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:
157 comments Page 4 of 16.

Anita Chettri said:   5 years ago
N = HCF of (4665-1305),(6905-4665) and (6905-1305).
= HCF of 3360,2240 and 5600.
= 2*2*2*2*2*5*7
= 1120.
Sum of digits in N = (1+1+2+0) = 4.
(4)

Anita Chettri said:   5 years ago
N = HCF of (4665-1305),(6905-4665)and (6905-1305).
= HCF of 3360,2240 and 5600,
= 2 * 2 * 2 * 2 * 2 * 5 * 7.
= 1120
Sum of digits in N = (1+1+2+0) = 4.
(1)

Haq ahmed said:   5 years ago
How we get the 1120 HCF of those number, can someone explain? please.

Vihaan said:   5 years ago
I can't understand why are we taking the numbers (4665 - 1305), (6905 - 4665) and (6905 - 1305)?

Can anyone please explain to me this in brief?
(1)

Jack said:   5 years ago
I can't understand why we are taking H.C.F. Of (4665 - 1305), (6905 - 4665) and (6905 - 1305).

Anyone please explain in detail.
(1)

Prince Khan said:   5 years ago
Thanks for the answer @Saraswati.

Jaya said:   5 years ago
Thanks @Saraswati.

Anom said:   5 years ago
Great explanation, Thanks @M. Harish.

Vishnu Sai said:   5 years ago
I have a question.

Can anyone please explain why they are taking the HCF of (4665-1305), (6905-4665) and (6905-1305) i.e. the HCF of 3360, 2240 and 5600.

Also in the question it states the remainder should be same, it does not state the remainder should only be zero.

So I wish some of the other clarifies my doubts.

K.manoj said:   6 years ago
Thanks @Saraswati.


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