41st International Mathematical OlympiadIMO 2000 Problems and SolutionsTaejon, KoreaJuly 13-25, 2000 Problem 4100 cards are numbered 1 to 100 (each card different) and placed in 3 boxes (at least one card in each box). How many ways can this be done so that if two boxes are selected and a card is taken from each, then the knowledge of their sum alone is always sufficient to identify the third box? Back to Competition Problems page SolutionAnswer: 12. Place 1, 2, 3 in different boxes (six possibilities) and then place n in the same box as its residue mod 3. Or place 1 and 100 in different boxes and 2-99 in the third box (six possibilities). An elegant solution communicated (in outline) by both Mohd Suhaimi Ramly and Fokko J van de Bult is as follows: Let H_{n} be the corresponding result that for cards numbered 1 to n the only solutions are by residue mod 3, or 1 and n in separate boxes and 2 to n - 1 in the third box. It is easy to check that they are solutions. H_{n} is the assertion that there are no others. H_{3} is obviously true (although the two cases coincide). We now use induction on n. So suppose that the result is true for n, and consider the case n + 1. Suppose n + 1 is alone in its box. If 1 is not also alone, then let N be the sum of the largest cards in each of the boxes not containing n + 1. Since n + 2 <= N <= n + (n - 1) = 2 n - 1, we can achieve the same sum N as from a different pair of boxes as (n + 1) + (N - n - 1). Contradiction. So 1 must be alone, and we have one of the solutions envisaged in H(n + 1). If n + 1 is not alone, then if we remove it, we must have a solution for n. But that solution cannot be the n, 1, 2 to n - 1 solution. For we can easily check that none of the three boxes will then accomodate n + 1. So it must be the mod 3 solution. We can easily check that in this case n + 1 must go in the box with matching residue, which makes the (n + 1) solution the other solution envisaged by H_{n + 1}. That completes the induction. My much more plodding solution (which I was quite pleased with until I saw the more elegant solution above) follows. It took about half an hour and shows the kind of kludge one is likely to come up with under time pressure in an exam! With a suitable labeling of the boxes as A, B, C, there are four cases to consider:
Case 2: A contains 1, 2 Case 3: A contains 1, 3; B contains 2 Case 4: A contains 1; B contains 2, 3. We show that Cases 1 and 4 each yield just one possible arrangement and Cases 2 and 3 none. In Case 1, it is an easy induction that n must be placed in the same box as its residue (in other words numbers with residue 1 mod 3 go into A, numbers with residue 2 go into B, and numbers with residue 0 go into C), for (n + 1) + (n - 2) = n + (n - 1). Hence n + 1 must go in the same box as n - 2 (if they were in different boxes, then we would have two pairs from different pairs of boxes with the same sum). It is also clear that this is a possible arrangement. Given the sum of two numbers from different boxes, take its residue mod 3. A residue of 0 indicates that the third (unused) box was C, a residue of 1 indicates that the third box was A, and a residue of 2 indicates that the third box was B. Note that this unique arrangement gives six ways for the question because there are six ways of arranging 1, 2, and 3 in the given boxes. In Case 2, let n be the smallest number not in box A. Suppose it is in box B. Let m be the smallest number in the third box, C. m - 1 cannot be in C. If it is in A, then m + (n - 1) = (m - 1) + n. Contradiction (m is in C, n - 1 is in A, so that pair identifies B as the third box, but m - 1 is in A and n is in B, identifying C). So m - 1 must be in B. But (m - 1) + 2 = m + 1. Contradiction. So Case 2 is not possible. In Case 3, let n be the smallest number in box C, so n - 1 must be in A or B. If n - 1 is in A, then (n - 1) + 2 = n + 2. Contradiction (a sum of numbers in A and B equals a sum from C and A). If n - 1 is in B, then (n - 1) + 3 = n + 2. Contradiction (a sum from B and A equals a sum from C and B). So Case 3 is not possible. In Case 4, let n be the smallest number in box C. n - 1 cannot be in A, or (n - 1) + 2 = 3 + n (pair from A, B with same sum as pair from B, C), so n - 1 must be in B. Now n + 1 cannot be in A (or (n + 1) + 2 = 3 + n), or in B or C (or 1 + (n + 1) = 2 + n). So n + 1 cannot exist, and hence n = 100. It is now an easy induction that all of 4, 5, ... , 98 must be in B. For given that m is in B, if m + 1 were in A, we would have 100 + m = 99 + (m + 1). But this arrangement (1 in A, 2-99 in B, 100 in C) is certainly possible: sums 3-100 identify C as the third box, sum 101 identifies B as the third box, and sums 102-199 identify A as the third box. Finally, as in Case 1, this unique arrangement corresponds to six ways of arranging the cards in the given boxes. Back to Competition Problems page |