GMAT Question of the Day - Daily to your Mailbox; hard ones only

 It is currently 20 Jul 2018, 09:45

### 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

# M23-29

Author Message
TAGS:

### Hide Tags

Math Expert
Joined: 02 Sep 2009
Posts: 47152

### Show Tags

16 Sep 2014, 01:19
00:00

Difficulty:

95% (hard)

Question Stats:

51% (01:08) correct 49% (01:44) wrong based on 148 sessions

### HideShow timer Statistics

If 1 book was left when a pile of books was stacked in rows of 5, how many books are there in all?

(1) When the books were stacked in rows of 7, 1 book was left

(2) When the books were stacked in rows of 8, 4 books were left

_________________
Math Expert
Joined: 02 Sep 2009
Posts: 47152

### Show Tags

16 Sep 2014, 01:19
1
1
Official Solution:

Statements (1) and (2) combined are insufficient. There is no limitation on the number of books. Without this limitation, the question cannot be answered: there is an infinite number of integers that satisfy the setup and the statements.

_________________
Intern
Joined: 06 Jun 2014
Posts: 5

### Show Tags

27 May 2015, 22:40
Hello,

Please explain stmt. 2 in detail - Is it possible to get more than one solution for stmt.2?

Regards,
Mahuya

Statements (1) and (2) combined are insufficient. There is no limitation on the number of books. Without this limitation, the question cannot be answered: there is an infinite number of integers that satisfy the setup and the statements.

Intern
Joined: 11 Nov 2014
Posts: 38
Concentration: Technology, Strategy
GMAT 1: 660 Q48 V31
GMAT 2: 720 Q50 V37
GPA: 3.6
WE: Consulting (Consulting)

### Show Tags

05 Jul 2015, 17:44
3
2
mahuya78

Basically the question is saying that X (being the number of books) divided by 5 gives you remainder of 1. Question is, can we determine X?

A) X/7 gives us remainder of 1. So X can be: 8, 15, 22, 29, 36, 43, 50, 57, 64, 71, 78, 85, 92, 99 etc (just keep adding 7 and you will get infinite options). In the list above, there are 2 numbers *that I listed* that will yield a remainder of 1, when divided by 5: 36 and 71. So insufficient.

B) X/8 gives us remainder of 4. So X can be: 12, 20, 28, 36, 44, 52, 60, 68, 76, 84, 92, 100, 108 etc (again, just keep adding 8 and you will get infinite options). In the list above, there are 2 numbers *that I listed* that will yield a remainder of 1, when divided by 5: 36 and 76. So again, insufficient.

C) Apparently there seems to be only one option: 36, which would make you believe you have enough information. But the truth is that if you keep on trying, you will find other options that will fit both restrictions (for instance: 316), making both together insufficient.

Since testing all these numbers is somewhat impossible in the given time, I recommend you to learn this type of question so whenever you are asked, you know how to tackle it.

Cheers!
Math Expert
Joined: 02 Aug 2009
Posts: 6258

### Show Tags

05 Jul 2015, 19:23
2
@michaelyb wrote:
@mahuya78

Basically the question is saying that X (being the number of books) divided by 5 gives you remainder of 1. Question is, can we determine X?

A) X/7 gives us remainder of 1. So X can be: 8, 15, 22, 29, 36, 43, 50, 57, 64, 71, 78, 85, 92, 99 etc (just keep adding 7 and you will get infinite options). In the list above, there are 2 numbers *that I listed* that will yield a remainder of 1, when divided by 5: 36 and 71. So insufficient.

B) X/8 gives us remainder of 4. So X can be: 12, 20, 28, 36, 44, 52, 60, 68, 76, 84, 92, 100, 108 etc (again, just keep adding 8 and you will get infinite options). In the list above, there are 2 numbers *that I listed* that will yield a remainder of 1, when divided by 5: 36 and 76. So again, insufficient.

C) Apparently there seems to be only one option: 36, which would make you believe you have enough information. But the truth is that if you keep on trying, you will find other options that will fit both restrictions (for instance: 316), making both together insufficient.

Since testing all these numbers is somewhat impossible in the given time, I recommend you to learn this type of question so whenever you are asked, you know how to tackle it.

Cheers!

Hi,
you have missed lot many numbers in between ...
after 36, the next will not be 316..
the way to find the next number after first number is .. first number +LCM of the two numbers..
in this case 36 + LCM of 7 & 8= 36 + 56 = 92..
and next 92 + 56 = 148 and so on..
_________________

1) Absolute modulus : http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372
2)Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html
3) effects of arithmetic operations : https://gmatclub.com/forum/effects-of-arithmetic-operations-on-fractions-269413.html

GMAT online Tutor

Intern
Joined: 11 Nov 2014
Posts: 38
Concentration: Technology, Strategy
GMAT 1: 660 Q48 V31
GMAT 2: 720 Q50 V37
GPA: 3.6
WE: Consulting (Consulting)

### Show Tags

17 Oct 2015, 15:40
1
chetan2u

You forgot to include 5 in your LCM calculation since X divided by 5 gives you a remainder of 1 according to the question stem, and 148 does not. So the next number after 36 is indeed 36 + the LCM of 7, 8 and 5, or 36+280 .
Manager
Joined: 14 Oct 2012
Posts: 177

### Show Tags

Updated on: 01 Apr 2017, 09:05
Given: 5x+1 -> 1,6,11,16...
S-1) 7x+1
Therefore the combine sequence will be x+35
Let's start with 7x+1, here LCM(7,5)=35 gives the repeating term "35". Now we need to figure out variable term.
7x+1
x=0 => 1 => Divide this output of 1 [7*0+1 = 1] by 5 => 1/5 => remainder(1).
Note that this remainder 1 matches with the repeating term of original sequence = 5x+1
That's how you know that we have received the variable term of the merged sequence => x + 35
Now common terms => x+35 => 35,36,37,38,39,40,41.....
5x+1 => 5*6+1=36, 5*8+1=41
Not sufficient

S-2) 8x+4
X=4 => 32+4 = 36 => 36/5 = r(1) => variable term = 36
Merged sequence = 36x+40
NOTE:initially you will take some time to figure out this merged sequence but with practice you will get better. I prefer this method because in using simply number from the sequence many times i miss some and am not able to get correct results as to if statement is sufficient or not.
36x+40 => 40,76,112,148,184,220,256
5x+1 -> 5*15+1=76, & 5*51+1=256
Not sufficient

S-(1+2): 5x+1 & 7x+1 & 8x+4 => common terms between x+35 & 36x+40
36x+40 = (35x+6)+(x+35)
Whenever, we have something as this => one sequence can be made part of other => there are ALWAYS common terms.

Also, x+35 is a sequence of adding 1 to all the numbers after 35 => 36,37,38...
So we can definitely say that 40, 76 and other terms of 36x+40 are also a part of x+35. So not sufficient.
E

Originally posted by manishtank1988 on 01 Apr 2017, 08:59.
Last edited by manishtank1988 on 01 Apr 2017, 09:05, edited 1 time in total.
Manager
Joined: 14 Oct 2012
Posts: 177

### Show Tags

01 Apr 2017, 09:01
Hello moderators,
Engr2012, Abhishek009, Skywalker18, Bunuel, mikemcgarry, VeritasPrepKarishma, Vyshak, msk0657, abhimahna
Please let me know if there is a faster method to solve these kinds of problems. It took me > 2.5 minutes in solving it. Any help appreciated.
Thanks
Board of Directors
Status: Stepping into my 10 years long dream
Joined: 18 Jul 2015
Posts: 3687

### Show Tags

01 Apr 2017, 09:22
1
1
manishtank1988 wrote:
Please let me know if there is a faster method to solve these kinds of problems. It took me > 2.5 minutes in solving it. Any help appreciated.
Thanks

Hey, the best way to solve such questions is as follows:

Question is saying N = 5k + 1; Number could be 1,6,...

Statement 1: N = 7k+1

If I want to find the common numbers, I will use First Common Number + LCM method.

So, N = 1, 1st Term+(LCM of 5,7), 2nd Term + (LCM of 5,7), and so on.

or N = 1, 36,72. Since, we can multiple values of N. It means Insufficient.

Statement 2: N = 8k+4

First common N = 36.

Next term = 36 + LCM(5,8), and so on.

or N = 36 + 40 = 76

So, we have more than 1 common values, hence, Insufficient.

Combining:

N = 7k+1

and N = 8k'+4

First common = 36.

Second = 36 + LCM(7,8)=92. Hence, again more than 1. Insufficient. Hence, E
_________________

My GMAT Story: From V21 to V40
My MBA Journey: My 10 years long MBA Dream
My Secret Hacks: Best way to use GMATClub | Importance of an Error Log!
Verbal Resources: All SC Resources at one place | All CR Resources at one place

GMAT Club Inbuilt Error Log Functionality - View More.
New Visa Forum - Ask all your Visa Related Questions - here.

Find a bug in the new email templates and get rewarded with 2 weeks of GMATClub Tests for free

Manager
Joined: 14 Oct 2012
Posts: 177

### Show Tags

02 Apr 2017, 10:15
abhimahna wrote:
manishtank1988 wrote:
Please let me know if there is a faster method to solve these kinds of problems. It took me > 2.5 minutes in solving it. Any help appreciated.
Thanks

Hey, the best way to solve such questions is as follows:

Question is saying N = 5k + 1; Number could be 1,6,...

Statement 1: N = 7k+1

If I want to find the common numbers, I will use First Common Number + LCM method.

So, N = 1, 1st Term+(LCM of 5,7), 2nd Term + (LCM of 5,7), and so on.

or N = 1, 36,72. Since, we can multiple values of N. It means Insufficient.

Statement 2: N = 8k+4

First common N = 36.

Next term = 36 + LCM(5,8), and so on.

or N = 36 + 40 = 76

So, we have more than 1 common values, hence, Insufficient.

Combining:

N = 7k+1

and N = 8k'+4

First common = 36.

Second = 36 + LCM(7,8)=92. Hence, again more than 1. Insufficient. Hence, E

Thanks abhimahna
GMAT Club Legend
Joined: 16 Oct 2010
Posts: 8132
Location: Pune, India

### Show Tags

03 Apr 2017, 07:14
5
1
manishtank1988 wrote:
Hello moderators,
Engr2012, Abhishek009, Skywalker18, Bunuel, mikemcgarry, VeritasPrepKarishma, Vyshak, msk0657, abhimahna
Please let me know if there is a faster method to solve these kinds of problems. It took me > 2.5 minutes in solving it. Any help appreciated.
Thanks

Check out these divisibility posts:

https://www.veritasprep.com/blog/2011/0 ... unraveled/
https://www.veritasprep.com/blog/2011/0 ... y-applied/
https://www.veritasprep.com/blog/2011/0 ... emainders/
https://www.veritasprep.com/blog/2011/0 ... s-part-ii/

Now the question can be done in seconds.

Data in the Question stem: N = (5a + 1) or (5a' - 4)

Stmnt 1: N = (7b + 1) or (7b' - 6)

Using Stmnt 1 alone (and data in the question stem),
The common remainder is 1 so N = 35m + 1. (because 35 is the LCM of 5 and 7)
There will be infinite values of N such as 1, 36, 71 etc

Stmnt 2: N = (8c + 4) or (8c' - 4)

Using Stmnt 2 alone (and data in the question stem),
The common remainder is -4 so N = 40p - 4 (because 40 is the LCM of 5 and 8)
There will be infinite values of N such as 36, 76 etc

Using both stmnts, we see that the first common value is 36.
So N = 280q + 36 (because 280 is the LCM of 35 and 40). There will be infinite numbers of this form.

_________________

Karishma
Private Tutor for GMAT
Contact: bansal.karishma@gmail.com

Manager
Joined: 14 Oct 2012
Posts: 177

### Show Tags

16 Apr 2017, 18:56
VeritasPrepKarishma wrote:
manishtank1988 wrote:
Hello moderators,
Engr2012, Abhishek009, Skywalker18, Bunuel, mikemcgarry, VeritasPrepKarishma, Vyshak, msk0657, abhimahna
Please let me know if there is a faster method to solve these kinds of problems. It took me > 2.5 minutes in solving it. Any help appreciated.
Thanks

Check out these divisibility posts:

https://www.veritasprep.com/blog/2011/0 ... unraveled/
https://www.veritasprep.com/blog/2011/0 ... y-applied/
https://www.veritasprep.com/blog/2011/0 ... emainders/
https://www.veritasprep.com/blog/2011/0 ... s-part-ii/

Now the question can be done in seconds.

Data in the Question stem: N = (5a + 1) or (5a' - 4)

Stmnt 1: N = (7b + 1) or (7b' - 6)

Using Stmnt 1 alone (and data in the question stem),
The common remainder is 1 so N = 35m + 1. (because 35 is the LCM of 5 and 7)
There will be infinite values of N such as 1, 36, 71 etc

Stmnt 2: N = (8c + 4) or (8c' - 4)

Using Stmnt 2 alone (and data in the question stem),
The common remainder is -4 so N = 40p - 4 (because 40 is the LCM of 5 and 8)
There will be infinite values of N such as 36, 76 etc

Using both stmnts, we see that the first common value is 36.
So N = 280q + 36 (because 280 is the LCM of 35 and 40). There will be infinite numbers of this form.

Great solution: VeritasPrepKarishma
Thanks a lot
Director
Joined: 08 Jun 2015
Posts: 517
Location: India
GMAT 1: 640 Q48 V29
GMAT 2: 700 Q48 V38

### Show Tags

26 Mar 2018, 09:03
+1 for option E
_________________

" The few , the fearless "

Manager
Joined: 26 Dec 2017
Posts: 127

### Show Tags

06 Jun 2018, 01:35
Quote:
manishtank1988 wrote:
Hello moderators,
Engr2012, Abhishek009, Skywalker18, Bunuel, mikemcgarry, VeritasPrepKarishma, Vyshak, msk0657, abhimahna
Please let me know if there is a faster method to solve these kinds of problems. It took me > 2.5 minutes in solving it. Any help appreciated.
Thanks

Check out these divisibility posts:

https://www.veritasprep.com/blog/2011/0 ... unraveled/
https://www.veritasprep.com/blog/2011/0 ... y-applied/
https://www.veritasprep.com/blog/2011/0 ... emainders/
https://www.veritasprep.com/blog/2011/0 ... s-part-ii/

Now the question can be done in seconds.

Data in the Question stem: N = (5a + 1) or (5a' - 4)

Stmnt 1: N = (7b + 1) or (7b' - 6)

Using Stmnt 1 alone (and data in the question stem),
The common remainder is 1 so N = 35m + 1. (because 35 is the LCM of 5 and 7)
There will be infinite values of N such as 1, 36, 71 etc

Stmnt 2: N = (8c + 4) or (8c' - 4)

Using Stmnt 2 alone (and data in the question stem),
The common remainder is -4 so N = 40p - 4 (because 40 is the LCM of 5 and 8)
There will be infinite values of N such as 36, 76 etc

Using both stmnts, we see that the first common value is 36.
So N = 280q + 36 (because 280 is the LCM of 35 and 40). There will be infinite numbers of this form.

Hi VeritasPrepKarishma / Experts ,
I have a doubt regarding the marked portion above.
For numbers I am aware that Next Num=First Num+LCM(a,b)

From the explanation I inferred for 3 numbers Next Num=First Num+LCM(a,b,c)

Is the above exactly correct in other words we are not missing any other values in between 36 and 316 (As pointed in one of the previous posts here by chetan2u) right?
_________________

--If you like my post pls give kudos

Board of Directors
Status: Stepping into my 10 years long dream
Joined: 18 Jul 2015
Posts: 3687

### Show Tags

06 Jun 2018, 02:12
tejyr wrote:
I have a doubt regarding the marked portion above.
For numbers I am aware that Next Num=First Num+LCM(a,b)

From the explanation I inferred for 3 numbers Next Num=First Num+LCM(a,b,c)

Is the above exactly correct in other words we are not missing any other values in between 36 and 316 (As pointed in one of the previous posts here by chetan2u) right?

Hey tejyr ,

Yes, that is correct.

When I am saying N = 280q + 36, then minimum value of N is 36 and the next possible value will be 316. There won't be any other value between 36 and 316.

I hope that helps
_________________

My GMAT Story: From V21 to V40
My MBA Journey: My 10 years long MBA Dream
My Secret Hacks: Best way to use GMATClub | Importance of an Error Log!
Verbal Resources: All SC Resources at one place | All CR Resources at one place

GMAT Club Inbuilt Error Log Functionality - View More.
New Visa Forum - Ask all your Visa Related Questions - here.

Find a bug in the new email templates and get rewarded with 2 weeks of GMATClub Tests for free

Re: M23-29 &nbs [#permalink] 06 Jun 2018, 02:12
Display posts from previous: Sort by

# M23-29

Moderators: chetan2u, Bunuel

# Events & Promotions

 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®.