Aptitude - Numbers - Discussion

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.

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

Let 3 digit number is 'a'
2272/a => quotient = b and remainder =c --> (1)
875/a => quotient = d and remainder = c -->(2)

Now we know that;
Divisior * quotient + remainder = dividend.

In equation (1) and (2).
ab + c = 2272.
ad + c = 875.
Subtract (1) - (2).
a(b-d) = 1397.
So a x (b-d) = 1397.

If we divide 1397 by prime number by 2,3,5,7,11,13,17...
Check which divides it and give remainder=0.
So you get 11 divides it properly with remainder =0.
1397/11= 127.
a(b-d)= 127 x 11.
a = 127 and b-d = 11.
And we need 3 digit number which is 127.
So, 1+2+7= 10.
Thank you.

@Jatin.

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

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.

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.

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

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).

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.

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.