3. The pigeon holes are the remainders when divided by 11. The pigeons are the numbers. (See also the solution to Problem 21.) If two numbers have the same remainder when di,-ided by 11. their difference must be divisible by 11.
4. The pigeon holes here are the numbers of hairs on a person's head (from 1 to 1,000,000). The pigeoIlS are the citizens of Leningrad.
6. Let us sort the football players by team as they come off their airplanes. There will be 10M + 1 players to sort. The General Pigeon Hole Principle assures us that there will be one team which has 11 players, and this team is complete.
8. There are five possible numbers of acquaintances for any person: 0, 1, 2, 3, or 4. So it would seem that each could haw a different number of friends. However, if any person has four acquaintances. then no perSOll may have zero acquaintances.
Hence two people must ha,-e the 5aIIle number of acquaintances.
9. If there are k teams. then the number of games played by each team varies from 0 to k - 1. However. if any team has played k - 1 games, then it has played every other team, and no team has played 0 games. Hence we are fitting k teams into k - 1 pigeon holes, which are either the numbers from 0 through k ~ 2 or the numbers 1 through k - 1.
lOa. The answer is 32. Indeed, suppose that 33 or more squares are colored green.
Then, after we have divided the board into si:'{teen 2 x 2 squares, the Pigeon Hole Principle guarantees that at least one ofthese squares contains 3 or more small green squares. These 3 green squares form the ã-forbiddenãã tromino in some position, and we have a contradiction. On the other hand_ we can color all the black fields (of the usual coloring) green, and this is an example of 32 green squares with the property needed.
lOb. The answer is 32 (again!). Indeed, if 31 or fewer squares are colored green, then one of those sixteen 2 x 2 squaTes contains 1 or 0 green squares. Then the other 3 or 4 squares are not colored green, and they form the tromino without green
224 MATHEMATICAL CIRCLES (RUSSIAN EXPERIENCE)
squares in it. This contradiction (and the same construction as above) completes the proof.
11. At least 1 + 2 + 3 = 6 problems were solved by the students mentioned in the problem statement. Therefore, there are 29 problems left to be solved, and 7 students to account for them. If each student had solved only 4 problems, then there would have been only 28 problems solved. Therefore, one student must have solved at least 5 problems.
12. Answer: 12 kings. See the hint to Problem lOa.
13. Divide the cobweb into 4 sectors as shown in Figure 135, each of which can hold no more than one spider.
FIGURE 135
14. Each of the smaller triangles can cover only one vertex of the larger triangle.
18. Color all the dry land red, and color each point diametrically opposite dry land green. Then there must be a point which is both red and green. Start the tunnel at this point. Do you see why this is, in a way, a Pigeon Hole Principle?
19. There are only 1987 possible remainders when a number is divided by 1987. If we examine, for example, the first 1988 powers of 2, we find that two of them must have the same remainder when divided by 1987 These two powers then differ by a multiple of 1987
20. When divided by 100, a perfect square can give only 51 remainders, since the numbers x2 and (100 - x)2 give the same remaillder. Hence of 52 integers, the squares of two must have the same remainder when divided by 100. These two squares differ by a multiple of 100.
22. If 3m and 3n (where m > n) are two powers of 3 which give the same remainder when divided by 1000, then 3m - 3n = 3n(3m- n
- 1) is divisible by 1000. Now the prime factors of 1000 are 2 and 5, and neither divides 3n It follows that 1000 muSt divide 3m-n - 1, which means that 3m-n is a power of 3 ending in the digits 001.
23. This sum can take on only seven values: the numbers 'from -3 through 3.
24. Divide all the people into 50 pairs who are sitting diametrically opposite each other. Consider these pairs as the pigeon holes. Since there are more than 50 men.
one pair must include more than one man.
25. If the conclusion is false, then it is clear that the boys will have gathered at least 0 + 1 + 2 + + 14 = 105 nuts. which is a contradiction.
26. The product of the numbers in all the groups is 9!= 362880. If the product of each group were no greater than 71. the product of all the numbers could only be 713 = 357911. It should be noted here that this method of proof is, in a way, more general than the simple Pigeon Hole Principle.
27. One can move from any squa!"e to any other by passing through neighboring cells, and we can always choose a path such that the number of squares visited is less than 19. This means that if a is the smallest number on the board, all the numbers are included between a and a + 95. Therefore there can be no more than 96 different numbers among the 100 on the board, and two must be equaL 28. We choose anyone person in :he group. Let us call him Bob. We sort the others into two pigeon holes: those who know Bob and those who do not. There are at least three of the remaining five people in one of these categories. Suppose Bob has three acquaintances. If n..-o of these know each other, then they, together with Bob, form the required triple. If none of them knows each other, then they themselves form the required triple. A similar argument holds if there are three people whom Bob does not know.
29. Consider the parity (remainder upon division by 2) of the coordinates of the points. There are four possibilities: (odd, odd); (odd, even); (even, odd); (even, even). Since there are fiw points, we can cho~se two of them whose coordinates both match in parity. It is not hard to see that the midpoint of the line segment they determine has integer coordinates.
30. There are two categories into which we can fit the three sizes: those sizes for which there are more right boots than left boots, and those sizes for which there are more left boots than right boots (if there happens to be an equal number of right and left boots in one size. we put that size in the second category). It follows that two sizes lie in the same category. Let us say that sizes 41 and 42 have more right boots than left boots (an analogous argument will hold if two sizes have more left boots than right boots).
Now there are 300 left boots in all, and at most 200 left boots in anyone size.
Therefore, the sum of the left boots in any two sizes is at least 100. We have shown that there are at least 100 left boots in sizes 41 and 42 (taken together), and that each of these sizes contains more right boots than left boots. Hence each left boot has a match, and there are at lea....q 100 good pairs in the warehouse.
31. There are 11 more consonants than vowels in the alphabet. Therefore, if we add the differences between the number of consonants and the number of vowels in each of the six subsets, these differences must sum to 11. It follows that there must be at least one subset in which this difference is less than 2, and the letters of this subset must form a word.
32. Consider the ten sums: Xl, Xl + X2. Xl + X2 + X3, , Xl + X2 + + XIOã
Two of these must have the same remainder when divided by 10. The difference between these two sums gives a set whose sum is divisible by 10.
33. We can divide the numbers from 1 through 20 into ten disjoint sets, such that if a pair of numbers is selected from the same set, one of the pair divides the other:
{ll}, {I3}, {IS}, {I7}, {I9}, {I, 2, 4, 8, 16}, {3, 6, 12}, {S, 10, 20}, {7, 14},
226 MATHEMATICAL CIRCLES (RUSSIAN EXPERIENCE)
{9, I8}. Then, of any eleven numbers not greater than 20, two of them must fit in one of these pigeon holes, and one of these two divides the other.
34. We can number the study groups with the numbers 1 through 5. Then, instead of considering each student him or herself, we can consider the set of numbers belonging to the study groups he or she is part of. Each of these is a subset of the set {I, 2, 3, 4, 5}. We solve the problem by dividing the 32 subsets of this set into 10 collections such that if two subsets are chosen from the same collection, one of them contains the other (compare this with the solution to Problem 33).
The following is such a collection. The subsets in each collection are written as numerals:
[0,{I},{I,2},{I,2,3},{I,2,3,4},{I,2,3,4,5B, [{2},{2,5},{I,2,5},{I,2,3,5}],
[{3},{I,3},{I,3,4},{I,3,4,5}], [{4},{I,4},{I,2,4},{I,2,4,5}], [{5}, {l, 5}, {I, 3, 5}],
[{2, 4}, {2, 4, 5}, {2, 3, 4, 5}], [{3,4},{3,4,5}],
[{3, 5}, {2, 3, 5}], [{4, 5}, {l, 4, 5}], [{2, 3}, {2, 3, 4}].