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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

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

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

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08 May 2012, 01:30

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?

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.

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

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20 Dec 2012, 03:02

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.
_________________

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

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27 May 2013, 16:02

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.

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

Show Tags

28 May 2013, 01:36

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.

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

Show Tags

28 May 2013, 16:44

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]

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19 Jul 2014, 02:36

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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, 06:57

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

Answer: A.

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 \(\frac{N}{2^8}\) ?
_________________

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.

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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