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:
150 comments Page 15 of 15.

Akshitha said:   6 years ago
Thnks @Sanchay.
(1)

Kumar said:   1 decade ago
Thanks yogesh.

Supriya said:   7 years ago
Thanks @Priya.

Babu said:   4 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:   3 years ago
Thanks all.

Neha said:   8 years ago
How 1120?


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