It is currently 21 Nov 2017, 11:09

Close

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
Your Progress

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

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

Close

Request Expert Reply

Confirm Cancel

Events & Promotions

Events & Promotions in June
Open Detailed Calendar

If 2^98=256L+N, where L and N are integers and

  new topic post reply Question banks Downloads My Bookmarks Reviews Important topics  
Author Message
TAGS:

Hide Tags

2 KUDOS received
Manager
Manager
avatar
Joined: 14 Mar 2010
Posts: 84

Kudos [?]: 178 [2], given: 44

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

Show Tags

New post 07 May 2012, 20:57
2
This post received
KUDOS
14
This post was
BOOKMARKED
00:00
A
B
C
D
E

Difficulty:

  45% (medium)

Question Stats:

64% (01:12) correct 36% (01:15) wrong based on 289 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

_________________

MGMAT CAT MATH http://gmatclub.com/forum/mgmat-cat-math-144609.html
MGMAT SC SUMMARY: http://gmatclub.com/forum/mgmat-sc-summary-144610.html


Last edited by Bunuel on 08 May 2012, 00:02, edited 1 time in total.

Kudos [?]: 178 [2], given: 44

6 KUDOS received
VP
VP
User avatar
S
Status: Top MBA Admissions Consultant
Joined: 24 Jul 2011
Posts: 1354

Kudos [?]: 651 [6], given: 20

GMAT 1: 780 Q51 V48
GRE 1: 1540 Q800 V740
Re: m06#07 gmatclub tests [#permalink]

Show Tags

New post 07 May 2012, 21:28
6
This post received
KUDOS
1
This post was
BOOKMARKED
Quote:
If 2^98+L, where L and N are integers and 0<=N<=4, what is the value of N?



The question has not been stated in the correct form. Looking into the tests, I see that the actual question is:

If 2^98= 256L + N and 0<=N<=4, what is the value of N?

To solve this, note that 2^8 = 256, so 2^98 = 2^[(8*12)+2]=4*[256*256*......12 times]

As this is a multiple of 256, N can only be 0.

Choice A
_________________

GyanOne | Top MBA Rankings and MBA Admissions Blog

Top MBA Admissions Consulting | Top MiM Admissions Consulting

Premium MBA Essay Review|Best MBA Interview Preparation|Exclusive GMAT coaching

Get a FREE Detailed MBA Profile Evaluation | Call us now +91 98998 31738

Kudos [?]: 651 [6], given: 20

Expert Post
8 KUDOS received
Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 42283

Kudos [?]: 132935 [8], given: 12391

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

Show Tags

New post 08 May 2012, 00:02
8
This post received
KUDOS
Expert's post
5
This post was
BOOKMARKED
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.
_________________

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 [?]: 132935 [8], given: 12391

Manager
Manager
avatar
Joined: 02 Jun 2011
Posts: 149

Kudos [?]: 106 [0], given: 11

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

Show Tags

New post 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?

Kudos [?]: 106 [0], given: 11

Expert Post
2 KUDOS received
Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 42283

Kudos [?]: 132935 [2], given: 12391

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

Show Tags

New post 08 May 2012, 01:33
2
This post received
KUDOS
Expert's post
kashishh wrote:
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?


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.

Hope it's clear.
_________________

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 [?]: 132935 [2], given: 12391

Manager
Manager
avatar
Joined: 02 Jun 2011
Posts: 149

Kudos [?]: 106 [0], given: 11

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

Show Tags

New post 08 May 2012, 01:44
@ Bunuel
yes, got it! thanx.

its integer - its value could only between >0 and <4!

Kudos [?]: 106 [0], given: 11

1 KUDOS received
Director
Director
User avatar
Joined: 03 May 2007
Posts: 867

Kudos [?]: 271 [1], given: 7

Schools: University of Chicago, Wharton School
Re: If 2^98=256L+N, where L and N are integers and [#permalink]

Show Tags

New post 09 May 2012, 13:39
1
This post received
KUDOS
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


Another way to solve this question:

Points to be remember:

1. L and N are integers
2. N could be any integer from 0 to 4.

Since 2^98 and 256L are even, N must be even. So N cannot be 1 and 3.

2^98 = 256L+N
2^98 = 2^8L+N
L = (2^98 - N)/2^8
L = 2^90 - N/2^8

If N = 0, L has integer value. Keep it.
If N = 2, L does not have integer value. Eliminate it.
If N = 4, L does not have integer value. Eliminate it.

Therefore, L has to be zero (A).

Kudos [?]: 271 [1], given: 7

Intern
Intern
avatar
Joined: 25 Jun 2012
Posts: 36

Kudos [?]: 30 [0], given: 4

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

Show Tags

New post 27 Nov 2012, 17:18
256= 2^8 and it is being multiplied by L. We don't really have to worry about L.

Just focus on 2^8 + N.

If N is 2, then 2(2^7 + 1) isn't going to result in a factor of 2, nor would N = 4 b/c 2^8+2^2 = 2^2(2^6+1). Same problem

Thus N has to be 0.

Kudos [?]: 30 [0], given: 4

VP
VP
User avatar
Status: Been a long time guys...
Joined: 03 Feb 2011
Posts: 1378

Kudos [?]: 1703 [0], given: 62

Location: United States (NY)
Concentration: Finance, Marketing
GPA: 3.75
GMAT ToolKit User Premium Member
Re: If 2^98=256L+N, where L and N are integers and [#permalink]

Show Tags

New post 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.
_________________

Prepositional Phrases Clarified|Elimination of BEING| Absolute Phrases Clarified
Rules For Posting
www.Univ-Scholarships.com

Kudos [?]: 1703 [0], given: 62

Intern
Intern
avatar
Joined: 29 Jul 2012
Posts: 28

Kudos [?]: 19 [0], given: 7

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

Show Tags

New post 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.

Kudos [?]: 19 [0], given: 7

Intern
Intern
User avatar
Status: Currently Preparing the GMAT
Joined: 15 Feb 2013
Posts: 30

Kudos [?]: 26 [0], given: 11

Location: United States
GMAT 1: 550 Q47 V23
GPA: 3.7
WE: Analyst (Consulting)
Re: If 2^98=256L+N, where L and N are integers and [#permalink]

Show Tags

New post 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.

Kudos [?]: 26 [0], given: 11

1 KUDOS received
Intern
Intern
User avatar
Joined: 13 May 2013
Posts: 12

Kudos [?]: 14 [1], given: 27

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

Show Tags

New post 28 May 2013, 02:17
1
This post received
KUDOS
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


256=2^8=> 2^98 can be written as 2^90*2^8 such that L can be treated as 2^90. hence N=0.
Ans A

Kudos [?]: 14 [1], given: 27

Intern
Intern
avatar
Joined: 29 Jul 2012
Posts: 28

Kudos [?]: 19 [0], given: 7

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

Show Tags

New post 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?

Kudos [?]: 19 [0], given: 7

Non-Human User
User avatar
Joined: 09 Sep 2013
Posts: 15576

Kudos [?]: 283 [0], given: 0

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

Show Tags

New post 19 Jul 2014, 02:36
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________

GMAT Books | GMAT Club Tests | Best Prices on GMAT Courses | GMAT Mobile App | Math Resources | Verbal Resources

Kudos [?]: 283 [0], given: 0

Senior Manager
Senior Manager
avatar
S
Joined: 18 Aug 2014
Posts: 260

Kudos [?]: 71 [0], given: 78

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

Show Tags

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

Please help me find my lost Kudo's bird

Kudos [?]: 71 [0], given: 78

Expert Post
1 KUDOS received
Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 42283

Kudos [?]: 132935 [1], given: 12391

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

Show Tags

New post 26 Nov 2015, 07:10
1
This post received
KUDOS
Expert's post
redfield wrote:
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 \(\frac{N}{2^8}\) ?


\(2^{98}=2^8*L+N\)

Dividing both parts by \(2^8\):

\(\frac{2^{98}}{2^8}=\frac{2^8*L+N}{2^8}\);

\(2^{98-8}=\frac{2^8*L}{2^8}+\frac{N}{2^8}\)

\(2^{90}=L+\frac{N}{2^8}\).

Hope it's clear.
_________________

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 [?]: 132935 [1], given: 12391

Expert Post
Senior Manager
Senior Manager
User avatar
Joined: 20 Aug 2015
Posts: 396

Kudos [?]: 349 [0], given: 10

Location: India
GMAT 1: 760 Q50 V44
If 2^98=256L+N, where L and N are integers and [#permalink]

Show Tags

New post 26 Nov 2015, 09:13
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


The first thing to note here is that LHS is an even number, hence RHS should also be even.
This elements N = 1,3

Now,
\(2^{98} = 2^8L + N\), Diving both the sides by \(2^8\)

\(2^{90} = L + N/2^8\),
Out of 0, 2 and 4, N can only be 0 for RHS to be an even number.

Option A
_________________

Reach out to us at bondwithus@gmatify.com

Image

Kudos [?]: 349 [0], given: 10

Non-Human User
User avatar
Joined: 09 Sep 2013
Posts: 15576

Kudos [?]: 283 [0], given: 0

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

Show Tags

New post 19 Jan 2017, 07:09
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________

GMAT Books | GMAT Club Tests | Best Prices on GMAT Courses | GMAT Mobile App | Math Resources | Verbal Resources

Kudos [?]: 283 [0], given: 0

Re: If 2^98=256L+N, where L and N are integers and   [#permalink] 19 Jan 2017, 07:09
Display posts from previous: Sort by

If 2^98=256L+N, where L and N are integers and

  new topic post reply Question banks Downloads My Bookmarks Reviews Important topics  


GMAT Club MBA Forum Home| About| Terms and Conditions| GMAT Club Rules| Contact| Sitemap

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®.