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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]

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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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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]

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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 $$0\leq{N}\leq4$$).

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Re: If 2^98=256L+N, where L and N are integers and [#permalink]

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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 $$0\leq{N}\leq4$$).

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]

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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 $$0\leq{N}\leq4$$).

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]

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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]

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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]

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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]

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20 Dec 2012, 02:02
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]

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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]

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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]

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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]

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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?
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Re: If 2^98=256L+N, where L and N are integers and [#permalink]

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19 Jul 2014, 01:36
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Re: If 2^98=256L+N, where L and N are integers and [#permalink]

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26 Nov 2015, 05:57
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 $$0\leq{N}\leq4$$).

I'm sorry but if you're dividing both sides of the equation by $$2^8$$ won't it cancel on the right side of the equation? So how do we come to $$\frac{N}{2^8}$$ ?
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Re: If 2^98=256L+N, where L and N are integers and [#permalink]

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26 Nov 2015, 06:10
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Expert's post
redfield 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 $$0\leq{N}\leq4$$).

I'm sorry but if you're dividing both sides of the equation by $$2^8$$ won't it cancel on the right side of the equation? So how do we come to $$\frac{N}{2^8}$$ ?

$$2^{98}=2^8*L+N$$

Dividing both parts by $$2^8$$:

$$\frac{2^{98}}{2^8}=\frac{2^8*L+N}{2^8}$$;

$$2^{98-8}=\frac{2^8*L}{2^8}+\frac{N}{2^8}$$

$$2^{90}=L+\frac{N}{2^8}$$.

Hope it's clear.
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If 2^98=256L+N, where L and N are integers and [#permalink]

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26 Nov 2015, 08:13
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

The first thing to note here is that LHS is an even number, hence RHS should also be even.
This elements N = 1,3

Now,
$$2^{98} = 2^8L + N$$, Diving both the sides by $$2^8$$

$$2^{90} = L + N/2^8$$,
Out of 0, 2 and 4, N can only be 0 for RHS to be an even number.

Option A
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Re: If 2^98=256L+N, where L and N are integers and [#permalink]

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19 Jan 2017, 06:09
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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Re: If 2^98=256L+N, where L and N are integers and   [#permalink] 19 Jan 2017, 06:09
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