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Math Expert V
Joined: 02 Sep 2009
Posts: 56303

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2
8 00:00

Difficulty:   45% (medium)

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

### HideShow timer Statistics 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

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Math Expert V
Joined: 02 Sep 2009
Posts: 56303

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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 $$0 \le N \le 4$$).

_________________
Manager  Status: GMAT Date: 10/08/15
Joined: 17 Jul 2014
Posts: 85
Location: United States (MA)
Concentration: Human Resources, Strategy
GMAT 1: 640 Q48 V35 GPA: 3.5
WE: Human Resources (Consumer Products)

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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 $$0 \le N \le 4$$).

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.

_________________
Thanks,
aimtoteach

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Please give Kudos if you find this post useful.
Intern  Joined: 12 Mar 2014
Posts: 5
Schools: Stanford '20

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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
Math Expert V
Joined: 02 Sep 2009
Posts: 56303

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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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Intern  Joined: 23 Sep 2015
Posts: 37

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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
Manager  Joined: 21 Sep 2015
Posts: 75
Location: India
GMAT 1: 730 Q48 V42 GMAT 2: 750 Q50 V41 ### Show Tags

I think this is a high-quality question and I agree with explanation.
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Appreciate any KUDOS given ! Intern  B
Joined: 15 Jan 2014
Posts: 33
Location: India
Concentration: Technology, Strategy
Schools: Haas '19
GMAT 1: 650 Q49 V30 GPA: 2.5
WE: Information Technology (Consulting)

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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
Senior Manager  S
Joined: 15 Jan 2017
Posts: 348

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Its great question and I'd like to practice a few more of these!
Senior Manager  S
Joined: 15 Jan 2017
Posts: 348

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Its great question and I'd like to practice a few more of these!
Intern  B
Joined: 24 Jun 2017
Posts: 34
Location: Singapore
GMAT 1: 660 Q46 V34 GPA: 3.83

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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 $$0 \le N \le 4$$).

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
Math Expert V
Joined: 02 Sep 2009
Posts: 56303

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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 $$0 \le N \le 4$$).

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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Intern  B
Joined: 07 Jul 2018
Posts: 47

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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?
Manager  B
Joined: 02 Jan 2016
Posts: 124

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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
Math Expert V
Joined: 02 Sep 2009
Posts: 56303

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hero_with_1000_faces wrote:
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

Because we are given that $$0 \le N \le 4$$, no other value of N nut 0 will make N/2^28 an integer.
_________________ Re: M06-07   [#permalink] 15 Nov 2018, 11:44
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# M06-07

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