Forum Home > GMAT > Quantitative > Problem Solving (PS)
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.
Jan 1507:30 AM PST
-08:30 AM PST
Studying for the GMAT can feel like a long journey, but having the right plan can keep you on track. In this live session, Julia, who scored a perfect 805, will share her step-by-step study plan for 2025. Join for help with key strategies and tips...
Jan 1610:00 AM EST
-12:00 PM EST
The Target Test Prep team is proud to introduce GMAT OnDemand, a video course created by Scott Woodbury-Stewart, founder and CEO, designed to provide you with everything you need to master the GMAT and achieve your score goals.- Jan 16
10:00 AM PST
-11:00 AM PST
Meet Kseniia, a former Public Relations Head at Google in Russia, whose inspiring journey defied the odds and landed her admissions to Wharton, Chicago Booth, and Yale SOM. - Jan 17
10:00 AM PST
-11:00 AM PST
Keshav who scored 99 percentile with near-perfect quant score shares his strategies and techniques to identify and overcome the weaknesses in his GMAT preparation. Learn the preparation tips and study strategies that worked for him. - Jan 18
10:00 AM PST
-11:00 AM PST
In this video, we talked to Rishi, an ambitious Indian student who successfully secured admission into HEC Paris and ESSEC Business School, two of the top Masters in Management (MiM) programs in Europe. 
Jan 1811:00 AM IST
-01:00 PM IST
Do RC/MSR passages scare you? e-GMAT is conducting a masterclass to help you learn – Learn effective reading strategies Tackle difficult RC & MSR with confidence Excel in timed test environment- Jan 19
11:00 AM IST
-01:00 PM IST
Attend this session to evaluate your current skill level, learn process skills, and solve through tough Arithmetic questions
E
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
35%
(medium)
Question Stats:
73% (02:03) correct
27%
(02:20)
wrong
based on 3806
sessions
History
Date
Time
Result
Not Attempted Yet
Positive integer n leaves a remainder of 4 after division by 6 and a remainder of 3 after division by 5. If n is greater than 30, what is the remainder that n leaves after division by 30?
A. 3
B. 12
C. 18
D. 22
E. 28
A. 3
B. 12
C. 18
D. 22
E. 28
06 May 2010, 04:18
Kudos
Bookmarks
To elaborate more.
Suppose we are told that:
Positive integer n leaves a remainder of 4 after division by 6 and a remainder of 2 after division by 8. What is the remainder that n leaves after division by 12?
The statement "positive integer n leaves a remainder of 4 after division by 6" can be expressed as: \(n=6p+4\). Thus according to this particular statement \(n\) could take the following values: 4, 10, 16, 22, 28, 34, 40, 46, 52, 58, 64, ...
The statement "positive integer n leaves a remainder of 2 after division by 8" can be expressed as: \(n=8q+2\). Thus according to this particular statement \(n\) could take the following values: 2, 10, 18, 26, 34, 42, 50, 58, 66, ...
The above two statements are both true, which means that the only valid values of \(n\) are the values which are common in both patterns. For example \(n\) can not be 16 (from first pattern) as the second formula does not give us 16 for any value of integer \(q\).
So we should derive general formula (based on both statements) that will give us only valid values of \(n\).
How can these two statement be expressed in one formula of a type \(n=kx+r\)? Where \(x\) is divisor and \(r\) is a remainder.
Divisor \(x\) would be the least common multiple of above two divisors 6 and 8, hence \(x=24\).
Remainder \(r\) would be the first common integer in above two patterns, hence \(r=10\).
Therefore general formula based on both statements is \(n=24k+10\). Thus according to this general formula valid values of \(n\) are: 10, 34, 58, ...
Now, \(n\) divided by 12 will give us the reminder of 10 (as 24k is divisible by 12).
Hope it helps.
Suppose we are told that:
Positive integer n leaves a remainder of 4 after division by 6 and a remainder of 2 after division by 8. What is the remainder that n leaves after division by 12?
The statement "positive integer n leaves a remainder of 4 after division by 6" can be expressed as: \(n=6p+4\). Thus according to this particular statement \(n\) could take the following values: 4, 10, 16, 22, 28, 34, 40, 46, 52, 58, 64, ...
The statement "positive integer n leaves a remainder of 2 after division by 8" can be expressed as: \(n=8q+2\). Thus according to this particular statement \(n\) could take the following values: 2, 10, 18, 26, 34, 42, 50, 58, 66, ...
The above two statements are both true, which means that the only valid values of \(n\) are the values which are common in both patterns. For example \(n\) can not be 16 (from first pattern) as the second formula does not give us 16 for any value of integer \(q\).
So we should derive general formula (based on both statements) that will give us only valid values of \(n\).
How can these two statement be expressed in one formula of a type \(n=kx+r\)? Where \(x\) is divisor and \(r\) is a remainder.
Divisor \(x\) would be the least common multiple of above two divisors 6 and 8, hence \(x=24\).
Remainder \(r\) would be the first common integer in above two patterns, hence \(r=10\).
Therefore general formula based on both statements is \(n=24k+10\). Thus according to this general formula valid values of \(n\) are: 10, 34, 58, ...
Now, \(n\) divided by 12 will give us the reminder of 10 (as 24k is divisible by 12).
Hope it helps.
29 Nov 2010, 04:40
Kudos
Bookmarks
Friends,
IT IS VERY COMMON IN GMAT to solve this kind of qtns using "NEGATIVE REMAINDER" theory.
The theory says:
if a # x is devided by y and leave the positive # r as the remainder then it can also leave negative # (r-y) as the remainder.
e.g:
9 when devided by 5 leves the remainder 4 : 9=5*1+4
it can also leave the remainder 4-5 = -1 : 9=5*2 -1
back to the original qtn:
n leaves a remainder of 4 after division by 6 and a remainder of 3 after division by 5
==> n leaves a remainder of -2 (i.e. 4-6) after division by 6 and a remainder of -2 (i.e. 3-5) after division by 5
==> n when devided by 5 and 6 leaves the same remainder -2.
what is n?
LCM (5,6)-2 = 30-2 = 28
CHECK: 28 when devided by 6 leaves the remainder 4 and when devided by 5 leaves the remainder 3
However, the qtn says n > 30
so what is the nex #, > 28, that can give the said remainders when devided by 6 and 5
nothing but 28 + (some multiple of 6 and 5) as this "some multiple of 6 and 5" will not give any remainder when devided by 5 or 6 but 28 will give the required remainders.
hence n could be anything that is in the form 28 + (some multiple of 6 and 5)
observe that "some multiple of 6 and 5" is always a multiple of 30 as LCM (5,6) = 30.
hence when n (i.e. 28 + some multiple of 6 and 5) is devided by 30 gives the remainder 28.
ANSWER "E"
Regards,
Murali.
Kudos?
IT IS VERY COMMON IN GMAT to solve this kind of qtns using "NEGATIVE REMAINDER" theory.
The theory says:
if a # x is devided by y and leave the positive # r as the remainder then it can also leave negative # (r-y) as the remainder.
e.g:
9 when devided by 5 leves the remainder 4 : 9=5*1+4
it can also leave the remainder 4-5 = -1 : 9=5*2 -1
back to the original qtn:
n leaves a remainder of 4 after division by 6 and a remainder of 3 after division by 5
==> n leaves a remainder of -2 (i.e. 4-6) after division by 6 and a remainder of -2 (i.e. 3-5) after division by 5
==> n when devided by 5 and 6 leaves the same remainder -2.
what is n?
LCM (5,6)-2 = 30-2 = 28
CHECK: 28 when devided by 6 leaves the remainder 4 and when devided by 5 leaves the remainder 3
However, the qtn says n > 30
so what is the nex #, > 28, that can give the said remainders when devided by 6 and 5
nothing but 28 + (some multiple of 6 and 5) as this "some multiple of 6 and 5" will not give any remainder when devided by 5 or 6 but 28 will give the required remainders.
hence n could be anything that is in the form 28 + (some multiple of 6 and 5)
observe that "some multiple of 6 and 5" is always a multiple of 30 as LCM (5,6) = 30.
hence when n (i.e. 28 + some multiple of 6 and 5) is devided by 30 gives the remainder 28.
ANSWER "E"
Regards,
Murali.
Kudos?
