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 15 of 16.

K.manoj said: 6 years ago

Thanks @Saraswati.

Jaya said: 5 years ago

Thanks @Saraswati.

Subhasis said: 4 years ago

Thanks @Saraswati.

Vishal said: 1 decade ago

Thanks saraswati.

Diya said: 8 years ago

Thanks @Prakash.

Kashi said: 7 years ago

Thanks @Prakash.

Aman Srivastava said: 7 years ago

Thanks @Yogesh.

Akshitha said: 7 years ago

Thnks @Sanchay.

(1)

Kumar said: 1 decade ago

Thanks yogesh.

Supriya said: 8 years ago

Thanks @Priya.

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