Aptitude - Numbers - Discussion

@Jatin.

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

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.

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.

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.

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.

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