Question

2. A two digit number has a remainder of 1 when it is divided by 3. It also has a remainder of 1 when it is divided by 5. What is the least two digit number satisfying the above conditions?​

Answer 1

Generally, let the number be N.

Since N leaves remainder 1 on division by 3, let N = 3a + 1.

Since N leaves remainder 1 on division by 5, let N = 5b + 1.

So,

N = 3a + 1 = 5b + 1

On subtracting 1 from each,

N - 1 = 3a = 5b

So we get that N - 1 is a multiple of both 3 and 5. To get least value for N which satisfies the condition, we have to take N - 1 as the LCM of 3 and 5.

Therefore, N - 1 = LCM(3, 5) = 15

Well, N - 1 = 15 is a two digit number.

So,

N = 16