Events & Promotions
|
|
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
Track
Your Progress
Practice
Pays
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
Apr 2610:00 AM EDT
-11:00 AM EDT
The Target Test Prep course represents a quantum leap forward in GMAT preparation, a radical reinterpretation of the way that students should study. Try before you buy with a 5-day, full-access trial of the course for FREE!
Apr 2711:00 AM EDT
-12:00 PM EDT
Prefer video-based learning? The Target Test Prep OnDemand course is a one-of-a-kind video masterclass featuring 400 hours of lecture-style teaching by Scott Woodbury-Stewart, founder of Target Test Prep and one of the most accomplished GMAT instructors
Apr 2808:00 AM PDT
-11:00 AM PDT
Whether you are just beginning your MBA journey or fine-tuning your path to a 700+ score, GMAT Day is designed to give you a competitive edge.
16 Sep 2014, 01:00
Kudos
Bookmarks
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. \(a+2=b+1\)
A. \(a\) is even
B. \(x+b\) is divisible by \(a\)
C. \(x-1\) is divisible by \(a\)
D. \(b=a-1\)
E. \(a+2=b+1\)
16 Sep 2014, 01:00
Kudos
Bookmarks
Official Solution:
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 \(x\) is divided by \(a\), the remainder is \(b\) can be expressed as \(x=aq+b\), where \((remainder=b) \lt (a=divisor)\) (as the remainder must be less than divisor);
When \(x\) is divided by \(b\), the remainder is \(a-2\) can be expressed as \(x=bp+(a-2)\), where \((remainder=a-2) \lt (b = divisor)\).
From these equations, we can deduce that: \(a - 2 < b < a\). Since \(a\) and \(b\) are integers, it must be true that \(b = a - 1\). This is because there is only one integer between \(a - 2\) and \(a\), which is \(a - 1\), and we know that this integer is equal to \(b\). Therefore, \(b = a - 1\).
Answer: D
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 \(x\) is divided by \(a\), the remainder is \(b\) can be expressed as \(x=aq+b\), where \((remainder=b) \lt (a=divisor)\) (as the remainder must be less than divisor);
When \(x\) is divided by \(b\), the remainder is \(a-2\) can be expressed as \(x=bp+(a-2)\), where \((remainder=a-2) \lt (b = divisor)\).
From these equations, we can deduce that: \(a - 2 < b < a\). Since \(a\) and \(b\) are integers, it must be true that \(b = a - 1\). This is because there is only one integer between \(a - 2\) and \(a\), which is \(a - 1\), and we know that this integer is equal to \(b\). Therefore, \(b = a - 1\).
Answer: D
05 Jul 2016, 06:48
Kudos
Bookmarks
gmatprepeugene2014
When you say "factor out q's" where does q go? It cannot just disappear.
But more importantly, the quotient should not be the same. If you check the solution above you'll see that it's x=aq+b in one case and x=bp+(a-2). We don't know whether q = p.
