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. 26)
26.
The greatest number which on dividing 1657 and 2037 leaves remainders 6 and 5 respectively, is:
Answer: Option
Explanation:
Required number = H.C.F. of (1657 - 6) and (2037 - 5)
= H.C.F. of 1651 and 2032 = 127.
Discussion:
38 comments Page 2 of 4.
Mini said:
1 decade ago
@Ramesh
It is better explained in the following link...check it out
http://www.ilovemaths.com/1lcmandhcf.asp
It is better explained in the following link...check it out
http://www.ilovemaths.com/1lcmandhcf.asp
Nikita said:
4 years ago
HCF = 2037-1657 = 381 and there factors answer is127.
Because we have to find greatest common factor.
Because we have to find greatest common factor.
(6)
Anu said:
1 decade ago
No .it is not possible .because 1657-6 gives the remainder 0,but 1657+6 gives the remainder 1.
Ashwarya said:
1 decade ago
Ho we know that 1657 - with 6 and how we know that 2037 is minus with 5 please explain.
Harpreet kaur said:
3 years ago
1651 = 13 * 127.
2032 = 2 * 2 * 2 * 2 * 127.
So common is 127.
That's why H.C.f 127.
2032 = 2 * 2 * 2 * 2 * 127.
So common is 127.
That's why H.C.f 127.
(16)
Prabhakaran said:
1 decade ago
@Thayagarajan. How we know 127 is the correct answer?, that's we need tell clearly.
Vivek said:
6 years ago
How can we find the HCF of Three co-prime numbers? please anyone explain.
Rajan gupta said:
9 years ago
Please explain the method to calculate the HCF with an example.
Kirti said:
1 decade ago
Can we do 1657+6 or 1657-6 both will give same answer. ?
Thyagarajan said:
1 decade ago
Answer:
13*127 = 1651+6 = 1657.
16*127 = 2032+5 = 2037.
13*127 = 1651+6 = 1657.
16*127 = 2032+5 = 2037.
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