# 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:
148 comments Page 1 of 15.

Surya said:   1 month ago
Why should we subtract the Numbers with the smallest number and so HCF for that?
(2)

Suraj Chaudhary said:   4 months ago
I can't understand this question properly.

So, please If anybody knows properly it. Please explain it.
(4)

Gourav said:   5 months ago
@Anomi

Here, we need to find the greatest number. So 5 is not the greatest number.
(2)

Agondeze milly said:   7 months ago
How do you get 4? I am not getting it clearly.
(9)

Krishnakanth said:   10 months ago
Given numbers 1305, 4665, 6905.

Now take difference;
4665 - 1305 = 3360,
6905 - 4665 = 2240,
6905 - 1305 = 5600.

3360 factor it will give you 4,4,10,7 all divisible by the other 2 numbers.
So, multiple them and you will get 1120 add them because in question they said digits give you 4 i.e answer.
(12)

Sumit said:   2 years ago
If we only take two differences i.e. 6905 - 4665 = 2240, 4665 - 1305 = 2360.
HCF of 2240 and 2360 is 40.
4+0 = 4.
(15)

Ashu said:   2 years ago
Thanks for explaining @Saraswati.
(3)

Komal said:   2 years ago
Thank you for explaining the answer @Greeshma.
(1)

Barath dasari said:   2 years ago
(1305 - 4665) = 3360.
(4665 - 6905 ) = 2240.
( 6905 - 1305) = 5600.

Then HCF Of 3360 - 3 * 1120,
HCF Of 2240 - 2 * 1120,
HCF Of 5600 - 5 * 1120,
Sum Of Digits in N = 1 + 1 + 2 + 0 = 4.

I hope it's easy to understand.
(28)

Aswini said:   2 years ago
@All.

So most of you people asked how (4665- 1305) (4665-6905) ( 6905- 1305).

So, By using (b-a) (b-c) (c-a) formula.
Given. a= 1305 , b=4665 c= 6905.
(5)

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