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M06-07

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M06-07 [#permalink]

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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
guilopesmartins wrote:
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


If it were a four-digit number then it would be explicitly mentioned.
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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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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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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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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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New to the Math Forum?
Please read this: Ultimate GMAT Quantitative Megathread | All You Need for Quant | PLEASE READ AND FOLLOW: 12 Rules for Posting!!!

Resources:
GMAT Math Book | Triangles | Polygons | Coordinate Geometry | Factorials | Circles | Number Theory | Remainders; 8. Overlapping Sets | PDF of Math Book; 10. Remainders | GMAT Prep Software Analysis | SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) | Tricky questions from previous years.

Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


What are GMAT Club Tests?
Extra-hard Quant Tests with Brilliant Analytics

Kudos [?]: 128890 [0], given: 12183

Re: M06-07   [#permalink] 06 Sep 2017, 21:09
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