Question

Find the smallest three digit number that divides 229 and 568 leaving remainder 3 in each case

Answer 1

Suppose the 3 digit number be n

Now, according to the question:

229=n*a+3 & 568=n*b+3

Subtracting both equations, we get:

568-229=n(b-a)

i.e. n(b-a)=339

It means n divides 339

But the divisors of 339 are 3, 113

Therefore, the smallest 3 digit number is 113.

Answer 2

Answer:

113 is smallest three digit number which divides 229 and 568 leaving remainder 3.

Explanation:

We need to find smallest three digit number that divides 229 and 568 leaves remainder 3 in each case.

Let us suppose three digit number be n

If we divide 229 by n leaves remainder 3.

If we subtract 3 from 229 then it would be exactly divisible by n.

New number is 229-3 = 226

226 is exactly division by n

If we divide 568 by n leaves remainder 3.

If we subtract 3 from 229 then it would be exactly divisible by n.

New number is 568-3 = 565

565 is exactly division by n

Two new number 226 and 565 is exactly divisible by n.

As we know if two numbers is exactly divisible by other number then other number would be HCF of both numbers.

Now we find the HCF of 226 and 565

Factors of 226 = 2 x 113

Factors of 565 = 5 x 113

HCF of (226,565) = 113

Thus, 113 is smallest three digit number which divides 229 and 568 leaving remainder 3.