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M06-07

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M06-07  [#permalink]

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New post 16 Sep 2014, 00:27
2
8
00:00
A
B
C
D
E

Difficulty:

  45% (medium)

Question Stats:

72% (01:09) correct 28% (01:46) wrong based on 184 sessions

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Re M06-07  [#permalink]

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New post 16 Sep 2014, 00:27
2
4
Official Solution:

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 \(0 \le N \le 4\)).


Answer: A
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Re: M06-07  [#permalink]

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New post 15 Jan 2015, 05:48
Bunuel wrote:
Official Solution:

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 \(0 \le N \le 4\)).


Answer: A



Hello Bunuel,

Could you please explain - I am not clear on this.

The way I was approaching it was as follows:

Step 1: Find out last digit of 2 to the power 98 --> 4
Step 2: Approximate values for L and N that will give 4
Step 3: 256*L can result in numbers ending with 6/2/8/4/0, using this with values of N (0/1/2/3/4)
Step 4: Answers can be L=4 and N=0

Now here is where I get stuck - the other option is L ends in 2 and N is 2 as well. It will give 4 as the last digit.

Whats wrong with my approach? Is it even correct.

Please help!!
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Re M06-07  [#permalink]

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New post 06 Feb 2015, 15:47
In my opinion it should have a multiplication sign between 256 and L. I lost a lot of time on this question, because I thought L was the last digit of a number, which is common on GMAT
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Re: M06-07  [#permalink]

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New post 07 Feb 2015, 08:01
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Re M06-07  [#permalink]

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New post 02 Mar 2016, 06:46
I think this is a high-quality question and the explanation isn't clear enough, please elaborate. i want to know if we can solve it using usual binomial method or not
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Re M06-07  [#permalink]

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New post 17 Jun 2016, 00:21
I think this is a high-quality question and I agree with explanation.
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M06-07  [#permalink]

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New post 13 Feb 2017, 11:09
4
1
Another way to approach this problem in terms of ODD and EVEN.

Given : \(2^{98}=256L+N\)

==> \(2^{98}=2^8L+N\) .Now , Left hand expression is EVEN and we know E+E or O+O only gives output as EVEN But since L is multiplied by 2^8 so ODD+ODD case is not possible.

Now , N can take only even values i.e. 0,2,4
If we Put N=2 then one 2 from N and L will cancel out from Right and Left side leaving below expression :

\(2^{97}=2^7L+1\)

But then Right side expression will give ODD value which is incorrect.Same is the case with N= 4 .

So only possible value is 0 . Ans A
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Re: M06-07  [#permalink]

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New post 01 Sep 2017, 03:11
Its great question and I'd like to practice a few more of these!
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Re: M06-07  [#permalink]

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New post 01 Sep 2017, 03:39
Its great question and I'd like to practice a few more of these!
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Re: M06-07  [#permalink]

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New post 06 Sep 2017, 19:15
Bunuel wrote:
Official Solution:

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 \(0 \le N \le 4\)).


Answer: A


Hi Brunel, so the sum or product of 2 integers is always an integer? is this the reason why you assumed \(\frac{N}{2^8}\) as integer? thanks
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Re: M06-07  [#permalink]

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New post 06 Sep 2017, 21:09
giuliab3 wrote:
Bunuel wrote:
Official Solution:

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 \(0 \le N \le 4\)).


Answer: A


Hi Brunel, so the sum or product of 2 integers is always an integer? is this the reason why you assumed \(\frac{N}{2^8}\) as integer? thanks


Yes.

Integer + Integer = Integer
(Not integer) + Integer = (Not integer)
(Not integer) + (Not integer) could be Integer as well as (Not integer).
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Re: M06-07  [#permalink]

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New post 26 Sep 2018, 23:37
Hi, In the first step 2^98=2^8∗L+N

2^98 is INTEGER
2^8 is INTEGER
L IS INTEGER
So, 2^8 * L is INTEGER

Hence this is Integer = Integer + N
so N can be any integer 0,1,2,3,4

Where am i thinking wrong in this approach?
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Re: M06-07  [#permalink]

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New post 15 Nov 2018, 10:21
Hi Bunuel,

I dont understand why you have to assume that N has to be "0",

Now, as both 2^90 and L are an integers then N/2^28, must also be an integer, which is only possible for N=0
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Re: M06-07  [#permalink]

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New post 15 Nov 2018, 11:44
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Re: M06-07   [#permalink] 15 Nov 2018, 11:44
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