Aptitude - Problems on H.C.F and L.C.M - Discussion
Discussion Forum : Problems on H.C.F and L.C.M - General Questions (Q.No. 1)
1.
Find the greatest number that will divide 43, 91 and 183 so as to leave the same remainder in each case.
Answer: Option
Explanation:
Required number = H.C.F. of (91 - 43), (183 - 91) and (183 - 43)
= H.C.F. of 48, 92 and 140 = 4.
Discussion:
210 comments Page 4 of 21.
Subhronil Biswas said:
3 years ago
@All.
According to me,
the solution is;
Factors of 43 are: 1, 43.
Factors of 91 are: 1, 7, 13, 91.
Factors of 180 are: 1, 2, 3, 4, 5, 6, 9, 10, 12, 15, 18, 20, 30, 36, 45, 60, 90, 180.
So, I go with my answer i.e., 1. None of the given options is correct.
According to me,
the solution is;
Factors of 43 are: 1, 43.
Factors of 91 are: 1, 7, 13, 91.
Factors of 180 are: 1, 2, 3, 4, 5, 6, 9, 10, 12, 15, 18, 20, 30, 36, 45, 60, 90, 180.
So, I go with my answer i.e., 1. None of the given options is correct.
(32)
Madhukumar E said:
1 decade ago
2(48
2(24
2(12
3(6
(2
So now 48 = 2*2*2*3*2.
Use the same method for 92 and 140.
Then they give = 2*2*23 and,
= 2*2*5*7.
Now, take the common factors. Here we go,
2*2*2*3*2.
2*2*23.
2*2*5*7.
2*2 are common factors.
So, H.C.F. of 48, 92 and 140 = 4.
2(24
2(12
3(6
(2
So now 48 = 2*2*2*3*2.
Use the same method for 92 and 140.
Then they give = 2*2*23 and,
= 2*2*5*7.
Now, take the common factors. Here we go,
2*2*2*3*2.
2*2*23.
2*2*5*7.
2*2 are common factors.
So, H.C.F. of 48, 92 and 140 = 4.
Priti Maurya said:
1 year ago
In the question, this line is mentioned that "find the greatest number which leaves the same remainder in each case".
When u divide all given numbers by option A;
Then you will get the same remainder "3".
So, that the correct answer is "4".
When u divide all given numbers by option A;
Then you will get the same remainder "3".
So, that the correct answer is "4".
(28)
VIGNESH KUMAR said:
1 decade ago
Simple way to calculate HCF:
2|48 92 140
-----------
2|24 46 70
-----------
|12 23 35
------------
Stop if you can't divide by common divisor, so answer = 2*2 = 4.
2|48 92 140
-----------
2|24 46 70
-----------
|12 23 35
------------
Stop if you can't divide by common divisor, so answer = 2*2 = 4.
Parthiban said:
1 decade ago
It is very simple
consider
43,91,183 are the 3 numbers
first ,
find the diff between them
91-43=48;
183-91=92;
183-43=140;
let divide the 48,92,140,by 2
we have
48=2*2*2*2*3
92=2*2*23
140=2*2*149
here common factor is 2*2
so
answer is 4
consider
43,91,183 are the 3 numbers
first ,
find the diff between them
91-43=48;
183-91=92;
183-43=140;
let divide the 48,92,140,by 2
we have
48=2*2*2*2*3
92=2*2*23
140=2*2*149
here common factor is 2*2
so
answer is 4
Mani said:
8 years ago
Why can't we think like this, In given four option one is HCF of 43, 91, 183. And leave the same reminder as per the question.
Let me consider 4 is HCF then divide the numbers by four. It shows the same reminder hence 4 is the answer.
Let me consider 4 is HCF then divide the numbers by four. It shows the same reminder hence 4 is the answer.
Nam said:
1 decade ago
Acc to the formulae H.C.F. of [(H.C.F. of any two) and (the third number)] gives the H.C.F. of three given number.
So as per the first formulae the hcf of 183 & 43 is comming 10
n the third no. is 91 so how will 4 be the ans
So as per the first formulae the hcf of 183 & 43 is comming 10
n the third no. is 91 so how will 4 be the ans
Habib said:
1 decade ago
The difference is taken to analyze the pattern. Amongst the numbers !
Ie each difference tells us that these numbers are multiples of 4 and have the same remainder because all numbers can be divided by 4.
Hence you get 4.
Ie each difference tells us that these numbers are multiples of 4 and have the same remainder because all numbers can be divided by 4.
Hence you get 4.
Daniel said:
1 decade ago
13 and 7 don't leave remainder in 91 while they leave in other and 9 clearly leave different remainder in 43 and 91. So it is 4 you can think logically or use hit and trial method or simply use formulas given by others.
Ahmad Zaidi said:
7 years ago
@Samir.
First find the LCM of 60, 80, 90, which is 720. Then divide it by 99999 which is highest 5 digit number, and less the reminder which is 639 and answer is 99360. Which is equally divisible by 60, 80, 90. Thanks.
First find the LCM of 60, 80, 90, which is 720. Then divide it by 99999 which is highest 5 digit number, and less the reminder which is 639 and answer is 99360. Which is equally divisible by 60, 80, 90. Thanks.
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