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:
123
127
235
305
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 4 of 4.

Ashwarya said:   1 decade ago
Ho we know that 1657 - with 6 and how we know that 2037 is minus with 5 please explain.

M.V.KRISHNA/PALVONCHA said:   1 decade ago
Here 6 & 5 are remainders of 1657(say A) & 2037(say B) respectively.

To make A and B(dividends) perfectly divisible by a divisor, the remainders should be subtracted from dividends.

Now if we divide the dividend by a divisor, the remainder will be zero, because we made dividends perfectly divisible.

Now the numbers are 1657-6=1651 and 2037-5=2032.

HCF of 1651,2032.

1651)2032(1
1651
-------
381)1651(4
1524
------
127)381(3
381
------
0
-------
So HCF is 127.

Hope you understand.

Venkat said:   1 decade ago
THANKS krishna your explanation is too easy.

Kirti said:   1 decade ago
Can we do 1657+6 or 1657-6 both will give same answer. ?

Anu said:   1 decade ago
It is not possible .
the term 1657+6 is not a perfet multiplier it gives a remainder 1.the term 1657-6 is a perfect multiplier it gives a remainder 0.that's why they doesnot give same answer.

Anu said:   1 decade ago
No .it is not possible .because 1657-6 gives the remainder 0,but 1657+6 gives the remainder 1.

Loganathan said:   1 decade ago
Thank you mr. M. V. Krishna/Palvoncha.

Thyagarajan said:   1 decade ago
Answer:

13*127 = 1651+6 = 1657.
16*127 = 2032+5 = 2037.


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