AIME 2026 II · 第 13 题

Solution 1 (Grouping Numbers)

Firstly, notice that if we have an element in $S$, there will be exactly $2$ options for the element in $T$ that can pair with the element in $S$. However, if $2$ elements in $S$ have a difference of $2$, then their pairs will not be independent.

For example, consider $1$ and $3$: $1$ can pair with $0$ or $2$ while $3$ can pair with $2$ or $4$. However, $2$ cannot pair with both $1$ and $3$, so we have $3$ possibilities.

As a result, we can show that if we have a group of $x$ elements in set $S$ that form an arithmetic sequence with common difference $2$, then there are exactly $x+1$ ways to create pairs with elements in set $T$. This result can be proved by induction. We can model this idea by thinking about separating each "set" of number that differ with each neighbor by $1$ and each set separated by a difference of $2$. Then we can easily control what factors we need to get $4040$.

We know $4040 = 101 \cdot 2^3 \cdot 5$, so our groups of $x$ elements will result in $x+1 = 101, 2, 2, 2, 5$. Thus, the lengths of our groups are $100, 1, 1, 1, 4$.

This gives a total of $100 + 1 + 1 + 1 + 4 = 107$ elements.

Therefore, the least number of elements that set $S$ can have is $\boxed{107}$.

~pl246631