Weronika Mysliwiec, klasa 8W, rok szkolny 2018/2019

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Wojciech Staniszewski
4 lat temu
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1 Poniższy zbiór zadań został wykonany w ramach projektu Mazowiecki program stypendialny dla uczniów szczególnie uzdolnionych - najlepsza inwestycja w człowieka w roku szkolnym 2018/2019. Tresci zadań rozwiązanych i przetłumaczonych na język angielski pochodzą z Internetowego Koła Matematycznego. Są to zestawy zadań na poziomie Olimpiady Matematycznej Juniorów, mające przygotować młodzież do podejmowania trudnych matematycznych wyzwań, w szczególnoci do startu w Olimpiadzie Matematycznej. Praca wykonana pod kierunkiem mgr Adrianny Żołnierczuk. Szkoła Podstawowa nr 112 Przymierza Rodzin im. Jana Pawła II

2 SET I Excercise 1 (answer) Let s call all the 2010 numbers respectively a 1, a 2,..., a 2009, a We know, that the first number is 7: 7, a 2, a 3,..., a 7, a 8, a 9,..., a 14, a 15, a 16,..., a 21, a 22,... The sum of the 7 consecutive numbers must equal 77, so and also: 7 + a a 7 = 77 a a 7 + a 8 = 77 therefore a 8 = 7. Similarly for the numbers a 15 and a 22 : The sums of numbers from a 8 to a 14 and from a 9 to a 15 must equal 77 and a 8 = 7, therefore a 15 = 7. Now we know that every number in a form a 7n=1 (when n is a natural number) equals = therefore a 2010 equals 7. Excercise 3 (answer) If x 4 = y , then x 4 y 4 = (x y)(x 3 +x 2 y +y 2 x+y 3 ) = (x y)(x+y)(x 2 +y 2 ) = We have three possibilities, either both x and y are even, or they are both odd, or one of them is even and the other is odd. 1. If they are both even or both odd, all the three expressions (x y), (x+ y), and(x 2 + y 2 ) are even, which means that (x y)(x + y)(x 2 + y 2 ) is divisible by 8, which is a contradiction, because isn t divisible by If one of them is even and the other is odd, all the three expressions (x y), (x+y), and(x 2 +y 2 ) are odd, which means that (x y)(x+y)(x 2 +y 2 ) is odd, which is a contradiction, because is even. Szkoła Podstawowa nr 112 Przymierza Rodzin im. Jana Pawła II 2

3 We proved that x and y can t be neither even nor odd, which means that no such numbers exist. Excercise 5 (answer) a = 9b + a+c 2 b = 9c + b+a 2 c = 9a + c+b 2 Let s sum up all three equations a 2 + b 2 + c = 10b + 10c + 10a Now, let s substract the left side of the equation from the right side a 2 10a b 2 10b c 2 10c + 25 = 0 (a 5) 2 + (b 5) 2 + (c 5) 2 = 0 A square is never a negative number, so all of the above squares equal 0, therefore a 5 = b 5 = c 5 = 0, so a = b = c = 5 Answer: The only real numbers which satisfy this equations are a = b = c = 5. Exercise 7 (answer) Firstly, let s prove that the square of a solar number is also a solar number. Let s take n = a 2 + 5b 2, then we have n 2 = (a 2 + 5b 2 ) 2 = a b a 2 b 2 = a b 4 10a 2 b a 2 b 2 = = (a 2 5b 2 ) 2 + 5(4ab) 2. Obviously (a 2 5b 2 ) is an integer number, because both a and b are integer, so their squares also are. Moreover we know that it doesn t equal 0, because a 2 can t equal 5b 2, because if it did, it would mean that a = 5b, which is a contradiction, since a is an integer number. 4ab is also a integer number, as both a and b are integer. Since both (a 2 5b 2 ) and 4ab are integer, we can say that n 2 is a solar number. We ve proven that a square of a solar number is also a solar number, therefore if n is a solar number, then n 4 is also a solar number, because n 4 = (n 2 ) 2, which ends the proof. Szkoła Podstawowa nr 112 Przymierza Rodzin im. Jana Pawła II 3

4 SET II Excercise 1 (answer) Let s call this number x The number of 1 and 2 is finite, so from a certain point there are only threes until the end. That means that we can describe x as the sum of a decimal fraction with a finite expansion* and a fraction 0, 0...0(3). Let s call the number of digits in the finite expansion n x =(the fraction with a finite expansion)+ 1 1,therefore x is a rational 3 10 n number, wchich ends the proof. Excercise 3 (answer) Let s assume that this number is integer and that p < q < r. r is a prime number, consequently it divides one of these sums: p+q, q+r, r+p If r divided q + r, r would also divide q, wchich would be a contradiction, because q is a prime number. If r divided r + p, it would also be a contradiction because it would mean that r divides p. Therefore, r p + q. 1 p+q p+q r < 2r r = 2 is prime, so it must equal 1, which means that p + q = r. r One of the numbers p, q must be even, because otherwise r would be both prime and even, wchich would mean that it equals 2 and that would be a contradiction because 2 is the smallest prime number and p < q < r. Therefore p=2. Now we have q + 2 = r so let us substitute this equation: (2+q)(2q+2)(q+4) 2q(q+2) = q q The only q wchich fulfils the foregoing equation is 2, because 4 must be integer q and q must be a prime number, but it s a contradiction because p = 2 = q and all the numbers p, q, r must be different. Therefore there aren t such numbers which fulfils this eqution. = 2q2 +10q+8 2q Excercise 5 (answer) Let s find the biggest number and call it x. Let s call all the numbers, beggining from x, a 1, a 2,..., a n. All the numbers a 1, a 2,..., a n are absolute values, therefore they are all nonnegative. If x equaled 0, all the numbers wolud equal 0, so let s assume that x > 0. Szkoła Podstawowa nr 112 Przymierza Rodzin im. Jana Pawła II 4

5 x = a 2 a 3 and a 2 a 3 is from the interval x, x (because 0 a 2 x and 0 a 3 x) so there are 2 possibilities: (1) a 2 a 3 = x and at that time a 2 = x and a 3 = 0, which means that: a 4 = x, a 5 = x, a 6 = 0, a 7 = x, a 8 = x, a 9 = 0 To continue let us note, that among the written numbers, all those with a numer divisible by 3 equals 0, and the rest equals x. a n = a 1 a 2 = x x = 0, therefore n is divisible by 3. (2)a 2 a 3 = x, so a 2 = 0 and a 3 = x, therefore: a 4 = x, a 5 = 0, a 6 = 0, a 7 = x, a 8 = 0, a 9 = 0 To continue let us note, that among the written numbers, all those which number divided by 3 give the reaminder of 2 equal 0, and the rest equal x. a n = a 1 a 2 = x 0 = x and a ( n 1) = a n a 1 = x x = 0, so in this case n is aslo divisible by 3. a) From there are 670 numbers divisible by 3 and, as we previously prooved, they equal 0, so the sum of all the 2010 number will be ( )x = 1340x We have 1340 x = 1340, so x = 1, which means that on the circle are written numbers (1, 1, 0). b) 1000 isn t divisible by 3, so on the cirlcle are written only 0, therefore the sum of the numbers equals 0. Excercise 7 (answer) Let s assume that there exist such numbers, that x + y = In that situation there exist also such integral numbers x 0, x 1, x 2, x 3, x 4, x 5 andy 0, y 1, y 2, y 3, y 4, y 5 from the interval 1; 9 that: x = x x x x x x 0 (the numbers x 0 x 5 are the consequent digits of x) Then: y = y y y y y y 0 And: x 0 + x 1 + x 2 + x 3 + x 4 + x 5 = x 0 + x 1 + x 2 + x 3 + x 4 + x 5 We can illustrate the addition x = y in a written way: x 5 x 4 x 3 x 2 x 1 x 0 + y 5 y 4 y 3 y 2 y 1 y 0 Szkoła Podstawowa nr 112 Przymierza Rodzin im. Jana Pawła II 5

6 Szkoła Podstawowa nr 112 Przymierza Rodzin im. Jana Pawła II 6

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