Aptitude - Numbers - Discussion

Consider that if two nos if divided by a same divisor gets same reminders. Then the difference between the nos will be exactly divisible by the divisor.

For example:

5/2 remainder is 1.

9/2 remainder is 1.

9-5 = 4, which is exactly divisible by 2.

I did not understand.

Here is an shortcut:

= 2+2+7+2-8+7+5 = 7.

= 7+3 = 10.

Thanks @Sravan.

If a Two numbers are divisible by same number & gives the same remainder that means they belong to same table.

And their difference will also be divisible by same number.

e.g.
Let 4 & 10 be the numbers.
let N be 2.
So (10 - 4) = 6 is also divisible by N = 2.

Once we get to 1397 & we know it is completely divided by N then we can use the options provided to find out the value of N.

Because 1397 must be completely divisible by one of them.

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).

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

1)If we divide two different number A and B which when divided by Q gives the same remainder then we can say that A-B is completely divisible by Q.

2) In the above question, we get 1397 after subtracting 875 from 2272.Therefore 1397 should be divided by N.

3) Now when I see the number 1397. I can see that the number is divisible by 11 because (sum of digits at odd place) - (Sum of digits at even) is 0. i.e (7+3) - (9-1) = 0. Therefore it is divisible by 11. Hence we get 127.

4) 1+2+7=10.

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.