# 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
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.

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)

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)

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)

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

Greeshma said:   3 years ago
Given numbers 1305, 4665, 6905.

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

Take the smallest difference i.e 2240 find factors of it;
1-> 2240
2-> 1120
4-> 560
5-> 448
and so on;

Check from highest factors if any one of them leaves the same remainder for all given numbers
So, 1st one 2240 it can't divide 1305. So, go for the second one 1120.

1305Ã·1120 leaves 185 as the remainder
4665Ã·1120 leaves 185 as the remainder
6905Ã·1120 leaves 185 as the remainder.

As all given conditions fulfilled :)
We got it;
1120.
1+1+2+0= 4.
Edits are welcome
(6)

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)

Anomi said:   2 years ago
When these numbers are divided by 5 then the remainder is zero right. Then the Answer should be 5 right. Please explain to me.
(4)

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

(4)

Anita Chettri said:   3 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.
(3)

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