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

We subtracted 1305 from 4665, 4665 from 6905 and 1305 from 6905 because we needed n as a common factor which can be obtained this way.

Let us consider 1305 as q1
4665 as q2
And 6905 as we
Then,

4665-1305 = N(q2 -q1).
6905-4665 = N(q3-q2 ).
6905-1305 = N(q3-q2 ).

Now we get here N as the factor of all 3 digits and we need to find the greatest factor so, take out the HCF of all three numbers i.e 1305, 4665, 6905 = 1120.

And this is your answer if your question asks no. Of digits in the number simply add the digits i.e 1+1+2+0=4.

Hope this helped.

Good explanation, Thanks @Yogesh.

Thanks @Saraswati.

@Anchit.

We should subtract remainder. i.e 12-2=10.

Thanks for explaining @Priya.

Thanks @Saraswati.

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.

Great explanation, Thanks @M. Harish.

Thanks @Saraswati.

Thanks for the answer @Saraswati.