|
|
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.
Updated on: 24 Aug 2021, 07:21
Originally posted by AkhilAggarwal on 26 May 2021, 16:26.
Last edited by Bunuel on 24 Aug 2021, 07:21, edited 3 times in total.
Last edited by Bunuel on 24 Aug 2021, 07:21, edited 3 times in total.
Kudos
Bookmarks
C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
65%
(hard)
Question Stats:
58% (02:11) correct
42%
(02:25)
wrong
based on 91
sessions
History
Date
Time
Result
Not Attempted Yet
What is the remainder when a positive integer n is divided by 39?
(1) The remainder when n is divided by 12 is 5.
(2) The remainder when n is divided by 65 is 7.
(1) The remainder when n is divided by 12 is 5.
(2) The remainder when n is divided by 65 is 7.
Pure algebraic questions like this one are no longer part of the GMAT Data Sufficiency syllabus.
For relevant Data Sufficiency questions, please check the Data Sufficiency forum:
Data Sufficiency Forum
Data Sufficiency Forum
24 Aug 2021, 06:57
Kudos
Bookmarks
Bunuel
Using stmt. 1,we can write
n=12a+5
=>not sufficient as we can't comment, what its remainder will be on dividing by 39
Similarily, from stmt. 2, we can write
n=65b+7
=>again,not sufficient as we can't comment, what its remainder will be on dividing by 39
Combining 1 and 2,
n=12a+5=65b+7
=>12a=65b+2----------(eqn1)
lowest number satisfying "eqn 1" is "132" that we will get on putting b=2
So,n=137 or in other words "n" will be of form LCM(12,65)+137
Now,we can easily find remainder of n on dividing by 39
So,the answer is C
PS: Bunuel ,could you please update the tags for this problem,so that this problem is accessible to many users.I don't have the rights to update the tags.
Using stmt. 1,we can write
n=12a+5
=>not sufficient as we can't comment, what its remainder will be on dividing by 39
Similarily, from stmt. 2, we can write
n=65b+7
=>again,not sufficient as we can't comment, what its remainder will be on dividing by 39
Combining 1 and 2,
n=12a+5=65b+7
=>12a=65b+2----------(eqn1)
lowest number satisfying "eqn 1" is "132" that we will get on putting b=2
So,n=137 or in other words "n" will be of form LCM(12,65)+137
Now,we can easily find remainder of n on dividing by 39
So,the answer is C
PS: Bunuel ,could you please update the tags for this problem,so that this problem is accessible to many users.I don't have the rights to update the tags.
22 Jan 2024, 03:41
Kudos
Bookmarks
From statement 1
equation will n=12a+5 by formula dividend= divisor x quotient + remainder
nothing is possible so eliminate this statement
from statement 2 again the equation is n = 65b+7
Now combining both equations n=12a+5=65b+7
n=12a=65b+7-5
n=12a=12b+53b+2 we will break 65b in terms of the 12b+53b so that we can easily
find out the value which gets divisible by 12.
n=12a=12b+(53b+2) For this 53b+2 we have to check what we can put the value in
this so if try for b=2 in 53b+2 put b=2 you will get 108 which is
divisible by 12. Now put the value of 2 in equation 2 which is
65b+7 it will come 137
Now take the LCM of 65 and 12 to make the equation
n=LCM(65,12)+137 NOTE: ALWAYS DO THIS FOR REMAINDER QUESTION IF 2 EQUATION IS GIVEN HAVING REMAINDER AND DIVISOR ASKING TO FIND OUT THE DIVIDEND OR REMAINDER
SO n=780I+137 I is the number you can take I=0,1,2,3,4,.....Put any number of I you will get the remainder 20
For example 780x0+137 divided by 39 remainder =20
780x1+137 divided by 39 remainder=20
equation will n=12a+5 by formula dividend= divisor x quotient + remainder
nothing is possible so eliminate this statement
from statement 2 again the equation is n = 65b+7
Now combining both equations n=12a+5=65b+7
n=12a=65b+7-5
n=12a=12b+53b+2 we will break 65b in terms of the 12b+53b so that we can easily
find out the value which gets divisible by 12.
n=12a=12b+(53b+2) For this 53b+2 we have to check what we can put the value in
this so if try for b=2 in 53b+2 put b=2 you will get 108 which is
divisible by 12. Now put the value of 2 in equation 2 which is
65b+7 it will come 137
Now take the LCM of 65 and 12 to make the equation
n=LCM(65,12)+137 NOTE: ALWAYS DO THIS FOR REMAINDER QUESTION IF 2 EQUATION IS GIVEN HAVING REMAINDER AND DIVISOR ASKING TO FIND OUT THE DIVIDEND OR REMAINDER
SO n=780I+137 I is the number you can take I=0,1,2,3,4,.....Put any number of I you will get the remainder 20
For example 780x0+137 divided by 39 remainder =20
780x1+137 divided by 39 remainder=20
