Question

Three numbers 5920, 7976 and 10803 when divided by a three digit number n' leaves the same remainder in each case

Answer 1

Answer:

257

Step-by-step explanation:

Three numbers 5920, 7976 and 10803 when divided by a three digit number n' leaves the same remainder in each case

Let say number = X  and remainder = R

Xa + R = 5920

Xb + R = 7976

Xc + R = 10803

X(b-a)  = 2056

X(c-a) = 4883

X(c-b) = 2827

2056 = 257 * 8

4883 = 257 * 19

2827 = 257 * 11

HCF of 2056 , 4883 & 2827 = 257

X = 257

257 * 23 + 9 =  5920

257*31 + 9 = 7976

257*42 + 9 = 10803