It is currently 20 Oct 2017, 07:58

### 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

# M21-35

Author Message
TAGS:

### Hide Tags

Math Expert
Joined: 02 Sep 2009
Posts: 41891

Kudos [?]: 128980 [0], given: 12185

### Show Tags

16 Sep 2014, 01:15
Expert's post
4
This post was
BOOKMARKED
00:00

Difficulty:

55% (hard)

Question Stats:

67% (01:27) correct 33% (01:29) wrong based on 45 sessions

### HideShow timer Statistics

If $$x$$ is a positive integer, what is the remainder when $$x$$ is divided by 3?

(1) The remainder when $$x$$ is divided by 4 is 1

(2) The remainder when $$x$$ is divided by 7 is 6
[Reveal] Spoiler: OA

_________________

Kudos [?]: 128980 [0], given: 12185

Math Expert
Joined: 02 Sep 2009
Posts: 41891

Kudos [?]: 128980 [0], given: 12185

### Show Tags

16 Sep 2014, 01:15
Expert's post
2
This post was
BOOKMARKED
Official Solution:

(1) The remainder when $$x$$ is divided by 4 is 1. Given: $$x=4q+1$$, so $$x$$ cold be: 1, 5, 9, 13, 17, 21, 25, 29, 33, 37, 41, ... Not sufficient.

(2) The remainder when $$x$$ is divided by 7 is 6. Given: $$x=7p+6$$, so $$x$$ cold be: 6, 13, 20, 27, 34, 41, ... Not sufficient.

(1)+(2) If $$x=13$$ then the remainder is 1 but if $$x=41$$ then the remainder is 2. Not sufficient.

_________________

Kudos [?]: 128980 [0], given: 12185

Current Student
Joined: 11 Nov 2014
Posts: 11

Kudos [?]: 2 [0], given: 17

### Show Tags

03 Jul 2015, 22:40
Bunuel wrote:
Official Solution:

(1) The remainder when $$x$$ is divided by 4 is 1. Given: $$x=4q+1$$, so $$x$$ cold be: 1, 5, 9, 13, 17, 21, 25, 29, 33, 37, 41, ... Not sufficient.

(2) The remainder when $$x$$ is divided by 7 is 6. Given: $$x=7p+6$$, so $$x$$ cold be: 6, 13, 20, 27, 34, 41, ... Not sufficient.

(1)+(2) If $$x=13$$ then the remainder is 1 but if $$x=41$$ then the remainder is 2. Not sufficient.

Thanks for the explanation Bunuel. I too used the same method.

However, I was wondering if there is any faster method to determine the two numbers (13 and 41).

In the exam, my intent is to save as much time as possible and move on to the next question. In that state of mind, I will be inclined to just test the (say) first 5 numbers of the sequence. In the case of (1) above, that will be just 1, 5, 9, 13, 17
and for (2), this will be 6,13, 20, 27, 34.

Clearly, I will have missed the important numbers (13 and 41) which are key to swinging towards the correct answer.
Any suggestions?

Kudos [?]: 2 [0], given: 17

Math Forum Moderator
Joined: 02 Aug 2009
Posts: 4977

Kudos [?]: 5481 [5], given: 112

### Show Tags

03 Jul 2015, 23:03
5
KUDOS
Expert's post
5
This post was
BOOKMARKED
samuraijack256 wrote:
Bunuel wrote:
Official Solution:

(1) The remainder when $$x$$ is divided by 4 is 1. Given: $$x=4q+1$$, so $$x$$ cold be: 1, 5, 9, 13, 17, 21, 25, 29, 33, 37, 41, ... Not sufficient.

(2) The remainder when $$x$$ is divided by 7 is 6. Given: $$x=7p+6$$, so $$x$$ cold be: 6, 13, 20, 27, 34, 41, ... Not sufficient.

(1)+(2) If $$x=13$$ then the remainder is 1 but if $$x=41$$ then the remainder is 2. Not sufficient.

Thanks for the explanation Bunuel. I too used the same method.

However, I was wondering if there is any faster method to determine the two numbers (13 and 41).

In the exam, my intent is to save as much time as possible and move on to the next question. In that state of mind, I will be inclined to just test the (say) first 5 numbers of the sequence. In the case of (1) above, that will be just 1, 5, 9, 13, 17
and for (2), this will be 6,13, 20, 27, 34.

Clearly, I will have missed the important numbers (13 and 41) which are key to swinging towards the correct answer.
Any suggestions?

Hi,
the fatser method in these Q is to find the first common multiple and thereafter add the LCM of two numbers..
in this case when we have got the fiest number as 13.. the next number will be 13 + lcm of 4 and 7=13+28=41..
the next 41+28=69 and so on...
so 13 gives remainder 1
41 gives remainder 2
69 gives remainder 0... and so on
hope it helped..
_________________

Absolute modulus :http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372
Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html

Kudos [?]: 5481 [5], given: 112

Math Forum Moderator
Joined: 02 Aug 2009
Posts: 4977

Kudos [?]: 5481 [2], given: 112

### Show Tags

03 Jul 2015, 23:10
2
KUDOS
Expert's post
samuraijack256 wrote:
Bunuel wrote:
Official Solution:

(1) The remainder when $$x$$ is divided by 4 is 1. Given: $$x=4q+1$$, so $$x$$ cold be: 1, 5, 9, 13, 17, 21, 25, 29, 33, 37, 41, ... Not sufficient.

(2) The remainder when $$x$$ is divided by 7 is 6. Given: $$x=7p+6$$, so $$x$$ cold be: 6, 13, 20, 27, 34, 41, ... Not sufficient.

(1)+(2) If $$x=13$$ then the remainder is 1 but if $$x=41$$ then the remainder is 2. Not sufficient.

Thanks for the explanation Bunuel. I too used the same method.

However, I was wondering if there is any faster method to determine the two numbers (13 and 41).

In the exam, my intent is to save as much time as possible and move on to the next question. In that state of mind, I will be inclined to just test the (say) first 5 numbers of the sequence. In the case of (1) above, that will be just 1, 5, 9, 13, 17
and for (2), this will be 6,13, 20, 27, 34.

Clearly, I will have missed the important numbers (13 and 41) which are key to swinging towards the correct answer.
Any suggestions?

another way would be to equate two eq from two statements..

$$x=4q+1$$, $$x=7p+6$$...
$$4q+1=7p+6$$....
$$4q=7p+5$$....

now substitute p as 1,3,5... as p has to be odd to make the eq even ..
where we getan integer value of q... those are the numbers we rae looking for..
for ex
p=1,q=3... so number=4q+1=13..
p=3, q is not an int..
p=5, q=10 number =4q+1=41.. and so on
_________________

Absolute modulus :http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372
Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html

Kudos [?]: 5481 [2], given: 112

Current Student
Joined: 11 Nov 2014
Posts: 11

Kudos [?]: 2 [0], given: 17

### Show Tags

03 Jul 2015, 23:22
chetan2u wrote:
samuraijack256 wrote:
Bunuel wrote:
Official Solution:

(1) The remainder when $$x$$ is divided by 4 is 1. Given: $$x=4q+1$$, so $$x$$ cold be: 1, 5, 9, 13, 17, 21, 25, 29, 33, 37, 41, ... Not sufficient.

(2) The remainder when $$x$$ is divided by 7 is 6. Given: $$x=7p+6$$, so $$x$$ cold be: 6, 13, 20, 27, 34, 41, ... Not sufficient.

(1)+(2) If $$x=13$$ then the remainder is 1 but if $$x=41$$ then the remainder is 2. Not sufficient.

Thanks for the explanation Bunuel. I too used the same method.

However, I was wondering if there is any faster method to determine the two numbers (13 and 41).

In the exam, my intent is to save as much time as possible and move on to the next question. In that state of mind, I will be inclined to just test the (say) first 5 numbers of the sequence. In the case of (1) above, that will be just 1, 5, 9, 13, 17
and for (2), this will be 6,13, 20, 27, 34.

Clearly, I will have missed the important numbers (13 and 41) which are key to swinging towards the correct answer.
Any suggestions?

Hi,
the fatser method in these Q is to find the first common multiple and thereafter add the LCM of two numbers..
in this case when we have got the fiest number as 13.. the next number will be 13 + lcm of 4 and 7=13+28=41..
the next 41+28=69 and so on...
so 13 gives remainder 1
41 gives remainder 2
69 gives remainder 0... and so on
hope it helped..

This is exactly what i had been looking for. Thanks Chetan!

Earlier, I too had tried looking at this from an LCM perspective, but after a few steps, I got confused and let it go.

I looked through your other proposed method as well (the one involving equating the two expressions), but I would prefer your first suggestion for the ease of application ( identifying if the variable has to be even/odd will again take a few 10s of seconds ).

Kudos [?]: 2 [0], given: 17

Current Student
Joined: 12 Aug 2015
Posts: 301

Kudos [?]: 545 [0], given: 1474

Concentration: General Management, Operations
GMAT 1: 640 Q40 V37
GMAT 2: 650 Q43 V36
GMAT 3: 600 Q47 V27
GPA: 3.3
WE: Management Consulting (Consulting)

### Show Tags

22 Nov 2015, 07:15
am i correct that as long as no limit is set on how far we can go calculating the potential values of n the answer will always be E? e.g. if we were set some limit: say 2 digit numbers or even etc - then we need to be accurate and find the values?
_________________

KUDO me plenty

Kudos [?]: 545 [0], given: 1474

Intern
Joined: 13 Jul 2016
Posts: 48

Kudos [?]: 6 [0], given: 309

GMAT 1: 770 Q50 V44

### Show Tags

14 Oct 2016, 11:59
Thinking in retrospect how the time for this question could have been reduced.

Clearly either statement is not sufficient on its own, which can be quickly tested..

Now for together:
x = LCM of (7,4) * (some variable integer K) + some constant remainder, lets say R

=> x = 28*K + R

We can find R by finding the common number as shown in the official solution.

Now remainder of 28K/3 will depend on the value of K, since 28 is not a multiple of 3.
So no matter what the value of R is, when we divide x by 3, the reminder obtained will depend on K hence we do not have a unique solution.

Bunuel - do you think it is valid line of reasoning. If yes, then we would not need to find R and can save some time.

Kudos [?]: 6 [0], given: 309

Intern
Joined: 12 Dec 2016
Posts: 2

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

### Show Tags

28 Aug 2017, 09:34
I think this is a high-quality question and I agree with explanation.

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

Re M21-35   [#permalink] 28 Aug 2017, 09:34
Display posts from previous: Sort by

# M21-35

Moderators: Bunuel, Vyshak

 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®.