Question

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 $f(n) = n.$

Answer

Solution:

If $n$ has a single digit $d,$ then the numeral $d$ has value $d$ in every base, so $f(n) = n:$ all $9$ one-digit numbers work. If $n$ has digits $d_{k-1} \ldots d_1 d_0$ with $k \ge 2,$ then $b \le 10$ always, and if $b \lt 10$ then $f(n) = \sum d_i b^i \lt \sum d_i 10^i = n$ because the leading digit satisfies $d_{k-1} b^{k-1} \lt d_{k-1} 10^{k-1}.$ So a multi-digit $n$ satisfies $f(n) = n$ exactly when $b = 10,$ that is, when some digit of $n$ equals $9.$

Two-digit numbers containing a $9:$ the numbers $90$ through $99$ plus $19, 29, \ldots, 89,$ for $10 + 8 = 18.$ Three-digit numbers containing a $9:$ $900 - 8 \cdot 9 \cdot 9 = 252,$ subtracting the numbers with no $9$ (leading digit $1$ – $8,$ others $0$ – $8$ ).

The total is $9 + 18 + 252 = 279.$

2026 AIME II Problem 4

Solution:

Problem 4 in Other Years