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07 Jan 2017, 05:31
Kudos
Bookmarks
Hi All,
I have a very specific doubt, please help me understand this
Self created DS question and statement :
Is p divisible by 16?
a) when p is divided by 3, remainder is 1
b) when p is divided by 7, remainder is 2
I go on to deduce the following relation:
From a -
p = 3m+1 => p = 1,4,7 ...in increments of 3. ofcourse not sufficient.
From b -
p = 7n+2 => p = 2,9,16 ...in increments of 7. ofcourse not sufficient.
From a+b -
Next, the first common number in both the set is the required number.
because earlier i didnt write the entire list, for C i have to list all the elements. So ->
From a, we have list A = 1,4,7,10,13,16,19,22,25,28,31,34,37,40,43,46,49,52,55,58,61,...
From b, we have list B = 2,9,16,23,30,37,44,51,58,65,72,...
Okay, so now we have list C = common numbers from both = 16,37,58,...
As per the question, if I had stopped at 16 in list C, my answer would have been C. but because I checked for next numbers as well, I can definitely say answer is E.
Doubt :
Trend I notice here is:
37 - 16 = 21
58 - 16 = 21
I dont need to list down all the elements in list A and list B, as soon as I get first common number I can stop and add 3*7 (quotient from statement a and b) to get the next common number in list C. This will definitely help me save time.
Can you please confirm - Is difference same i.e. quotient 1 * quotient 2, each and every time?
If so, any logic underlying this?
I have a very specific doubt, please help me understand this
Self created DS question and statement :
Is p divisible by 16?
a) when p is divided by 3, remainder is 1
b) when p is divided by 7, remainder is 2
I go on to deduce the following relation:
From a -
p = 3m+1 => p = 1,4,7 ...in increments of 3. ofcourse not sufficient.
From b -
p = 7n+2 => p = 2,9,16 ...in increments of 7. ofcourse not sufficient.
From a+b -
Next, the first common number in both the set is the required number.
because earlier i didnt write the entire list, for C i have to list all the elements. So ->
From a, we have list A = 1,4,7,10,13,16,19,22,25,28,31,34,37,40,43,46,49,52,55,58,61,...
From b, we have list B = 2,9,16,23,30,37,44,51,58,65,72,...
Okay, so now we have list C = common numbers from both = 16,37,58,...
As per the question, if I had stopped at 16 in list C, my answer would have been C. but because I checked for next numbers as well, I can definitely say answer is E.
Doubt :
Trend I notice here is:
37 - 16 = 21
58 - 16 = 21
I dont need to list down all the elements in list A and list B, as soon as I get first common number I can stop and add 3*7 (quotient from statement a and b) to get the next common number in list C. This will definitely help me save time.
Can you please confirm - Is difference same i.e. quotient 1 * quotient 2, each and every time?
If so, any logic underlying this?
Archived Topic
Hi there,
This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
Where to now? Join ongoing discussions on thousands of quality questions in our Quantitative Questions Forum
Still interested in this question? Check out the "Best Topics" block below for a better discussion on this exact question, as well as several more related questions.
Thank you for understanding, and happy exploring!
07 Jan 2017, 23:36
Kudos
Bookmarks
namratakt
Hi
After you have found the first number, add LCM of the two divisors.. here it is 3 and 7..
So 16 +21 and next 16+21+21 and so on..
Reason is 16 leaves a certain remainder, and if we add a number to 16 which is divisible by both 3 and 7, the remainder will still be the one we had initially..
Here let the new number be 16 + 21n where n is a positive integer.
So when you divide by 3..( 16+21n)/3= 16/3 + 7n so the remainder will be 1..
Similarly when you divide by 7 ( 16+21n)/7= 16/7+ 3n so the remainder will be 2..
Similarly
11 Jan 2017, 03:38
Kudos
Bookmarks
namratakt
Check HERE to know how to derive general formula for such cases.
Hope it helps.
Archived Topic
Hi there,
This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
Where to now? Join ongoing discussions on thousands of quality questions in our Quantitative Questions Forum
Still interested in this question? Check out the "Best Topics" block above for a better discussion on this exact question, as well as several more related questions.
Thank you for understanding, and happy exploring!
