Aptitude - Numbers - Discussion

Here we need to find the sum of the digits of the required 3-digit number.
As the given two numbers when divided by 3-digit numbers give the same remainder then that means when we find the difference between them then that difference will be divisible by that three-digit number.

So, we will first find the difference between the given numbers, i.e. 2272 and 875.

Difference=2272-875=1397 --->(1)

1397 will be divisible by a three-digit number i.e. N.
.
We know that the factors of the number 1397 are 11 and 127.
The only three-digit number which has a factor of 1397 is 127.
So, we can say that the value of the three-digit number i.e. N.
It's equal to 127.
Now, we will find the sum of the digits of the N.
The sum of the digits of N.
=1+2+7=10.
Hence, the correct option is option A.

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.

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).

hello buddys!

I think the following could be one of the methods:

we know, dividend=divisor*quosent+remainder
divisor=d,quosen=1(if not given assume it to be 1)+remainder=r

so, 2272=d*1+r
875=d*1+r
since remainder are equal from the question,
r=2272-d -equ1
r=875-d -equ2
equ=equ2

2272-d=875-d
we get 1397 this our N.and sum of the digits 20.
becoz we r dealing with two equ same remainder, so divide it by 2.
i.e 20/2= 10.

I dont know whether is it wright but any ways am getting the answer.

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 us suppose that, 2272=x+r and 875=y+r then x-y=2272-875=1397 this is the difference between the two numbers. This 1397 should be completely divisible by divisor (as 1397 is free from remainder and is the actual difference between the (2272-r) & (875-r) hence should be completely divisible by the divisor. Now 1397=11*127. As divisor should be a 3 digit number, so 127 can be considered as the remainder which fulfills our criteria. Therefore sum of digits=1+2+7=10.

2272 = N*Q1 + R -->eq1.
875 = N*Q2 + R -->eq2.
2272 - 875 = N(Q1-Q2) -->eq1 - eq2.
1397 = N(Q1-Q2) -->eq3.

Since there is no remainder in eq3, so 1397 is completely divisible by N.
Therefore, N should be a factor of 1397.
Factors of 1397 = 11 and 127.
Since N should be a 3 - digit number. Therefore N = 127.
Hence answer = sum of digits of N = 1 + 2 + 7 = 10.

As 2272 and 875 is divided by same number(x), obviously they are the factors of (x).

So when we subtract 2272-875 = 1397.

Now, this 1397 is also factor of (x).

Now just simply divide 1397 by options given in number.

i.e. (10, 11, 12, 13) then we find that 11 is the only number divisible by 1397.

Hence 1397/11 = 127 i.e. 11*127 = 1397.

1+2+7 = 10.

Lets take the remainder as x,

2272-x is the quotient which is divisible by n.

875-x is the quotient which is divisible by n.

[2272-x]-[875-x] = a number which is divisible by n.

= 1397.

The factors of 1397 are 11 and 127.

127 is n as it is a three digit number and fills all conditions.

1+2+7 = 10.

10 is the answer.

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.