olymmath

NL-NWO 2017

Dutch Mathematical Olympiad Final 2017.

Other problems

Problem 1

We consider positive integers written down in the (usual) decimal system. Within such an integer, we number the positions of the digits from left to right, so the leftmost digit (which is never a 00 ) is at position 11 . An integer is called even-steven if each digit at an even position (if there is one) is greater than or equal to its neighbouring digits (if these exist). An integer is called oddball if each digit at an odd position is greater than or equal to its neighbouring digits (if these exist). For example, 31223122 is oddball but not even-steven, 77 is both even-steven and oddball, and 123123 is neither even-steven nor oddball. (a) Prove: every oddball integer greater than 99 can be obtained by adding two oddball integers. (b) Prove: there exists an oddball integer greater than 99 that cannot be obtained by adding two even-steven integers.

Problem 2

A parallelogram ABCDABCD with ∣AD∣=∣BD∣|AD| =|BD| has been given. A point EE lies on line segment ∣BD∣|BD| in such a way that ∣AE∣=∣DE∣|AE| = |DE| . The (extended) line AEAE intersects line segment BCBC in FF . Line DFDF is the angle bisector of angle CDECDE . Determine the size of angle ABDABD .

AA BB CC DD EE FF

Problem 3

Six teams participate in a hockey tournament. Each team plays exactly once against each other team. A team is awarded 33 points for each game they win, 11 point for each draw, and 00 points for each game they lose. After the tournament, a ranking is made. There are no ties in the list. Moreover, it turns out that each team (except the very last team) has exactly 22 points more than the team ranking one place lower. Prove that the team that fi nished fourth won exactly two games.

Problem 4

If we divide the number 1313 by the three numbers 5,75, 7 , and 99 , then these divisions leave remainders: when dividing by 55 the remainder is 33 , when dividing by 77 the remainder is 66 , and when dividing by 99 the remainder is 4. If we add these remainders, we obtain 3+6+4=133 + 6 + 4 = 13 , the original number. (a) Let nn be a positive integer and let aa and bb be two positive integers smaller than nn . Prove: if you divide nn by aa and bb , then the sum of the two remainders never equals nn . (b) Determine all integers n>229n > 229 having the property that if you divide nn by 99,13299, 132 , and 229229 , the sum of the three remainders is nn .

Problem 5

The eight points below are the vertices and the midpoints of the sides of a square. We would like to draw a number of circles through the points, in such a way that each pair of points lie on (at least) one of the circles. Determine the smallest number of circles needed to do this.