Aptitude - Numbers - Discussion

I can't understand the steps. Can anyone please explain with an easy method?

Let;
a=pn+r.....(1)
b=qn+r......(2)

(1)-(2)

a-b==n(p-q)
or a-b=nd

So, a-b also exactly divisible by n.

@Jatin.

Is it applicable to all problems? How did you take 3?

Why do we subtract the two numbers, here? Please tell me.

For 1397 =11*127.

Prime factorization : You should divide the number (1397) with prime numbers till no remainder come.

i.e:prime nos 2,3,5,7,9,11,13,17...
Now 1397/2 --> Remainder.
.
1397/11 no reminder and quotient is 127.

So 11*127.
The answer is 127.
1+2+7=10.

Both leaving same remainder(let Remainder is R) hence,
2272-R and 875-R Both are completely divisible by number.

Their difference will also divisible by the number,
(2272-R)-(875-R) = 1397.

Factors of 1397 = 11*127.
Asked for 3 Digit number.
Hence sum of digits is 1+2+7 = 10.

2272-875 = 1397.
1397/11 = 127.
1+2+7 = 10.

Thank you @Queeen.

Any other easy way?

We know that, dividend = (divisor * quotient) + remainder.

From the problem statement we have,
2272 = (N*q1) + R (here we don't know the quotients, let them be any values q1 and q2.
875 = (N*q2) + R N is a 3-digit divisor and in both cases we get the same remainder R).

Now solving these two equations we get,
1397 = N * (q1-q2) (here we factor 1397 into 11 and 127..as we are looking for a 3-digit divisor).
127 * 11 = N * (q1-q2).

Hence (N=127) and (the sum of the digits of N = 10).