How many three-digit numbers can be formed when the leading digit cannot be 0 and the number must be a multiple of 3?

Answer 1

These numbers can form an arithmetic progression.

The 1st element t₁ = 102, the last element t_n = 999, common difference d = 3.

Let's find n - total number of elements.

Substitute given numbers into formula t_n = t₁ + d*(n-1).

999 = 102 + 3 * (n-1)

Answer: n= 300