It is currently 21 Feb 2018, 08:47

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

# M06-07

Author Message
TAGS:

### Hide Tags

Math Expert
Joined: 02 Sep 2009
Posts: 43851

### Show Tags

15 Sep 2014, 23:27
Expert's post
15
This post was
BOOKMARKED
00:00

Difficulty:

45% (medium)

Question Stats:

72% (01:04) correct 28% (01:40) wrong based on 162 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
[Reveal] Spoiler: OA

_________________
Math Expert
Joined: 02 Sep 2009
Posts: 43851

### Show Tags

15 Sep 2014, 23:27
1
KUDOS
Expert's post
4
This post was
BOOKMARKED
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 $$0 \le N \le 4$$).

_________________
Current Student
Status: GMAT Date: 10/08/15
Joined: 17 Jul 2014
Posts: 93
Location: United States (MA)
Concentration: Human Resources, Strategy
GMAT 1: 640 Q48 V35
GPA: 3.5
WE: Human Resources (Consumer Products)

### Show Tags

15 Jan 2015, 04:48
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 $$0 \le N \le 4$$).

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

~~~~~~~~~~~~~~~~~

Please give Kudos if you find this post useful.

Intern
Joined: 12 Mar 2014
Posts: 5
Schools: Stanford '20

### Show Tags

06 Feb 2015, 14:47
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
Math Expert
Joined: 02 Sep 2009
Posts: 43851

### Show Tags

07 Feb 2015, 07:01
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.
_________________
Intern
Joined: 23 Sep 2015
Posts: 44

### Show Tags

02 Mar 2016, 05:46
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
Manager
Joined: 21 Sep 2015
Posts: 82
Location: India
GMAT 1: 730 Q48 V42
GMAT 2: 750 Q50 V41

### Show Tags

16 Jun 2016, 23:21
I think this is a high-quality question and I agree with explanation.
_________________

Appreciate any KUDOS given !

Intern
Joined: 15 Jan 2014
Posts: 22
Location: India
Concentration: Technology, Strategy
Schools: Haas '19
GMAT 1: 650 Q49 V30
GPA: 2.5
WE: Information Technology (Consulting)

### Show Tags

13 Feb 2017, 10:09
2
KUDOS
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
Joined: 15 Jan 2017
Posts: 359

### Show Tags

01 Sep 2017, 02:11
Its great question and I'd like to practice a few more of these!
Senior Manager
Joined: 15 Jan 2017
Posts: 359

### Show Tags

01 Sep 2017, 02:39
Its great question and I'd like to practice a few more of these!
Intern
Joined: 24 Jun 2017
Posts: 39
Location: Singapore
GMAT 1: 660 Q46 V34
GPA: 3.83

### Show Tags

06 Sep 2017, 18:15
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 $$0 \le N \le 4$$).

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
Joined: 02 Sep 2009
Posts: 43851

### Show Tags

06 Sep 2017, 20:09
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 $$0 \le N \le 4$$).

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).
_________________
Re: M06-07   [#permalink] 06 Sep 2017, 20:09
Display posts from previous: Sort by

# M06-07

Moderators: chetan2u, Bunuel

 Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.