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07 Jan 2017, 05:31
Kudos
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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?
07 Jan 2017, 23:36
Kudos
Bookmarks
namratakt
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?
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
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?
Check HERE to know how to derive general formula for such cases.
Hope it helps.
