November 22, 2018 November 22, 2018 10:00 PM PST 11:00 PM PST Mark your calendars  All GMAT Club Tests are free and open November 22nd to celebrate Thanksgiving Day! Access will be available from 0:01 AM to 11:59 PM, Pacific Time (USA) November 23, 2018 November 23, 2018 10:00 PM PST 11:00 PM PST Practice the one most important Quant section  Integer properties, and rapidly improve your skills.
Author 
Message 
TAGS:

Hide Tags

Math Expert
Joined: 02 Sep 2009
Posts: 50711

Question Stats:
67% (01:18) correct 33% (02:05) wrong based on 145 sessions
HideShow timer Statistics



Math Expert
Joined: 02 Sep 2009
Posts: 50711

Re M2135
[#permalink]
Show Tags
16 Sep 2014, 00:15
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. Answer: E
_________________
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? Extrahard Quant Tests with Brilliant Analytics



Current Student
Joined: 11 Nov 2014
Posts: 11

Re: M2135
[#permalink]
Show Tags
03 Jul 2015, 21:40
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.
Answer: E 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?



Math Expert
Joined: 02 Aug 2009
Posts: 7039

Re: M2135
[#permalink]
Show Tags
03 Jul 2015, 22:03
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.
Answer: E 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..
_________________
1) Absolute modulus : http://gmatclub.com/forum/absolutemodulusabetterunderstanding210849.html#p1622372 2)Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html 3) effects of arithmetic operations : https://gmatclub.com/forum/effectsofarithmeticoperationsonfractions269413.html
GMAT online Tutor



Math Expert
Joined: 02 Aug 2009
Posts: 7039

Re: M2135
[#permalink]
Show Tags
03 Jul 2015, 22:10
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.
Answer: E 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
_________________
1) Absolute modulus : http://gmatclub.com/forum/absolutemodulusabetterunderstanding210849.html#p1622372 2)Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html 3) effects of arithmetic operations : https://gmatclub.com/forum/effectsofarithmeticoperationsonfractions269413.html
GMAT online Tutor



Current Student
Joined: 11 Nov 2014
Posts: 11

Re: M2135
[#permalink]
Show Tags
03 Jul 2015, 22:22
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.
Answer: E 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 ).



Current Student
Joined: 12 Aug 2015
Posts: 287
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)

Re: M2135
[#permalink]
Show Tags
22 Nov 2015, 06:15
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



Intern
Joined: 13 Jul 2016
Posts: 38

Re: M2135
[#permalink]
Show Tags
14 Oct 2016, 10:59
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.



Intern
Joined: 12 Dec 2016
Posts: 4

Re M2135
[#permalink]
Show Tags
28 Aug 2017, 08:34
I think this is a highquality question and I agree with explanation.



Intern
Joined: 26 Feb 2018
Posts: 4

Re: M2135
[#permalink]
Show Tags
29 Oct 2018, 16:33
Hi Bunuel Can you please confirm if my logic is right while doing this exercise?
n = 4k + 1 > Insuf
n = 7k + 6 > Insuf
n = 28k + 13 (13 is the frist common number of 4k+1 and 7k+6).
As 28k is not divide by 3, the Reminder will change.
Thanks in advance!



Math Expert
Joined: 02 Sep 2009
Posts: 50711

Re: M2135
[#permalink]
Show Tags
29 Oct 2018, 21:09



Intern
Joined: 26 Feb 2018
Posts: 4

Re: M2135
[#permalink]
Show Tags
30 Oct 2018, 08:27
Bunuel wrote: efr wrote: Hi Bunuel Can you please confirm if my logic is right while doing this exercise?
n = 4k + 1 > Insuf
n = 7k + 6 > Insuf
n = 28k + 13 (13 is the frist common number of 4k+1 and 7k+6).
As 28k is not divide by 3, the Reminder will change.
Thanks in advance! Correct with small but crucial correction: quotient in all three cases (highlighted) will not be the same so you should not sue one variable for them (use p, q, k instead). Perfect! Thanks for the explanation Bunuel!










