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:
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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