AIME 2026 II · 第 4 题

Solution 1

Notice that if we consider each number in base $b<10$, then the value strictly decreases. For instance, $f(257)=2\cdot 8^2+5\cdot 8^1+7\cdot 8^0$, strictly less than $257=2\cdot 10^2+5\cdot 10^1+7\cdot 10^0$. Therefore, $b=10$, so the number must contain a $9$. The only other cases are when the number has only one digit in which case the base doesn't really matter at all.

We have two cases to consider:

Case 1. Number containing $9$. We use complementary counting on three place values, where each place value can contain a digit $0-9$, where leading zeros are permitted, and $000$ is understood to be $1000$ (this produces a bijection between the positive integers from $1$ to $1000$ inclusive and the possibilities from these three place values). There are $9\cdot9\cdot9=729$ ways for none of the three place values to contain a $9$, so there are $1000-729=271$ total possibilities in this case.

Case 2. Other one-digit numbers: $8$ possibilities (namely $1,2,\ldots,8$).

The total is $8+271=\boxed{279}$.

~ Edited and elaborated by eevee9406

Solution 2

$n$ is either $1$ digit, $2$ digits, or $3$ digits So take $n = a$. So $b = a + 1$. What is $a$ in base $a + 1$? obviously $a$. Therefore all $1$ digit numbers work. $n = 10a + c$ for the case of $2$ digits. Case 1: $a$ is greater than or equal to $c$ so $b = a + 1$. We need two digit number $ac$ in base $a+1$ which is simply $(a + 1)a + c = 10a + c$ so $a + 1 = 10$ and $a = 9$. So the two digit number $9c$ works. Case 2: $a$ is less than or equal to $c$ so $b = c + 1$. We need two digit number $ac$ in base $c + 1$ which is simply $(c + 1)a + c = 10a + c$ or $c = 9$ so the two digit number $a9$ works. In short, any two digit number with $9$ as a digit works. $n = 100a + 10c + d$ for the case of $3$ digits. Case 1: $a$ is the greatest so $b = a + 1$ We have $(a + 1)^2 \cdot a + (a + 1)c + d = 100a + 10c + d$ or $(a + 1)^2 \cdot a + (a + 1)c = 100a + 10c$ therefore $a((a + 1)^2 - 100) + c(a - 9) = 0$ or in short the only way this could happen is if both summands are $0$ therefore $(a + 1)^2 = 100$ and $a - 9 = 0$ so $a = 9$ works in both cases. We can go through all the casework but by symmetry, we realize any three digit number that has $9$ has a digit also works So the number of one digit numbers are $1 to 9$ or $9$. The number of two digit numbers with $9$ in it could be all of $(1 to 9)9$ or $9(0 to 9)$ so $9 + 10 = 19$ ways subtracting $99$ since that's over counted so $18$ ways. Now for three digits and that could be $9(0 to 9)(0 to 9)$ so $10 \cdot 10 = 100$. Now $(1 to 9)9(0 to 9)$ so $9 \cdot 10 = 90$. And finally $(1 to 9)(0 to 9)9$ so $9 \cdot 10 = 90$. But we overcount $99(0 to 9)$ and $9(0 to 9)9$ and $(1 to 9)99$ and $999$ so we overcount $10 + 10 + 9 = 29$ numbers. But we're taking $999$ twice since $99(0 to 9)$ and $9(0 to 9)9$ and $(1 to 9)99$ all overcount three times, but adding the $999$ we overcount it twice so we need to add a further $1$ to only count it once. we have $9 + 18 + 100 + 90 + 90 - 29 + 1 = 27 + 280 - 29 + 1 = 280 - 2 + 1 = 280 - 1 = \boxed{279}$.

~ilikemath247365

Solution 3 (Basically formalized Solution 1)

For any positive integer $n < 1000$, let its digits be represented as $d_k d_{k-1} \dots d_0$ in base $10$. Let $M = \max(d_k, d_{k-1}, \dots, d_0)$ be the greatest digit in $n$. According to the problem, the base used for $f(n)$ is $b = M + 1$.

The condition $f(n) = n$ implies:

$$\sum_{i=0}^{k} d_i (M+1)^i = \sum_{i=0}^{k} d_i (10)^i$$ Case 1: $M < 9$

If the maximum digit $M$ is less than $9$, then the base $b = M+1$ satisfies $b \le 9$.

For single-digit numbers ($k=0$), the equation simplifies to $d_0 = d_0$, which is an identity. Thus, $n \in \{1, 2, \dots, 8\}$ all work (8 values).

For multi-digit numbers ($k \ge 1$), since $M+1 < 10$, it follows that $(M+1)^i < 10^i$ for all $i \ge 1$. Because $d_i \ge 0$ and at least one $d_i > 0$ for $i \ge 1$, the Left Hand Side (LHS) is strictly less than the Right Hand Side (RHS). Therefore, no multi-digit numbers work in this case.

Case 2: $M = 9$

If the maximum digit is $9$, then the base is $b = 9+1 = 10$.

The equation becomes:

$$\sum_{i=0}^{k} d_i (10)^i = \sum_{i=0}^{k} d_i (10)^i$$ This is an identity. Thus, any number $n < 1000$ that contains at least one digit $9$ will satisfy $f(n) = n$.

Using complementary counting: Integers with no digit $9$: Each of the three decimal places (treating $1$ as $001$, etc.) can be any of the $9$ digits $\{0, 1, \dots, 8\}$. This gives $9 \times 9 \times 9 = 729$ such numbers.

Total for Case 2: $1000 - 729 = 271$.

Adding the two cases together, we get $8 + 271 = \boxed{279}$.

~LI,CHENXI