Author 
Message 
TAGS:

Hide Tags

Manager
Joined: 16 Jul 2009
Posts: 246

If x, a, and b are positive integers such that when x is [#permalink]
Show Tags
29 Sep 2010, 21:27
6
This post received KUDOS
60
This post was BOOKMARKED
Question Stats:
53% (02:03) correct 47% (02:26) wrong based on 707 sessions
HideShow timer Statistics
If x, a, and b are positive integers such that when x is divided by a, the remainder is b and when x is divided by b, the remainder is a2, then which of the following must be true? A. a is even B. x+b is divisible by a C. x1 is divisible by a D. b=a1 E. a+2=b+1
Official Answer and Stats are available only to registered users. Register/ Login.



Math Expert
Joined: 02 Sep 2009
Posts: 43853

Re: m16 #35 [#permalink]
Show Tags
29 Sep 2010, 21:49
56
This post received KUDOS
Expert's post
33
This post was BOOKMARKED
If \(x\), \(a\), and \(b\) are positive integers such that when \(x\) is divided by \(a\), the remainder is \(b\) and when \(x\) is divided by \(b\), the remainder is \(a2\), then which of the following must be true?A. \(a\) is even B. \(x+b\) is divisible by \(a\) C. \(x1\) is divisible by \(a\) D. \(b=a1\) E. \(a+2=b+1\) When \(x\) is divided by \(a\), the remainder is \(b\) > \(x=aq+b\) > \(remainder=b<a=divisor\) (remainder must be less than divisor); When \(x\) is divided by \(b\), the remainder is \(a2\) > \(x=bp+(a2)\) > \(remainder=(a2)<b=divisor\). So we have that: \(a2<b<a\), as \(a\) and \(b\) are integers, then it must be true that \(b=a1\) (there is only one integer between \(a2\) and \(a\), which is \(a1\) and we are told that this integer is \(b\), hence \(b=a1\)). Answer: D.
_________________
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



Senior Manager
Status: Time to step up the tempo
Joined: 24 Jun 2010
Posts: 403
Location: Milky way
Schools: ISB, Tepper  CMU, Chicago Booth, LSB

Re: m16 #35 [#permalink]
Show Tags
29 Sep 2010, 22:00
1
This post received KUDOS
Awesome Bunuel. Was trying hard to figure out the answer. All the information was just readily available but was looking all around. +1 to you.
_________________
Support GMAT Club by putting a GMAT Club badge on your blog



Manager
Joined: 16 Jul 2009
Posts: 246

Re: m16 #35 [#permalink]
Show Tags
29 Sep 2010, 22:06
Thanks Bunnel!! Do you suggest using a particular strategy for these problems or using different strategy for every problem and whichever fits the bill for the given question..?



Retired Moderator
Joined: 03 Aug 2010
Posts: 234

Re: m16 #35 [#permalink]
Show Tags
29 Sep 2010, 22:18
Simply amazing bunuel...... you are a genius....
_________________
http://www.gmatpill.com/gmatpracticetest/
Amazing Platform



Manager
Joined: 25 Mar 2009
Posts: 54

Re: m16 #35 [#permalink]
Show Tags
29 Sep 2010, 22:29
cool, BUenel, i had go arround with the formulations but can not find that A2<B tks much



Math Expert
Joined: 02 Sep 2009
Posts: 43853

Re: m16 #35 [#permalink]
Show Tags
29 Sep 2010, 22:39
4
This post received KUDOS
Expert's post
8
This post was BOOKMARKED



Senior Manager
Joined: 18 Feb 2008
Posts: 494
Location: Kolkata

Re: m16 #35 [#permalink]
Show Tags
30 Sep 2010, 04:57
2
This post was BOOKMARKED
Bunuel wrote: Priceless info.Ton thanks



Manager
Joined: 19 Apr 2010
Posts: 205
Schools: ISB, HEC, Said

Re: m16 #35 [#permalink]
Show Tags
01 Oct 2010, 06:29
Awesome solution Bunuel. +1 Great question Kudos to GMAT club tests



Manager
Joined: 06 Nov 2009
Posts: 177
Concentration: Finance, Strategy

Re: m16 #35 [#permalink]
Show Tags
01 Oct 2010, 09:35
Awesome!!! I didnt get the fastest way, but through number plugin I was able to figure out the solution. But your solution is so much better!!



Senior Manager
Joined: 06 Jun 2009
Posts: 327
Location: USA
WE 1: Engineering

Re: m16 #35 [#permalink]
Show Tags
01 Oct 2010, 18:49
nice and efficient !
_________________
All things are possible to those who believe.



Manager
Joined: 16 Jul 2009
Posts: 246

Re: m16 #35 [#permalink]
Show Tags
02 Oct 2010, 19:02
Bunnel, thanks so much for the compilation! By 'these problems' I meant 'Must be true' questions in which at times you have more than 1 correct answers. My apologies for the lack of clarity their. Your compilation should be enough to practice. Thanks again!



Intern
Joined: 20 Jul 2010
Posts: 49

Re: m16 #35 [#permalink]
Show Tags
20 Oct 2010, 03:38
Nice approach. +1 kudos



Senior Manager
Joined: 11 May 2011
Posts: 356

Re: If x, a, and b are positive integers such that when x is div [#permalink]
Show Tags
04 Sep 2012, 00:14
2
This post received KUDOS
sanjoo wrote: If x, a, and b are positive integers such that when x is divided by a, the remainder is b and when x is divided by b, the remainder is a−2, then which of the following must be true?
A)a is even b)x+b is divisible by a c)x−1 is divisible by a d)b=a−1 e)a+2=b+1 When divided by A, remainder is B, this implies A > B When divided by B, remainder is A2, this implies B > A 2 Combining both, B < A < (B + 2) Since, A and B are integers, A = B + 1 Answer is (D) . Cheers!
_________________
 What you do TODAY is important because you're exchanging a day of your life for it! 



Manager
Joined: 28 Feb 2012
Posts: 114
Concentration: Strategy, International Business
GPA: 3.9
WE: Marketing (Other)

Re: If x, a, and b are positive integers such that when x is [#permalink]
Show Tags
05 Sep 2012, 22:56
Bunuel wrote: If \(x\), \(a\), and \(b\) are positive integers such that when \(x\) is divided by \(a\), the remainder is \(b\) and when \(x\) is divided by \(b\), the remainder is \(a2\), then which of the following must be true?
A. \(a\) is even B. \(x+b\) is divisible by \(a\) C. \(x1\) is divisible by \(a\) D. \(b=a1\) E. \(a+2=b+1\)
When \(x\) is divided by \(a\), the remainder is \(b\) > \(x=aq+b\) > \(remainder=b<a=divisor\) (remainder must be less than divisor); When \(x\) is divided by \(b\), the remainder is \(a2\) > \(x=bp+(a2)\) > \(remainder=(a2)<b=divisor\).
So we have that: \(a2<b<a\), as \(a\) and \(b\) are integers, then it must be true that \(b=a1\) (there is only one integer between \(a2\) and \(a\), which is \(a1\) and we are told that this integer is \(b\), hence \(b=a1\)).
Answer: D. Indeed very nice explanation, but for me, for the person who is not that strong in quants sometimes difficult to keep all that concepts in my head and i am jumping to different approaches. Whenever i see must be true questions i plug in some numbers and see which answer works, since it is must be true questions any numbers should work equally. For example in this problem: lets says x=5, a=3 then b=2, so check all the answers and we see that only d works, but if there will be two answers that work try different numbers till we get only one. It could be time consuming, but when we are asked simple expressions it is easy to find numbers that work well. Bunuel, do you think there are any pitfalls that i should be aware of?
_________________
If you found my post useful and/or interesting  you are welcome to give kudos!



Senior Manager
Joined: 23 Mar 2011
Posts: 460
Location: India
GPA: 2.5
WE: Operations (Hospitality and Tourism)

Re: If x, a, and b are positive integers such that when x is [#permalink]
Show Tags
20 Apr 2013, 11:57
My approach of plug in numbers was certainly not the best approach, in a zest of GMAT I dont know why I am loosing to think simple....this was a simple algebra which I complicated with numbers
_________________
"When the going gets tough, the tough gets going!"
Bring ON SOME KUDOS MATES+++  Quant Notes consolidated: http://gmatclub.com/forum/consoloditedquantguidesofforummosthelpfulinpreps151067.html#p1217652 My GMAT journey begins: http://gmatclub.com/forum/mygmatjourneybegins122251.html All about Richard Ivey: http://gmatclub.com/forum/allaboutrichardivey148594.html#p1190518



Math Expert
Joined: 02 Sep 2009
Posts: 43853

Re: If x, a, and b are positive integers such that when x is [#permalink]
Show Tags
19 Jun 2013, 03:52



Manager
Joined: 30 May 2013
Posts: 186
Location: India
Concentration: Entrepreneurship, General Management
GPA: 3.82

Re: If x, a, and b are positive integers such that when x is [#permalink]
Show Tags
26 Sep 2013, 19:26
Way bunuel approch the problem is fantastic. I dont know whether i would be able to think the way he does . . tat too in time pressure +1 kudos



Intern
Joined: 15 Dec 2007
Posts: 13

Re: If x, a, and b are positive integers such that when x is [#permalink]
Show Tags
03 Nov 2013, 00:50
2
This post received KUDOS
This problem can be solved by plugging numbers. For instance if x=5, a=3, b=2 we get: ...when x is divided by a, the remainder is b... > 5/3=1+2/3 ...and when x is divided by b, the remainder is a2... > 5/2=2+1/2
Both work. If we plug numbers into the answers, only D will work.
Please give kudos, if you like this.



Intern
Joined: 11 Aug 2011
Posts: 12
Location: India
GPA: 3
WE: Engineering (Other)

Re: m16 #35 [#permalink]
Show Tags
06 Nov 2013, 13:10
Bunuel wrote: If \(x\), \(a\), and \(b\) are positive integers such that when \(x\) is divided by \(a\), the remainder is \(b\) and when \(x\) is divided by \(b\), the remainder is \(a2\), then which of the following must be true?
A. \(a\) is even B. \(x+b\) is divisible by \(a\) C. \(x1\) is divisible by \(a\) D. \(b=a1\) E. \(a+2=b+1\)
When \(x\) is divided by \(a\), the remainder is \(b\) > \(x=aq+b\) > \(remainder=b<a=divisor\) (remainder must be less than divisor); When \(x\) is divided by \(b\), the remainder is \(a2\) > \(x=bp+(a2)\) > \(remainder=(a2)<b=divisor\).
So we have that: \(a2<b<a\), as \(a\) and \(b\) are integers, then it must be true that \(b=a1\) (there is only one integer between \(a2\) and \(a\), which is \(a1\) and we are told that this integer is \(b\), hence \(b=a1\)).
Answer: D. Hi Bunuel, I solved this problem with a bit different approach x = p*a + b..........eqn(1) and x = q*b + (a2)..............eqn(2) now, equating eqn(1) and eqn(2) p*a + b = q*b + (a2) a(p1) = b(q1)  2 if we put p = q = 3 we get, 2a = 2b  2 or a = b  1 or a + 2 = b + 1 which is option E would pl tell me where am i wrong with my approach?? Thanks.







Go to page
1 2
Next
[ 28 posts ]



