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If 2^98=256L+N, where L and N are integers and

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If 2^98=256L+N, where L and N are integers and [#permalink] New post 07 May 2012, 19:57
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If 2^{98} = 256L + N, where L and N are integers and 0 \le N \le 4 , what is the value of N ?

A. 0
B. 1
C. 2
D. 3
E. 4
[Reveal] Spoiler: OA

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Last edited by Bunuel on 07 May 2012, 23:02, edited 1 time in total.
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Re: m06#07 gmatclub tests [#permalink] New post 07 May 2012, 20:28
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If 2^98+L, where L and N are integers and 0<=N<=4, what is the value of N?



The question has not been stated in the correct form. Looking into the tests, I see that the actual question is:

If 2^98= 256L + N and 0<=N<=4, what is the value of N?

To solve this, note that 2^8 = 256, so 2^98 = 2^[(8*12)+2]=4*[256*256*......12 times]

As this is a multiple of 256, N can only be 0.

Choice A
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Re: If 2^98=256L+N, where L and N are integers and [#permalink] New post 07 May 2012, 23:02
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monir6000 wrote:
If 2^{98} = 256L + N, where L and N are integers and 0 \le N \le 4 , what is the value of N ?

A. 0
B. 1
C. 2
D. 3
E. 4


Given: 2^{98}=2^8*L+N --> divide both parts by 2^8 --> 2^{90}=L+\frac{N}{2^8}. Now, as both 2^{90} and L are an integers then \frac{N}{2^8} must also be an integer, which is only possible for N=0 (since 0\leq{N}\leq4).

Answer: A.
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Re: If 2^98=256L+N, where L and N are integers and [#permalink] New post 08 May 2012, 00:30
Bunuel wrote:
monir6000 wrote:
If 2^{98} = 256L + N, where L and N are integers and 0 \le N \le 4 , what is the value of N ?

A. 0
B. 1
C. 2
D. 3
E. 4


Given: 2^{98}=2^8*L+N --> divide both parts by 2^8 --> 2^{90}=L+\frac{N}{2^8}. Now, as both 2^{90} and L are an integers then \frac{N}{2^8} must also be an integer, which is only possible for N=0 (since 0\leq{N}\leq4).

Answer: A.


Dear Bunuel,

could you pls explain why N/2^8 = 0 ?
why it could not be another multiple of 2?
L is also there? why we are sure L+N/2^8 do not = 2^90?
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Re: If 2^98=256L+N, where L and N are integers and [#permalink] New post 08 May 2012, 00:33
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kashishh wrote:
Bunuel wrote:
monir6000 wrote:
If 2^{98} = 256L + N, where L and N are integers and 0 \le N \le 4 , what is the value of N ?

A. 0
B. 1
C. 2
D. 3
E. 4


Given: 2^{98}=2^8*L+N --> divide both parts by 2^8 --> 2^{90}=L+\frac{N}{2^8}. Now, as both 2^{90} and L are an integers then \frac{N}{2^8} must also be an integer, which is only possible for N=0 (since 0\leq{N}\leq4).

Answer: A.


Dear Bunuel,

could you pls explain why N/2^8 = 0 ?
why it could not be another multiple of 2?
L is also there? why we are sure L+N/2^8 do not = 2^90?


Sure. Given that 0\leq{N}\leq4, so N can only be 0, 1, 2, 3, or 4. So, N/2^8 is an integer only for N=0.

Hope it's clear.
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Re: If 2^98=256L+N, where L and N are integers and [#permalink] New post 08 May 2012, 00:44
@ Bunuel
yes, got it! thanx.

its integer - its value could only between >0 and <4!
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Re: If 2^98=256L+N, where L and N are integers and [#permalink] New post 09 May 2012, 12:39
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monir6000 wrote:
If 2^{98} = 256L + N, where L and N are integers and 0 \le N \le 4 , what is the value of N ?

A. 0
B. 1
C. 2
D. 3
E. 4


Another way to solve this question:

Points to be remember:

1. L and N are integers
2. N could be any integer from 0 to 4.

Since 2^98 and 256L are even, N must be even. So N cannot be 1 and 3.

2^98 = 256L+N
2^98 = 2^8L+N
L = (2^98 - N)/2^8
L = 2^90 - N/2^8

If N = 0, L has integer value. Keep it.
If N = 2, L does not have integer value. Eliminate it.
If N = 4, L does not have integer value. Eliminate it.

Therefore, L has to be zero (A).
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Re: If 2^98=256L+N, where L and N are integers and [#permalink] New post 27 Nov 2012, 16:18
256= 2^8 and it is being multiplied by L. We don't really have to worry about L.

Just focus on 2^8 + N.

If N is 2, then 2(2^7 + 1) isn't going to result in a factor of 2, nor would N = 4 b/c 2^8+2^2 = 2^2(2^6+1). Same problem

Thus N has to be 0.
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Re: If 2^98=256L+N, where L and N are integers and [#permalink] New post 20 Dec 2012, 02:02
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Another approach to tackle this problem:
The given relation is:
2^98=256L+N
This is actually written in "remainder format" i.e. X=Quotient*some integer + Remainder.
Here X=2^98, Quotient=256 and Remainder=0.

We just have to check whether 256 OR 2^8 divides 2^ 98. If yes, then the value of N is straightaway 0. No need to check choices even.
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Re: If 2^98=256L+N, where L and N are integers and [#permalink] New post 27 May 2013, 15:02
Marcab wrote:
Another approach to tackle this problem:
The given relation is:
2^98=256L+N
This is actually written in "remainder format" i.e. X=Quotient*some integer + Remainder.
Here X=2^98, Quotient=256 and Remainder=0.

We just have to check whether 256 OR 2^8 divides 2^ 98. If yes, then the value of N is straightaway 0. No need to check choices even.


Could you tell me if I'm understanding it correctly?

What if we had 12/5 = 2 remainder 2? Then in remainder format we would have 2 * 5 + 2 = 12.

So according to what you're saying, if we just have to check if 2 can divide 12? In this case it does, but the remainder is 2 not 0.

Please, help me in understanding where I'm going wrong.
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Re: If 2^98=256L+N, where L and N are integers and [#permalink] New post 28 May 2013, 00:36
youngkacha wrote:
Marcab wrote:
Another approach to tackle this problem:
The given relation is:
2^98=256L+N
This is actually written in "remainder format" i.e. X=Quotient*some integer + Remainder.
Here X=2^98, Quotient=256 and Remainder=0.

We just have to check whether 256 OR 2^8 divides 2^ 98. If yes, then the value of N is straightaway 0. No need to check choices even.


Could you tell me if I'm understanding it correctly?

What if we had 12/5 = 2 remainder 2? Then in remainder format we would have 2 * 5 + 2 = 12.

So according to what you're saying, if we just have to check if 2 can divide 12? In this case it does, but the remainder is 2 not 0.

Please, help me in understanding where I'm going wrong.


Your reasoning unfortunately is incorrect since 5 is the divisor in question. The first 2 obtained is the quotient (which is the integer result obtained when dividing the numerator with the denominator) and cannot be used to question whether 2 divides 12 or not.

To further illustrate, take a look at the problem itself : 2^98 = 256*L + N which can be rephased as the following : does 256 divide 2^98 ? (Since in this instance 2^98 is the numerator, 256 is the denominator, L is the quotient and N is the remainder). To which the answer, obviously, is yes, since 256 can be written as a power of two, in this case 2^8.
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Re: If 2^98=256L+N, where L and N are integers and [#permalink] New post 28 May 2013, 01:17
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monir6000 wrote:
If 2^{98} = 256L + N, where L and N are integers and 0 \le N \le 4 , what is the value of N ?

A. 0
B. 1
C. 2
D. 3
E. 4


256=2^8=> 2^98 can be written as 2^90*2^8 such that L can be treated as 2^90. hence N=0.
Ans A
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Re: If 2^98=256L+N, where L and N are integers and [#permalink] New post 28 May 2013, 15:44
Virgilius wrote:
youngkacha wrote:
Marcab wrote:
Another approach to tackle this problem:
The given relation is:
2^98=256L+N
This is actually written in "remainder format" i.e. X=Quotient*some integer + Remainder.
Here X=2^98, Quotient=256 and Remainder=0.

We just have to check whether 256 OR 2^8 divides 2^ 98. If yes, then the value of N is straightaway 0. No need to check choices even.


Could you tell me if I'm understanding it correctly?

What if we had 12/5 = 2 remainder 2? Then in remainder format we would have 2 * 5 + 2 = 12.

So according to what you're saying, if we just have to check if 2 can divide 12? In this case it does, but the remainder is 2 not 0.

Please, help me in understanding where I'm going wrong.


Your reasoning unfortunately is incorrect since 5 is the divisor in question. The first 2 obtained is the quotient (which is the integer result obtained when dividing the numerator with the denominator) and cannot be used to question whether 2 divides 12 or not.

To further illustrate, take a look at the problem itself : 2^98 = 256*L + N which can be rephased as the following : does 256 divide 2^98 ? (Since in this instance 2^98 is the numerator, 256 is the denominator, L is the quotient and N is the remainder). To which the answer, obviously, is yes, since 256 can be written as a power of two, in this case 2^8.


Sorry, but I'm still lost. Marcab has the quotient as 256, but you're saying 256 is the divisor and L is the quotient?
Re: If 2^98=256L+N, where L and N are integers and   [#permalink] 28 May 2013, 15:44
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