60-odd years of moscow mathematical
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Moscow olympiad problems
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100 |. 32.1.10.3. A pack of cards with their backs down is arranged in a row. If two cards of the same suit are next to one another, or have just one card between them, then the Rule allows us to remove the extreme left one. Besides, the Rule allows us to add any number of cards from other packs to the right hand side of the row. Prove that it is possible to add or remove cards so that in the end only 4 cards are left. 32.1.10.4. Is there a real number h such that [h · 1969 n ] is not divisible by [h · 1969 n−1 ] for any positive integer n? 32.1.10.5. Given square ABCD, find the locus of points M such that ∠AM B = ∠CM D. Tour 32.2 Grade 7 32.2.7.1. m and n are two positive integers. All different divisors of m — the numbers a, b, . . . , k — and all different divisors of n — the numbers s, t, . . . , z — are written out (m, and n, and 1 are included). It turns out that a + b + · · · + k = s + t + · · · + z and 1 a + 1 b + · · · + 1 k = 1 s + 1 t + · · · + 1 z . Prove that m = n. 32.2.7.2. We strike out two consecutive digits a and b (a preceding b) of the number N = 123456789101112 . . . 9989991000 and replace them with the number a + 2b; the number a may be an unwritten zero if b is the first digit of N . (Clearly, there are many ways to perform this operation.) The same operation is repeated with the numbers obtained, and so on. (For example, in one step the numbers 218307, 38307, 117307, 111407, 11837, 118314 may be obtained from 118307.) Prove that several application of this operation turn the given number into 1. 32.2.7.3. A crook acquired a square lot, fenced it in and got permission from the credulous president of his collective farm to perform a few times the following operation: draw a straight line through any two points of the fence, destroy the part of the fence between these two points on one side of the line, and build a new part of the fence symmetric to the destroyed part with respect to the line. Can the crook increase the area of his patch with such manipulations? (See Fig. 69) Figure 69. (Probl. 32.2.7.3) 94 MOSCOW MATHEMATICAL OLYMPIADS 1 – 59 32.2.7.4. Two players play the following game. Every player, in turn, chooses 9 numbers in the sequence 1, 2, 3, . . . , 100, 101 and strikes them out. After eleven turns there are two numbers left. The second player then pays the first one the difference between the two numbers in roubles. Prove that the first player can always win at least 55 roubles, no matter how the second one plays. Note. Students who play this game will be fired from the school. 32.2.7.5. A pearl of radius 3 mm is baked inside a round pudding of radius 10 cm. The Rule allows us to cut the pudding along a straight line with a sharp knife into two (equal or unequal) parts. If the pearl is not found in one cut, (does not occur under the knife), one of the parts may be cut again; if this does not help, it is allowed to cut one of the three obtained parts and so on. Prove that it is possible not to find the pearl after 32 cuts, no matter how they are made. Prove that it is possible to make 33 cuts so that the pearl will be found, no matter where it is. Grade 8 32.2.8.1. See Problem 32.2.7.2. 32.2.8.2. A white knight is on square a1 of a chessboard. Two players take turns daubbing one square of the chessboard at a time with bauxite glue. They must do this in such a way that the knight could move according to usual rules onto any clean square without getting stuck. The loser is the player who cannot make a move. Who wins provided both play optimally? 32.2.8.3*. Two regular pentagons have one common vertex. The vertices of both pentagons are num- bered clockwise 1 to 5, the number of the common vertex being 1. The vertices with the same numbers are connected by straight lines. Prove that these four lines meet at one point. (See Fig. 70) Figure 70. (Probl. 32.2.8.3) 32.2.8.4. Finite sequences of positive integers are composed so that every next number is greater than the square of the preceding one, and the last number of each sequence is equal to 1969 (sequences may have different lengths). Prove that there are fewer than 1969 different such sequences. 32.2.8.5*. 100 cubes are arranged in a row, 77 black and 23 white among them. They are arranged Download 1.08 Mb. Do'stlaringiz bilan baham: |
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