2001 USAMO Problems/Problem 1

Problem

Each of eight boxes contains six balls. Each ball has been colored with one of $n$ colors, such that no two balls in the same box are the same color, and no two colors occur together in more than one box. Determine, with justification, the smallest integer $n$ for which this is possible.

Solution 1

We claim that $n=23$ is the minimum. Consider the following construction (replacing colors with numbers) which fulfills this: $\left[ \begin{array}{cccccccc} 1 & 1 & 1 & 2 & 3 & 4 & 5 & 6 \\ 2 & 7 & 12 & 7 & 8 & 9 & 10 & 11 \\ 3 & 8 & 13 & 12 & 13 & 14 & 15 & 16 \\ 4 & 9 & 14 & 17 & 17 & 17 & 18 & 19 \\ 5 & 10 & 15 & 18 & 20 & 22 & 20 & 21 \\ 6 & 11 & 16 & 19 & 21 & 23 & 22 & 23 \end{array} \right]$ Suppose a configuration exists with $n \le 22$ .

Suppose a ball appears $5$ or more times. Then the remaining balls of the $5$ boxes must be distinct, so that there are at least $n \ge 5 \cdot 5 + 1 = 26$ balls, contradiction. If a ball appears $4$ or more times, the remaining balls of the $4$ boxes must be distinct, leading to $5 \cdot 4 + 1 = 21$ balls. The fifth box can contain at most four balls from the previous boxes, and then the remaining two balls must be distinct, leading to $n \ge 2 + 21 = 23$ , contradiction.

However, by the Pigeonhole Principle, at least one ball must appear $3$ times. Without loss of generality suppose that $1$ appears three times, and let the boxes that contain these have balls with colors $\{1,2,3,4,5,6\},\{1,7,8,9,10,11\},\{1,12,13,14,15,16\}$ . Each of the remaining five boxes can have at most $3$ balls from each of these boxes. Thus, each of the remaining five boxes must have $3$ additional balls $> 16$ . Thus, it is necessary that we use $\le 22 - 16 = 6$ balls to fill a $3 \times 5$ grid by the same rules.

Again, no balls may appear $\ge 4$ times, but by Pigeonhole, one ball must appear $3$ times. Without loss of generality, let this ball have color $17$ ; then the three boxes containing $17$ must have at least $2 \cdot 3 + 1 = 7$ balls, contradiction.

Therefore, $n = 23$ is the minimum.

Solution 2

Similar to the above solution, no color can appear 4 times or more.

We can use PIE. Let $S_1,S_2,\cdots ,S_8$ be the colors in each of the $8$ boxes. Then $n=|\cup_{i=1}^8S_i|$ . By PIE we know that $|\cup_{i=1}^8S_i|=\sum_{i=1}^8 |S_i|-\sum_{1\leq i<j\leq 8}|S_i\cap S_j|+\sum_{1\leq i<j<k\leq 8}|S_i\cap S_j\cap S_k|-\sum_{1\leq i<j<k<l\leq 8}|S_i\cap S_j\cap S_k\cap S_l|\cdots.$ Note however that no color can appear 4 times or more, so all the items after $\sum_{1\leq i<j<k\leq 8}|S_i\cap S_j\cap S_k|$ must be equal to 0, so $|\cup_{i=1}^8S_i|=\sum_{i=1}^8 |S_i|-\sum_{1\leq i<j\leq 8}|S_i\cap S_j|+\sum_{1\leq i<j<k\leq 8}|S_i\cap S_j\cap S_k|.$ Now note that for each $i$ , $|S_i|=6,$ so $\sum_{i=1}^8 |S_i|=48.$ Also no pair of colors appears twice, so for any $i,j, |S_i\cap S_j|\leq 1,$ and so $\sum_{1\leq i<j\leq 8}|S_i\cap S_j|\leq \binom82=28.$ Thus, $n\geq 48-28+\sum_{1\leq i<j<k\leq 8}|S_i\cap S_j\cap S_k|.$ Given that there are $n$ colors, to minimize the number of colors in 3 boxes means each color is in either 2 or 3 boxes, and since there are 48 total spots, this means there are $48-2n$ colors in 3 boxes and $3n-48$ colors in 2. Thus $n\geq 20+48-2n,$ so $3n\geq 68$ , and $n\geq23$ and we are done.

Solution 3

Let $m_{i,j}$ be the number of balls which are the same color as the $j^{\text{th}}$ ball in box $i$ (including that ball). For a fixed box $i$ , $1\leq i\leq 8$ , consider the sums $S_i = \sum_{j=1}^6 m_{i,j}\quad\text{and}\quad s_i = \sum_{j=1}^6 \frac{1}{m_{i,j}}.$ For each fixed $i$ , since no pair of colors is repeated, each of the remaining seven boxes can contribute at most one ball to $S_i$ . Thus, $S_i\leq 13$ . It follows by the convexity of $f(x) = 1/x$ that $s_i$ is minimized when one of the $m_{i,j}$ is equal to 3 and the other five equal to 2. Hence $s_i\geq 17/6$ . Note that $n = \sum_{i=1}^8 \sum_{j=1}^6 \frac{1}{m_{i,j}}\geq 8\cdot\frac{17}{6} = \frac{68}{3}.$ Hence there must be 23 colors. The construction for $n = 23$ is above.

2001 USAMO (Problems • Resources)
Preceded by First question	Followed by Problem 2
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2001 USAMO Problems/Problem 1

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Problem

Solution 1

Solution 2

Solution 3

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