# Are TTM grades added to math

## Grade 7, 8th Department of Mathematics Mathematics Day November 9, 2013

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1 Department of Mathematics Mathematics Day November 9, 2013 Grades 7, 8 12 Task 1 (points). Master Hora has a curious clock: The hour hand does not jump every hour at 30, as usual, but at 150 (clockwise). (a) What should the dial look like if you should still be able to read the hour from 1 to 12 (The 12 should always be on top.) (b) Enter an angle other than 30 and 150, with one every hour clear position can be read and the hour hand is back in the same place after exactly 12 hours. (c) What other clocks can there still be in which the hour hand jumps every hour by a fixed angle clockwise so that every hour can be read at a clear place and in which the hour hand is in the same place even after exactly 12 hours? is (d) Why can there be no other clocks apart from the clocks mentioned and found by you in the previous parts of the exercise (Note: Consider angles as fractions, i.e. 30 = 1 V, = 5 V etc. with the full angle 12 V = 360.) (e) If the hour hand always reaches the starting position again at some point, if it jumps every hour by any fraction multiplied by 360 Explain your answer briefly. Additional part (if there is still time). Describes what happens when the hour hand jumps every hour through an angle that is not an arbitrary fraction multiplied by 360. (There are such numbers.) Universität Hamburg Gateway to the World of Science Day of Mathematics

2 Hamburg Mathematics Day, November 9, 2013 Grades 7, 8 2 ˇ (ˇ ˇ (ˇ ˇ (Task 2 (points). In music there are notes with note values: Note ˇ ˇ (ˇ) Note value You can also use a Place a period after a note. This lengthened these dotted notes by half of their value, for example the note ˇ (the value = 3. (For connoisseurs: Other note values such as triplets should not be used.) Several notes form a measure, For example, in a bar the note values of the notes contained must be added. In a rhythm, the order of the notes is important. Above is an example of the rhythm of a bar. (a) Write all eight possible rhythms of a bar in which the denominators of the note values are at most. (b) How many different rhythms of 2 measures are there in which the denominators of the note values are at most 8, if no dotted notes are used (c) How many different rhythms of 3 bars there are those in which the denominators of the note values are at most 8, if dotted notes are not allowed to be used (Note: Differentiate between rhythms that contain 0, 2, or 6 eighth notes.) In music, you can also combine several notes by using instead slurs, for example with the connected notes ˇ ˇ (ˇ) the total note value = 15. This makes the math easier, because you can now get all the multiples of the shortest note value. (Only the value decides: ˇ = = ˇ) (d) How many different rhythms of measures are there, if slurs are allowed and the denominator of the note values may not exceed 1 How many ... 2 may be How many ... How many ... 8 can be (e) How many different rhythms of bars are there, if slurs are allowed and the denominator of the note values is at most 1, 2, 8, 16, 32, 6, 128 etc. may be

3 Hamburg Mathematics Day, November 9, 2013 Grades 7, 8 3 Exercise 3 (points). (a) Give the twelve factors of 200 and show that 200 has no other factors. (b) Let p, q and r be different prime numbers. Proves that the number p 3 q 2 r has exactly 2 factors. (c) Find the five smallest numbers with exactly six factors. (d) Shows: Every square number has an odd number of divisors. If a number has an odd number of factors, it is always a square number. Note: It can be assumed that every natural number greater than 1 has a unique prime factorization (except for the order of the factors). For example 120 =

4 Hamburg Mathematics Day, November 9, 2013 Grade 7, 8 exercise (points). The aim is to determine angles between different points of a regular hexagon (each with a reason; measuring is not a reason): EDFCAB How big are (a) AEC, (b) ADC, (c) AF D, DAF, F DA, (d) CBA and BAC (e) Are there any other angle sizes that are smaller than 180, based on triangles consisting of corner points of the regular hexagon. Now we consider two circles, each going through the center of the other: SM 1 M 2 (f) Determines the angle at an intersection S of the circles to the two circle centers M 1 and M 2: M 1 SM 2 (g) Determines the Angle at the intersection point S between a tangent to the first circle and a tangent to the other circle.

5 Hamburger Tag der Mathematik, November 9, 2013 Solutions 7, 8 5 Solutions 7, 8 Solution 1. (a) The clock face should look like this: (b) With the angle 210 you can see the clock face from the previous part on the vertical Use mirrored axis because 210 =. (c) In addition to the angles 30 and 150 as well as 210 from the previous exercise part mentioned in the exercise, the angle may only be 330 so that the clock runs exactly mirrored to the normal clock. (d) On the one hand, the clock should be in the starting position again after twelve hours, so that the angle must be a multiple of. On the other hand, among the multiples 1, 2, ..., 12, the angle must not be two equal angles, otherwise the time could not be read off clearly. In particular, only 12 of the angle may again be a multiple of 360. This is not the case if the abbreviated fraction no longer has the denominator 12: 2 12 = 1 6, 3 12 = 1, 12 = 1 3, 6 12 = 1 2, 8 12 = 2 3, 9 12 = 3, = 5 6 If the denominator is 2, 3 or 6, the starting position is reached again after 2, 3 or 6 hours. Only 1, 5, 7 11 and cannot be shortened anymore, so 30, 150, 210 and 330 are the only possible angles. (e) If the clock jumps every hour by any fraction multiplied by 360, it always reaches its starting position at some point: at the latest after as many hours as the denominator of the fraction is large, it has covered a multiple of 360.

6 Hamburg Mathematics Day, November 9, 2013 Answers 7, 8 6 Additional part (if there is still time). If the hour hand jumps every hour through an angle that is not an arbitrary fraction multiplied by 360, it never points to the same place twice: if it had reached the same position n hours later and covered z 360, it would have to go zn 360 every hour have jumped. In addition, the hour hand comes as close as desired to any point on the clock face: Since it points to an infinite number of different points, there are two numbers on the clock face for every angle, no matter how small, with the angle between them even smaller. But if the positions of the numbers n and n + k differ only by the small angle α, the positions of the numbers n + 2k, n + 3k etc. are always shifted in the same direction by α on the dial. So there is no arc of a circle on the clock face, no matter how small, that the hour hand never points into. The hour hand does not hit a position twice, but it passes in every area, no matter how small (and infinitely often). With such a clock you can clearly read not only the hour, but also every future date and hour, and the infinite number of inscriptions are evenly distributed on the edge. Solution 2. (a), ˇ ,, ˇ ˇ, ˇ, ˇ ˇ, ˇ ˇ, ˇ ˇ ˇ ˇ. (b) There are six different such rhythms :, ˇ ˇ, ˇ ˇ (ˇ (, ˇ (ˇ ˇ (, ˇ (ˇ (ˇ, ˇ (ˇ (ˇ (ˇ (. (If you double the note values, you get Exactly one of the six rhythms without dotted notes from the first part of the exercise.) There are obviously exactly two such rhythms without eighth notes and one with four eighth notes. If such a rhythm contains two eighth notes, it can also contain only one quarter note, which is the first, second or third note. So there are no other such rhythms. (c) If such a measure does not contain eighth notes, it can contain either half and quarter notes or only quarter notes. In the first case there are two possibilities and in the second case only one, so there are three such rhythms. If such a measure contains exactly two eighth notes, the rest of the measure can consist of half or two quarter notes. In the first case, the half note can be in the first, second or third position. In the second case there are six possibilities chances, three each if the last grade is a quarter or

7 Hamburg Mathematics Day, November 9, 2013 Solutions 7, 8 7 is an eighth. (The other quarter or eighth note can then be in the first, second or third position.) So there are nine such rhythms with exactly two eighth notes. If there are exactly four eighth notes, the measure can also only contain a quarter, which can be played as the first to fifth note. There is also the rhythm of exactly six eighths. This gives a total of = 18 possible rhythms. (d) If the denominator may not exceed 1 or 2, there are one or two possible rhythms (only the one from a whole note or the one from two halves). From the first part of the exercise it is known that there are eight possible rhythms from note values with the maximum denominator, and with the help of the slurs, only notes can be formed here that also existed as normal and dotted notes, e.g. B. ˇ ˇ ˇ = ˇ =. The number of possible rhythms with a maximum denominator of 8 can be obtained, for example, by considering the following: You can write each of these rhythms as eight eighth notes, some of which are connected by slurs. There are two possibilities between two adjacent eighth notes: Either they are connected to one another or a new note begins. Since there are seven digits between adjacent eighths, you get 2 7 = 128 different possibilities. (e) With the considerations from the last part of the exercise, one finds 2 15 possibilities for the maximum denominator 16, 2 31 possibilities for the maximum denominator 32 and so on. Solution 3. (a) We consider the prime factorization 200 = Since prime numbers have no factors other than 1 and themselves, any divisor of 200 not equal to 1 can only be the product of prime factors of 200. Furthermore, no divisor can contain a prime factor more often than 200 itself. This means that the following list is a complete list of all divisors of 200: 1, 2, = 2 2, 5, 8 = 2 3, 10 = 2 5, 20 = 2 2 5, 25 = = 2 3 5, 50 = 2 5 2, 100 = and 200 = (b) We generalize the argument from part a. To do this, it is useful to remember that for every number a 0, a 0 = 1.

8 Hamburger Tag der Mathematik, November 9, 2013 Solutions 7, 8 8 Just as we argued in part a, every divisor of p 3 q 2 r is a product of the form piqjrk, where i is an integer between 0 and 3, j is an integer between 0 and 2, and k must be either 0 or 1. For i = j = k = 0 we get z. B. the divisor 1 and for i = 3, j = 2 and k = 1 the number p 3 q 2 r itself. So there are possible values for i, 3 possible values for j and 2 possible values for k. Each of these combinations gives a different factor. For this we use the uniqueness of the prime factorization given in the note. In total there are exactly 3 2 = 2 different divisors of p 3 q 2 r. (c) The reasoning from the previous part of the exercise can be generalized immediately: You can always determine the number of divisors of a number from its prime factorization by adding 1 to the frequency (to the exponent) of each prime factor and multiplying all these numbers together. We know that a number x has 6 factors if and only if it has only one prime factor that occurs exactly five times, i.e. H. x = p 5 for a prime number p, or if x has exactly two prime factors, one of which occurs once and the other occurs twice, i.e. H. x = pq 2. There cannot be other possibilities, since 6 = 2 3 only has the factors 1, 2, 3 and 6 and can only be written as a product in the form 1 6 and 2 3 (except for the order of the factors ). The smallest numbers x of the type x = p 5 for a prime number p are 2 5 = 32, 3 5 = 23 and 5 5 = we get the smallest numbers x of the type x = pq 2 for q = 2 or p = = 12, = 20, = 28 and = = 18, = 50 and = 98. The smallest number x = pq 2 for which neither p nor q is 2 is 5 = and so far we have already found five smaller numbers with exactly six factors (in fact, 5 is just the seventh smallest number with exactly six factors). So the five smallest numbers with six factors are the 12, the 18, the 20, the 28 and the 32.

9 Hamburg Mathematics Day, November 9, 2013 Solutions 7, 8 9 (d) If a is a divisor of x, then x / a is also a divisor of x. If x is not a square number, then we can divide the divisors into pairs (a, x / a), and so the number of divisors of x is then even. On the other hand, if x = y 2 is a square number, then all divisors except for y = x / y can be divided into pairs and therefore the number of divisors is then odd. Note: You can alternatively argue with the prime factorization: Square numbers are exactly the numbers in which all prime factors occur in an even number, which is why there are precisely many possibilities for the divisors of the square numbers for each prime factor as to how often it is used. Exactly products of only odd factors are odd, which is why exactly the square numbers in total have an odd number of divisors. Solution. Preliminary remark: It is presumably known that the sum of the interior angles in the hexagon = 720: To see this, split the hexagon into 6 triangles by adding the line to the center M from each corner point. The sum of the interior angles of the triangles together is 6 180, all of which except the full angle around the center point belong to interior angles of the hexagon. Therefore 360 still have to be deducted. In a regular hexagon, every interior angle is therefore = 120. At the center M, each angle between two adjacent corner points is equal, i.e. = 60. Since half of each interior angle = 60, the regular hexagon can be divided into six regular triangles. (a) AEC = 60, since AEC is an equilateral triangle for reasons of symmetry. (b) ADC = 60 since it is half of an interior angle of the regular hexagon. (c) AF D = 90, since F lies on the Thales circle above the diameter AD. DAF = 60 since it is half of an interior angle of the regular hexagon. F DA = 30 because it is half of an interior angle of the regular triangle BDF. Note: Of course you only have to know two of these angles and you can calculate the third using the inside angle sum in the triangle ADF, in particular you don't have to know the Thales theorem if you know the statement from the preliminary remark. (d) CBA = 120 is an interior angle of the regular hexagon. With the sum of the interior angles in the triangle ABC we get BAC + ACB = 180 CBA = 60. For reasons of symmetry, however, the angles must be the same, so BAC = ACB = 30.

10 Hamburger Tag der Mathematik, November 9, 2013 Solutions 7, 8 10 (e) Since rotations and reflections of already examined triangles made of hexagonal points do not provide new angles, one only has to examine one of each type of triangle: three neighboring hexagonal points form, for example, ABC and two adjacent and one opposing ADF. Without neighboring points, you can only take every second hexagon point, for example ACE. These are all types of triangles and all interior angles of each were determined in the previous parts of the exercise. Points of the regular hexagon can only form angles of 30, 60, 90 and 120. (f) M 1 S and M 1 M 2 are radii of the circle around M 1, so M 1 S = M 1 M 2. M 1 M 2 = M 2 S also applies, since these are radii of the circle around M 2 . M 1 M 2 S is therefore a regular triangle and M 1 SM 2 = 60. Note: If you draw the hexagon from the previous parts of the exercise in the circle around M 1, you can therefore choose, for example, M 2 = C and S = D and then M 1 SM 2 = ADC. (g) Each of the tangents is perpendicular to one of the radii M 1 S or M 2 S. To calculate the angle between the tangents, you can add the two right angles and subtract M 1 SM 2, which you then count twice: M 1 SM 2 = = 120

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