aime26-18

Proposed solution

I need to find the number of positive integers $n < 1000$ such that $f(n) = n$, where $f(n)$ is the value when the base-ten numeral $n$ is interpreted in base $b$, with $b$ being the smallest integer greater than the greatest digit in $n$.

Let me first understand what $f(n)$ means:
- For a number $n$, find its digits
- Let $d_{max}$ be the largest digit in $n$
- Set $b = d_{max} + 1$
- Interpret $n$ as a base-$b$ numeral and convert to base-10

Let me work through the example:
For $n = 72$: digits are 7, 2. So $d_{max} = 7$, thus $b = 8$.
Interpret 72 in base 8: $7 \cdot 8^1 + 2 \cdot 8^0 = 56 + 2 = 58$.
So $f(72) = 58 \neq 72$.

I want to find when $f(n) = n$.

Let me consider different cases based on the number of digits in $n$.

**Case 1: Single digit numbers ($1 \leq n \leq 9$)**
For single digit $n$, we have $d_{max} = n$, so $b = n + 1$.
But when we interpret $n$ as a base-$(n+1)$ numeral, it's just $n$ (since $n < n+1$).
So $f(n) = n$ for all single digit numbers.

That gives us 9 values: $n = 1, 2, 3, 4, 5, 6, 7, 8, 9$.

**Case 2: Two digit numbers ($10 \leq n \leq 99$)**
Let $n = 10a + b$ where $a, b$ are digits and $a \geq 1$.
Then $d_{max} = \max(a,b)$ and $b = d_{max} + 1$.

We want $f(n) = n$, which means:
$a \cdot b + b = 10a + b$
$a \cdot b = 10a$
$a(b-10) = 0$

Since $a \geq 1$, we need $b = 10$.

But if $b = 10$, then $b$ is not a single digit, which contradicts our setup. 

Wait, let me reconsider the problem more carefully.

Actually, when we interpret $n$ in base $b$, we're interpreting the digits of $n$ as coefficients in base $b$. So if $n = ab$ (where $a,b$ are digits), then:
$f(n) = a \cdot b + b$

But we also have that $n = 10a + b$.

So we want: $a \cdot b + b = 10a + b$
This gives us: $ab = 10a$, so $b = 10$ (since $a \neq 0$).

But if $b = 10$, then $d_{max} = \max(a,10) = 10$, so $b = 11$.
But we also need the interpretation to be valid with base $b = 11$.

Actually, let me restart with a clearer approach.

For a two-digit number $n = 10a + b$ where $a,b \in \{0,1,\ldots,9\}$ and $a \geq 1$:
- The greatest digit is $\max(a,b)$
- So $b = \max(a,b) + 1$
- We want $f(n) = n$, which means: $a \cdot b + b = 10a + b$

This gives us: $ab + b = 10a + b$, so $ab = 10a$, so $b = 10$.

But if $b = 10$, then $\max(a,b) = 10$, so $b = 11$. 

However, we can't have a digit of 10 in base-11 (digits must be less than the base). This is a contradiction.

So there are no two-digit solutions where $f(n) = n$.

Wait, I'm making an error. Let me re-read the problem.

If $n = 72$, then we interpret it as base-8: $7 \cdot 8 + 2 = 58$.
But if we have a number like $n = 10$, and the greatest digit is 1, then $b = 2$. 
Interpreting 10 in base-2: $1 \cdot 2 + 0 = 2$. So $f(10) = 2 \neq 10$.

Let me reconsider what happens with two-digit numbers more systematically.

For a two-digit number $n = 10a + b$ where $a,b$ are digits and $a \geq 1$:
- $d_{max} = \max(a,b)$
- $b = d_{max} + 1$
- $f(n) = a \cdot b + b$

We want: $a \cdot b + b = 10a + b$
This gives us: $ab + b = 10a + b$, so $ab = 10a$, so $b = 10$.

But if $b = 10$, then we need $d_{max} = \max(a,10) = 10$, so $b = 11$.
This means we're interpreting the digits in base-11, but we can't have a digit of 10 in base-11.

Actually, I think there's an issue with my understanding. Let me be more careful about what "interpretation" means.

If $n = ab$ (where $a,b$ are digits), and we interpret it in base $b$, then:
$f(n) = a \cdot b + b = b(a+1)$

But wait, that's not right either. If we have number $n = 10a + b$ and we want to interpret the digits $a,b$ as a base-$b$ numeral, then:
$f(n) = a \cdot b + b$

But we also need $b < b$, which is impossible.

Let me re-read carefully. The problem says "let $b$ be the least integer greater than the greatest digit in $n$."

So if $n = 72$, digits are 7,2, so $d_{max} = 7$, so $b = 8$.
Then we interpret 72 (the numeral) in base-8: $7 \cdot 8 + 2 = 58$.

So for a two-digit number $n = 10a + b$ with digits $a,b$:
$d_{max} = \max(a,b)$
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