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:

157 comments Page 16 of 16.

Babu said: 5 years ago

Thank you all.

Jinx said: 1 decade ago

Thanks sarsu.

Ramya said: 1 decade ago

Thankx priya.

Nagesh said: 1 decade ago

Yogesh thanks

Neha said: 1 decade ago

Thanx priya.

Minkel Aryan said: 4 years ago

Thanks all.

Neha said: 10 years ago

How 1120?

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