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.

Where did 11 and 127 comes from?

Please explain.

a=np+r
2272 = np + r ----------> (1)
b=nq+r
875=nq+r --------------> (2).

Subtracting (1)-(2).
2272-875 = n(p-q),
1397 = n(p-q),
1397 is only divisible by 11,
1397 = 11*127.

Hence n=127,
p-q = 11,
1+2+7=10.

@All.

Here is the solution.

2272-R=0(mod N).
875-R=0(mod N).
2272-875=0(mod N).
1397=0(mod N).

So, any factor of 1397 can be the answer for N which 127 will be correct according to the question given (3digit numbers).

1+2+7=10.

2272 - 875 = 1397 which is divisible by 11.
i.e. , 1397 = 11 X 127.
so, 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.

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.

2272-875 = 1397 difference,
1397/11 = 127 divisible by 11,
1+2+7 = 10.

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.

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.