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:
212 comments Page 12 of 22.
Sharmi said:
1 decade ago
Why you have to divide all the number with 4. What is the reason, please explain?
Afsa said:
1 decade ago
Some one help! What is the logic behind H. C. F?
Mani pandian said:
1 decade ago
Here are the list of prime factors of 24 and 36:
24 = 2 x 2 x 2 x 3.
36 = 2 x 2 x 3 x 3.
If we write down the numbers that are the same in both lists, they will give us the highest common factor of 24 and 36:
HCF of 24 and 36 is 2 x 2 x 3 = 12.
This is the meaning of H.C.F.
24 = 2 x 2 x 2 x 3.
36 = 2 x 2 x 3 x 3.
If we write down the numbers that are the same in both lists, they will give us the highest common factor of 24 and 36:
HCF of 24 and 36 is 2 x 2 x 3 = 12.
This is the meaning of H.C.F.
Rajvir singh said:
1 decade ago
Why do need to take difference three times? (91-43), (183-91) was enough. and then take h.c.f. of 48 and 92 please someone give another example why do we need to do (183-43)?
Tahir said:
1 decade ago
Just simple according to the question there is no compilation is 43/4 the reminder is 3 then 91/4 the remainder is 3 and now 183/4 the reminder is once again three. F look at question the reminder is same ie 3.
Sarath said:
1 decade ago
I think the smarter way will be to divide the numbers with options and check. But your speed has to be so high.
Kumari said:
1 decade ago
Dear friends, Why they took difference between them and is any other easy as well as short method ? please communicate this answer as soon as possible. Thanking you.
Prem said:
1 decade ago
I am unable to understand this problem. Can anyone explain all steps involved?
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.
Aditya said:
1 decade ago
Find the least number divisible by 12, 32, 42, and 63 and it must be a perfect square?
Please give me the solution.
Please give me the solution.
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