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. 4)
4.
Let N be the greatest number that will divide 1305, 4665 and 6905, leaving the same remainder in each case. Then sum of the digits in N is:
4
5
6
8
Answer: Option
Explanation:

N = H.C.F. of (4665 - 1305), (6905 - 4665) and (6905 - 1305)

  = H.C.F. of 3360, 2240 and 5600 = 1120.

Sum of digits in N = ( 1 + 1 + 2 + 0 ) = 4

Discussion:
158 comments Page 13 of 16.

Provith said:   7 years ago
Thanks all for explaining the answer.
(1)

Anom said:   5 years ago
Great explanation, Thanks @M. Harish.

Saniya taj said:   7 years ago
Easy to understand. Thanks @Suraj.
(1)

Nidhitiwari said:   11 months ago
Very good and helpful. Thanks all.
(3)

Kotresh said:   6 years ago
Good explanation, Thanks @Yogesh.

Prince Khan said:   5 years ago
Thanks for the answer @Saraswati.

Ashu said:   3 years ago
Thanks for explaining @Saraswati.
(4)

Mounika said:   9 years ago
Thank you all for explaining it.

Hanumant said:   8 years ago
Thank you all for explaining it.

GIRISH SINGH BISHT said:   6 years ago
Thanks for explaining @Priya.


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