How to understand the condition of the task?

Question

Java

Count the number of numbers from A to B (integer) such that the digits of the number, considered in
order from left to right, form an arithmetic or geometric progression. We
will assume that all one-digit and two-digit numbers form an arithmetic progression. To
determine whether a number fits the requirements of the task, implement the function. Also
implement a function to count the number of such numbers from 1 to n. It is forbidden to use strings
and arrays.

Answer 1

It seems to me that any number consisting of one digit, by condition, is a progression, i.e. it suits us. Same for doubles. Those. all one-digit and two-digit numbers that fall into the range [A, B] are suitable for us.
Consider 3-digit numbers and numbers of higher orders. In the case of a 3-digit number, for example 123 - an arithmetic progression with a difference of 1 (1, 2, 3); 124 - geometric progression with divisor 2 (1, 2, 4); 321 - arithmetic progression with difference -1 (3, 2, 1); 421 is a geometric progression with a divisor 1/2 (4, 2, 1). Those. you need to find all such numbers. Naturally, the difference can be any depending on the digits (for 3 digits from -4 to 4) and the divisor too (for 3 digits from -3 to 3). Well, this is already thinking about the decision.