AIME 2026 I · 第 2 题

Solution 1

To count possible numbers in a structured manner, we can split the casework depending on how many digits the number has. Only numbers with an odd number of digits are possible, because numbers with an even number of digits will have digits adding up to an even number, and 13 is odd.

13 digits

There is one case for this, $1111111111111$.

11 digits

We have $6$ possibilities: $5$ arrangements of $11112121111$ and one arrangement of $11111311111$.

9 digits

There are $15$ possibilities: $6$ arrangements of $112212211$, $4$ arrangements of $111313111$, $4$ arrangements of $111232111$, and $1$ arrangement of $111151111$.

7 digits

There are $20$ possibilities: $3$ arrangements of $1141411$, $1$ arrangement of $2221222$, $6$ arrangements of $1231321$, $3$ arrangements of $3113113$, $3$ arrangements of $1223221$, $3$ arrangements of $1125211$, and $1$ arrangement of $1117111$.

5 digits

There are $15$ possibilities: $2$ arrangements of $15151$, $2$ arrangements of $42124$, $1$ arrangement of $33133$, $2$ arrangements of $13431$, $2$ arrangements of $32323$, $2$ arrangements of $13531$, $1$ arrangement of $22522$, $2$ arrangements of $12721$, and $1$ arrangement of $11911$.

3 digits

There are $5$ possibilities: $292$, $373$, $454$, $535$, and $616$.

Since there are no other possibilities, the total number of possibilities is $1 + 6 + 15 + 20 + 15 + 5 = \boxed{62}$.

~ Logibyte, ~gb1falcon

Solution 2

In order for the sum of the digits of a palindrome to be odd, there must be an odd number of digits, and the middle digit must be odd.

We can construct all palindromes whose digits sum to 13 in the following way:

Start with the digit $\overline{1}$.

We define the following 2 moves:

$A$: Increase the digits left-most and right-most digit by $1$ each. Note that if we only have $1$ digit, it is considered both the left-most and right-most digit, so we increase it by $2$.

$$\overline{abc\dots cba} \stackrel{A}{\rightarrow}\overline{(a+1)bc\dots cb(a+1)}$$ $B$: Add two digits of $1$ to surround our number.

$$\overline{abc\dots cba} \stackrel{B}{\rightarrow} \overline{1abc\dots cba1}$$ $A$ and $B$ contribute $2$ to a palindrome's sum. Therefore, we must apply $A$ or $B$ $6$ times to create a palindrome whose digits sum to $13$.

For example, here is the palindrome created by $AABBAB$:

$$\overline{1} \stackrel{A}{\rightarrow} \overline{3} \stackrel{A}{\rightarrow} \overline{5} \stackrel{B}{\rightarrow} \overline{151} \stackrel{B}{\rightarrow} \overline{11511}\stackrel{A}{\rightarrow} \overline{21512} \stackrel{B}{\rightarrow} \overline{1215121}$$ The total number of distinct palindromes are the number of ways to choose $A$ or $B$ $6$ times, which is $2^6 = 64$ ways.

However, we must account for exceptions, namely when a digit exceeds $9$ (number is no longer base 10) This will happen in two scenarios: $AAAAAA$ and $AAAAAB$.

Therefore, we will have a total of $64-2 = \boxed{62}$ valid palindromes.

~Samueleb27

Solution 3

Let $d$ be the digit in the middle. The possible values are $d=1,3,5,7,9$. Considering the digits to the right of the middle digit, they must sum to $d'=6,5,4,3,2$ respectively. Notice this is the same as having $d'$ balls and putting a divider in any subset of the gaps between the balls, so in each case there are $2^{d'-1}$ possibilities. The answer is

$$2^5+2^4+2^3+2^2+2^1=\boxed{62}$$

Solution 4

The conclusions made at the beginning of Solution 2 are noted here: the number must have an odd number of digits and an odd number at its center. Let the central digit be $d$, and let the number have $2n + 1$ digits. Note that the sum of the digits at either side of the central digit must be equal to $\frac{13 - d}{2}$ Each of these configurations corresponds to a star-and-bars case with no nonzero components where there are $\frac{13 - d}{2}$ stars and $n$ bars. Therefore, the number of ways to distribute them is ${\frac{13-d}{2} - 1 \choose n - 1}$ The maximum value of $\frac{13-d}{2}$ is equal to $6$ where $d=1$, and the total number of cases in the case where there are $n$ digits can be expressed as: ${6 - 1 \choose n - 1} + {5 - 1 \choose n - 1} + ... + {n - 1 \choose n - 1}$ This is the case for only one such $d$ for one such $n$. The total number of valid palindromes for any given $n$ is equal to ${6 \choose n}$ by the Hockey-Stick Identity. The possible values of $n$ are $1$ through $6$ (the number can't be 1 - digit) However, the case where $n = 1$ and $d = 11$ is omitted because $11$ isn't a valid digit. Finally, we have a final total sum of ${6 \choose 1} +{6 \choose 2} + ... + {6 \choose 6} - 1 = \boxed{62}$ palindromes

~tikachaudhuri

Solution 5

Any palindrome with an odd sum must have an odd number of digits. For a positive integer $n$, the number of palindromic compositions of $n$ is

$$2^{\lfloor n/2 \rfloor}$$ For $n=13$, this gives

$$2^{\lfloor 13/2 \rfloor} = 2^6 = 64.$$ Since the digits are in base $10$, no part of the composition may be $10$ or greater. The digits on the sides cannot exceed $9$, since the maximum possible sum on each side is

$$\frac{13-1}{2} = 6.$$ Thus, the only possible violations occur when the middle digit is at least $10$.

Because the total sum is $13$, the middle digit must be odd. The only invalid odd middle digits are $11$ and $13$, giving the following invalid palindromes: - $(13)$ - $(1,11,1)$

Subtracting these from the total,

$$64 - 2 = 62.$$ Therefore, the final answer is $\boxed{62}$.

~matchas (made by ninth graders if any issues feel free to edit)

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2026 AIME I #2

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