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If n is a positive integer and r is the remainder when 4+7n

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If n is a positive integer and r is the remainder when 4+7n [#permalink] New post 15 Sep 2006, 17:15
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If n is a positive integer and r is the remainder when 4+7n is divided by 3, what is the value of r?

1) n + 1 is divisible by 3.
2) n > 20
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 [#permalink] New post 15 Sep 2006, 18:42
Answer A

if n+ 1 is divisible by 3 then n =2, 5, 8 ... etc

for those values 7n + 4 = 18, 39, 60 etc (divisible by 3 and hence r =0)

B does not tell anything
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 [#permalink] New post 16 Sep 2006, 00:21
If n is a positive integer and r is the remainder when 4+7n is divided by 3, what is the value of r?

1) n + 1 is divisible by 3.
2) n > 20

from stem

4+7n = 3x+r


from one

n+1 = 3y ie n= 3y+1 when n is devided by 3 it gives a remainder of 1


subestitute in the given

4+21y+21 = 3x+r and since each term on the left hand side is devisible by 3 therfore r results from deviding 4 by 3 giving a remainder of one all the time.....suff

from two

some the remainder changes with the values of N>20 ...INSUFF
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 [#permalink] New post 16 Sep 2006, 01:02
Folks, again this is simple if u know the basic concepts of remainders.
Let's look at this simple problem first.

Remainder obtained when 11xK is divided by 4 is nothing but the remainder obtained when 11 is divided by 4 x K divided by 4

ie the remainder of 11xK / 4 will be of the form 3xK (in case 3xK is greater than 4 it will be further divided by 4 to get the remainder)

Similarly when 11xK + 17 is divided by 9 the remainder will of the form 2xK + 8.

Using this concept to the given problem

4+7n/3 the remainder will be of the form 1 + n. ( So the remainder now depends upon the value of n)

Statement 1: Since it is given that n+1 is divisible by 3.
Hence 4 + 7n / 3 the remainder will be 0.
So statement 1 alone is sufficient.

Statement 2: n > 20.
For different values of n we get different remainders.
So statement 2 is not sufficient

Hence A.

Regards,
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Last edited by cicerone on 25 Sep 2008, 00:08, edited 1 time in total.
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 [#permalink] New post 16 Sep 2006, 06:53
cicerone wrote:
Folks, again this is simple if u know the basic concepts of remainders.
Let's look at this simple problem first.

Remainder obtained when 11xK is divided by 4 is nothing but the remainder obtained when 11 is divided by 4 x K divided by 4

ie the remainder of 11xK / 4 will be of the form 3xK (in case 3xK is greater than 4 it will be further divided by 4 to get the remainder)

Similarly when 11xK + 17 is divided by 9 the remainder will of the form 2xK + 8.

Using this concept to the given problem

4+7n/3 the remainder will be of the form 1 + n. ( So the remainder now depends upon the value of n)

Statement 1: Since it is given that n+1 is divisible by 3.
Hence 4 + 7n / 3 the remainder will be 0.
So statement 1 alone is sufficient.

Statement 2: n > 20.
For different values of n we get different remainders.
So statement 2 is not sufficient

Hence A.

Regards,

I have been looking for a good remainder rule. This seem to be a good one, however, I am not getting it. Can you help me understand it? Thanks
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 [#permalink] New post 16 Sep 2006, 07:26
cicerone wrote:
Folks, again this is simple if u know the basic concepts of remainders.
Let's look at this simple problem first.

Remainder obtained when 11xK is divided by 4 is nothing but the remainder obtained when 11 is divided by 4 x K divided by 4

ie the remainder of 11xK / 4 will be of the form 3xK (in case 3xK is greater than 4 it will be further divided by 4 to get the remainder)

Similarly when 11xK + 17 is divided by 9 the remainder will of the form 2xK + 8.

Using this concept to the given problem

4+7n/3 the remainder will be of the form 1 + n. ( So the remainder now depends upon the value of n)

Statement 1: Since it is given that n+1 is divisible by 3.
Hence 4 + 7n / 3 the remainder will be 0.
So statement 1 alone is sufficient.

Statement 2: n > 20.
For different values of n we get different remainders.
So statement 2 is not sufficient

Hence A.

Regards,


Cicerone, why are you taking 3 and 2? and not 2 and 1? Can you pls clarify?
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 [#permalink] New post 17 Sep 2006, 03:57
ak_idc wrote:
cicerone wrote:
Folks, again this is simple if u know the basic concepts of remainders.
Let's look at this simple problem first.

Remainder obtained when 11xK is divided by 4 is nothing but the remainder obtained when 11 is divided by 4 x K divided by 4

ie the remainder of 11xK / 4 will be of the form 3xK (in case 3xK is greater than 4 it will be further divided by 4 to get the remainder)

Similarly when 11xK + 17 is divided by 9 the remainder will of the form 2xK + 8.

Using this concept to the given problem

4+7n/3 the remainder will be of the form 1 + n. ( So the remainder now depends upon the value of n)

Statement 1: Since it is given that n+1 is divisible by 3.
Hence 4 + 7n / 3 the remainder will be 0.
So statement 1 alone is sufficient.

Statement 2: n > 20.
For different values of n we get different remainders.
So statement 2 is not sufficient

Hence A.

Regards,


Cicerone, why are you taking 3 and 2? and not 2 and 1? Can you pls clarify?


Hey dear,

when 11 is divided by 4 the reamainder is 3
So 11k divided by 4 the remainder will be of the form 3k
Similary 11 divided by 9 the remainder is 2
So 11k divided by 9 the remainder will be of the form 2k

Let me be more clear

Let's take 45. When 45 is divided by 8 the remainder is 5

Now 45 = 9 x 5

Now 9 leaves a remainder 1 when divided by 8
and 5 leaves a remainder 5 when divided by 5.

Since 45 = 9 x 5
The remainder also will be remainder of 9/8 x remainder of 5/8
ie 1 x 5 = 5

Let's take another example

2272/6 the raminder is 4
Now 2272 = 32 x 71
32/6 the remainder is 2
71/6 the remainder is 5

So according to our rule the remainder must be 2 x 5 = 10
Since 10 is greater than 6 it is further divided and hence the remainder becomes 4.

So all the three basic operations i.e additon, subtraction and multiplication of remainders can be done.

i.e Remainder of 45k + 23p - 17q when divided by 12 will take the form of 9k + 11p - 5q.

I think this is clear to everyone

Regards,
  [#permalink] 17 Sep 2006, 03:57
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