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)

Problems on H.C.F and L.C.M

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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