Aptitude - Problems on H.C.F and L.C.M - Discussion

Thanks all.

Thanks all for explaining.

Thanks @Saraswati.

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.

1305÷1120 leaves 185 as the remainder
4665÷1120 leaves 185 as the remainder
6905÷1120 leaves 185 as the remainder.

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

Thanks all for epxlaining.

Good, Thanks you @Greeshma.

I understood well, thanks to all for explaining.

When these numbers are divided by 5 then the remainder is zero right. Then the Answer should be 5 right. Please explain to me.

@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.

(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.