2026 AIME II Problem 4

For each positive integer n let f(n) be the value of the base-ten numeral n viewed in base b, where b is the least integer greater than the greatest digit in n. For example, if n=72, then b=8, and 72 as a numeral in base 8 equals 7\cdot 8+2=58; therefore f(72)=58. Find the number of positive integers n less than 1000 such that f(n)=n.

Leading zeroes must be inputted, so if your answer is 34, then input 034

Full credit goes to MAA for authoring these problems. These problems were taken on the AOPS website.