jan 2009, 09:35
Another approach.

Let's write out a few possible values of n and t.
n e {2,5,8,11,14,17,20...}
t e {3,8,13,18,23,28...}

1) first condition limits possible values of n: n e {2,17...}
Let's check remainders for a couple of nt numbers at different t: n=2,t=3 --> r=6; n=2,t=8 --> r=1 - insufficient

2) second condition limits possible values of t: t e {3,18...}
Let's check remainders for a couple of nt numbers at different n: n=2,t=3 --> r=6; n=5,t=3 --> r=0 - insufficient

1)&2) first condition limits possible values of n: n e {2,17...} and second condition limits possible values of t: t e {3,18...}

Let's check remainders for a couple of nt numbers at different n and t:
n=2,t=3 --> r=6;
n=2,t=18 --> r=6;
n=17,t=3 --> r=6; - sufficient

This approach is a bit longer but it is always better to have a few approaches in a belt...
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Re: gmatprep   [#permalink] 30 Jan 2009, 09:35
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