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 12 of 21.
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.
Sreyas Biju said:
1 decade ago
The smallest number divisible by 12,32,42 and 63 is its LCM which is 2016.
The prime factorization of 2016 is 2x2x2x2x2x3x3x7.
Pick out the pairs in this step. They are 2x2 2x2 3x3.
2 and 7 are not in pairs. But if there is one more 2 and one more 7, both 2 and 7 can also be in pairs.
So the required numbers are one more 2 and one more 7. So it is 2x7= 14. Therefore, 2016x14 = 28224.
So the required square number is 28224.
Thank you!
The prime factorization of 2016 is 2x2x2x2x2x3x3x7.
Pick out the pairs in this step. They are 2x2 2x2 3x3.
2 and 7 are not in pairs. But if there is one more 2 and one more 7, both 2 and 7 can also be in pairs.
So the required numbers are one more 2 and one more 7. So it is 2x7= 14. Therefore, 2016x14 = 28224.
So the required square number is 28224.
Thank you!
Rony said:
1 decade ago
HCF of two numbers is 12, their sum is 72, then the two numbers are?
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