2026 AMC 10A Problem 25

This is a "minimum" question, so it has two halves: show that fewer than $95$ days cannot work, then show that $95$ days can.

Setting up. Each day, $18$ of the $20$ rooms get one nut each and the other $2$ rooms get nothing; call those two rooms "skipped" that day. At the start the rooms hold $1 + 2 + \cdots + 20 = 210$ nuts, an average of $10.5$ per room.

Part 1: fewer than 95 days is impossible.

1. Room 1 is the furthest behind, so it is the room that limits everything. Compare it with the average.
2. Each day the total grows by $18$, so the average grows by $\frac{18}{20} = 0.9$. Room 1 grows by at most $1$. So each day room 1 gains at most $1 - 0.9 = 0.1$ on the average.
3. Room 1 starts at $1$ nut, which is $10.5 - 1 = 9.5$ below the average. At the end every room is equal, so room 1 is the average. It has to close a gap of $9.5$ at no more than $0.1$ per day, which takes at least
$$\frac{9.5}{0.1} = 95 \text{ days.}$$
4. The same argument with whole numbers: suppose that after $D$ days every room holds $T$ nuts. Counting all the nuts gives $20T = 210 + 18D$. Room 1 has at most $1 + D$ nuts, so $T \le 1 + D$. Then
$$210 + 18D = 20T \le 20 + 20D, \qquad \text{so} \qquad 190 \le 2D \quad\text{and}\quad D \ge 95.$$

Part 2: 95 days is enough.

5. The final amount. After $95$ days the total is $210 + 18 \cdot 95 = 210 + 1710 = 1920$, which is $20 \times 96$. So every room must end with $96$ nuts.
6. How often each room is skipped. Room $k$ needs $96 - k$ more nuts in $95$ days, so it is skipped on $95 - (96 - k) = k - 1$ days. Room 1 is never skipped, room 2 once, room 3 twice, and so on up to room 20, which is skipped $19$ times. The total number of skips is $0 + 1 + \cdots + 19 = 190 = 2 \times 95$, exactly two per day, as it should be.
7. Only the counts matter. The final amounts depend only on how many times each room is skipped, not on which days. So we just need to sort the skips into $95$ days, two different rooms per day.
8. Line the skips up and fold the line in half. Write a list of $190$ entries: room 2 once, then room 3 twice, then room 4 three times, and so on, ending with room 20 nineteen times:
$$2,\ 3, 3,\ 4, 4, 4,\ \ldots,\ 20, 20, \ldots, 20.$$
On day $1$ skip entries $1$ and $96$, on day $2$ skip entries $2$ and $97$, and in general on day $j$ skip entries $j$ and $j + 95$.
9. The two rooms on a day are always different. All the entries for one room sit next to each other in the list, at most $19$ of them in a row. Two entries that are $95$ places apart cannot both be in a run that short. So each day skips two different rooms and fills the other $18$, exactly as the rules require. For example, day $1$ skips rooms $2$ and $15$, and day $95$ skips rooms $15$ and $20$.
10. After these $95$ days, room $k$ has been skipped $k - 1$ times, so it holds $k + (95 - (k - 1)) = 96$ nuts. Every room has $96$ nuts.

Fewer than $95$ days is impossible and $95$ days works, so the minimum is $95$.

The answer is $\boxed{\textbf{(B)}\ 95}$.

When the goal is to make everything equal, the object furthest behind sets the pace, so compare its best possible speed with the speed of the average. For the construction, recording only how many times each room is skipped removes the order of the days from the problem. The fold-in-half trick then pairs up the skips with no repeats, and it works whenever no single room needs more than half of all the skips.

Test the answer choices with the two facts from Part 1: after $D$ days the common amount is $T = \frac{210 + 18D}{20}$, which must be a whole number, and room 1 needs $T \le 1 + D$.

So the smallest choice that works is $95$. Choice $75$ fails because room 1 cannot keep up, which is the same reason no number below $95$ works.