Question

Find the smallest 3 digits natural number which when divides 1496 and 948 leaves the same remainder.

Answer 1

According to question the divisor for both the numbers will be the same.

Let us assume it to be a and the quotient be p and q and remainder be b

Therefore we can write:

ap + b = 1496 and aq + b = 948

Subtracting these two we get, ap - aq = 548.

or, a(p - q) = 548

Now the smallest three digit factor of 548 = 137. There are no other factor of 548 smaller than this with 3 digit.

Thus the smallest 3 digits natural number which when divides 1496 and 948 leaves the same remainder is 137.