olymmath

AIME 2026b Problem 4

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For each positive integer nn let f(n)f(n) be the value of the base-ten numeral nn viewed in base bb, where bb is the least integer greater than the greatest digit in nn. For example, if n=72n=72, then b=8b=8, and 7272 as a numeral in base 88 equals 7⋅8+2=587\cdot 8+2=58; therefore f(72)=58f(72)=58. Find the number of positive integers nn less than 10001000 such that f(n)=nf(n)=n.

Dataset text AIME 1983 to 2026. Quoted with source.

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