AIME 2026 I · 第 11 题

Solution 1

As a general intuition, we want larger numbers to be next to smaller numbers (and vice versa) to maximize this sum. The best way to do this is in a checkerboard pattern, so that all numbers in the lower half are next to higher numbers.

We then split the numbers into two sets: Small numbers $\{1,2,\dots32\}$ and large numbers $\{33,34,\dots 64\}$ (note that all large numbers are greater than all small numbers). We wish to determine where to put these numbers to maximize the differences.

We now examine this problem from a local view: Suppose we have large numbers, $A,B$ and small numbers $a,b,c\dots h$ such that $A$ is surrounded by $a,b,c,d$ and $B$ is surrounded by $e,f,g,h$. Then the sum of the differences with $A,B$ is $4A-(a+b+c+d)+4B-(e+f+g+h)$.

Notice that if we swap $A$ with $B$, the sum is $4B-(a+b+c+d)+4A-(e+f+g+h)$ which is the same. The same logic goes for if a small number is surrounded by larger numbers. Therefore, the key observation is that swapping two numbers that are surrounded by the same amount of numbers yields the same sum.

However, if $A$ is only surrounded by $a,b,c$ (as it would be if it was in an edge position), while $B$ is surrounded by $e,f,g,h$, then the initial sum would be $3A-(a+b+c)+4B-(e+f+g+h)$, while swapping $A$ and $B$ would yield the different sum $3B-(a+b+c)+4A-(e+f+g+h)$.

As we wish to maximize the sum, we want the greater number to have a coefficient of $4$, and therefore we want the greater extremes (e.g. 1 and 64) to be bordering four numbers, with the rest being at the edges.

With this intuition, we may see that the best arrangement is in the aforementioned checkerboard formation, with $31,32,33,34$ being in the corners, $19,20,\dots30$ and $35,36\dots46$ being on the edges, and the rest $1,2,\dots18$ and $47,48,\dots64$ being in the middle.

We now calculate the sum for this which is

$$4((47+48+\dots+64)-(1+2+\dots+18))+3((35+36+\dots+46)-(19+20+\dots+30))+2((33+34)-(31+32))=3896.$$ The requested remainder is $\boxed{896}$.

~SilverRush

It isn't correct.

~metrixgo

fixed it, ty

~SilverRush

Solution 2

Like in Solution 1, make a checkerboard pattern with the numbers $\{1, 2, \dots, 32\}$ in the dark squares and $\{33, 34, \dots, 64\}$ in the light squares. Now, place $32.5$ on every line between adjacent squares. The absolute difference between any two adjacent squares is the sum of the difference between the first square and $32.5$ and the difference between $32.5$ and the second square. So, we can decompose our sum as

$$\sum_{x = 1}^{64} |x - 32.5| \cdot \#\text{neighbors of }x.$$ Most squares have $4$ neighbors, but squares on the edges have $3$ neighbors, and squares in the corners have $2$. Clearly, then, we want to put the numbers closest to $32.5$ in the corners and edges. There are $4$ corner squares, $24$ edge squares, and $36$ squares in the middle.

Let's put $\{31, 32, 33, 34\}$ in the corners, and $\{19, 20, \dots, 30\} \cup \{35, 36, \dots, 46\}$ along the edges. The remaining numbers $\{1, 2, \dots, 18\} \cup \{47, 48, \dots, 64\}$ go in the middle.

The average distance of the corner numbers from $32.5$ is $1$. The average distance of the edge numbers from $32.5$ is $8$. The average distance of the middle numbers from $32.5$ is $23$. This gives us a total of

$$36 \cdot 23 \cdot 4 + 24 \cdot 8 \cdot 3 + 4 \cdot 1 \cdot 2 = 3896,$$ giving an answer of $\boxed{896}$.

Video Solution

2026 AIME I #11

By piacademyus.org