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

(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.

If we only take two differences i.e. 6905 - 4665 = 2240, 4665 - 1305 = 2360.
HCF of 2240 and 2360 is 40.
4+0 = 4.

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.

How do you get 4? I am not getting it clearly.

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

@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.

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

I can't understand this question properly.

So, please If anybody knows properly it. Please explain it.

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.

Thanks for explaining @Saraswati.