Solution to Maths Challenge #30. For Years 6 to 9. For Years 10 to 13. In a 100m race, A beats B by 1m and C by 2m. By how much does B beat C?

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1 Solution to Maths Challenge #30 For Years 6 to 9 In a 100m race, A beats B by 1m and C by 2m. By how much does B beat C? When A crosses the line, B has covered 99m and C 98m. B runs 99m in the time it takes C to run 98m. For every 1m B runs, C runs 98/99m As B runs his last 1m, C only runs 98/99m, so C still has 1 1 / 99 m to run B beats C by 100/99m = 1.01m approx Correct answer from Emma (7D) For Years 10 to 13 Prove that the number is composite (i.e. not prime) can be written as or And this, in turn, can be written as ( ) 2 which is a square (and therefore composite) number.

2 Solution to Maths Challenge #31 For Y6 to Y7 The following number is formed by a special pattern and is the only one of its kind: What is the pattern? The digits from 0 to 9 have been written in alphabetic order (when written in English)! Correct Answers from: Dinky, Safaniya, Caliste and Joanne (6M) For Y10 to Y12 What is the greatest whole number that MUST be a factor of the sum of any four consecutive positive odd numbers? The four consecutive odd numbers are 2n 1, 2n + 1, 2n + 3, 2n + 5 The sum is (2n 1) + (2n + 1) + (2n + 3) + (2n + 5) = 8n + 8 = 8(n + 1) which is divisible by 8 Correct Answer from Anne (13D, Zanzibar)

3 Solution to Maths Challenge #32 Bananas Today, Bananas Tomorrow! A monkey has 75 bananas. Each day, he kept a fraction of his bananas, gave the rest away and ate one. These are the fractions he decided to keep: 1 2, 1 4, 3 4, 3 5, 5 6, In what order did he keep these fractions, given that he ended with one bananas? There s a certain amount of trial and error, but it soon becomes obvious when you make a mistake! Here is the correct order for the fractions:!!!",!!,!!,!!,!!,!! Check that this means he ended up with one banana!

4 Solution to Maths Challenge #33 The combination is: Correct answers from: Safaniya, Dinky, Zubeiba (all Y6) and Emma (7D, Zanzibar) who extended the problem to 8 digits (with 4 digits between the two 4s)

5 Solution to Maths Challenge #34 All Grades Find the difference between the sum of the first 1,000,000 positive even numbers and the sum of the first 1,000,000 positive odd numbers. There is a very easy pattern than can help us here, and here is a brief snapshot of that pattern: Sum of first 5 even numbers = = 30 Difference = 5 Sum of first 5 odd numbers = = 25 Sum of first 10 even numbers = = 110 Difference = 10 Sum of first 10 odd numbers = = 100 This leads to the (correct) assumption that the difference between the first n even numbers and the first n odd numbers is n. So the difference between the first even numbers and the first odd numbers is Correct Answers from: Dinky and Niharika (both 6M, Simba) as well as Emma (7D, Zanzibar)

6 Solution to Maths Challenge #35 Grades 6 to 9 A number is defined as upright if the sum of the first two digits equals the third digit. So, for example, 145 is upright (1 + 4 = 5). How many three digit upright numbers are there? If the first digit is 1, there are 9 possibilities (101, 112, 123, 134, 145, 156, 167, 178 and 189) If the first digit is 2, there are 8 possibilities (202, 213, 224, 235, 246, 257, 268, and 279) If the first digit is 3, there are 7 possibilities (3023, 314, 325, 336, 347, 358, and 369) This pattern continues all the way to where the first digit is 9 1 possibility (909) So there are or 45 upright three- digit numbers Correct Solution from Emma (7D, Zanzibar) and also from: Dinky, Naharika, Nivedita (6M, Simba) and Sonaa, Caliste (6M, Twiga) Grades 10 to 13 How many factors does 60 5 have? 60 5 can be written as In other words, 60 5 can be divided by 2 0 through 2 10, by 3 0 through 3 5 and by 5 0 through 5 5 This gives us 11 x 6 x 6 or 396 factors.

7 Solutions to Maths Challenge #23 Grades 6 to 8 Every time the angle between the hands of a clock makes 90 0 you are paid $8. How much money do you make in a day? Right angles occur twice each hour, except that the second right angle between 2 o'clock and 3 o'clock occurs exactly at 3 o'clock, which is also the first occurrence between 3 o'clock and 4 o'clock. The same thing happens at 9 o'clock. Therefore, only 22 right angles occur in a 12- hour period, or 44 in a 24- hour period. Thus, you will earn $352. Correct Answers from: Dinky, Safaniya and Shweta (all 6M) Grades 9 to 12 Show that is a composite (i.e. not prime) number. Let N represent Then N = (2 10 )(3 10 ) = (2 9 ) 2 + 2(2 9 )(3 10 )+ (3 10 ) 2 = ( ) 2 i.e. N is a square number, therefore it cannot be prime.

8 Solutions to Maths Challenge #37 Grades 6 to 9!!" =!! +!! where A and B are positive whole numbers. Find A and B.!!" =!! +!!! =!!!!"!" 3AB = 23(A + B) which means AB is divisible by 23 and A+B is divisible by 3. After a little trial and error, we get: 3 23 = Grades 10 to 13 A gambler has two coins in his pocket one fair coin and one two- headed coin. He selects a coin at random and flips it twice. If he gets two heads, what is the probability that he selected the fair coin? Sample Space: HH HH HT TH TT (where HH is the two- headed coin) Of the two times when he gets HH, one came from the fair coin Probability he d selected the fair coin is 1/2

9 Solution To Maths Challenge #38 For Years 6 to 9 The price of a dress is such that the profit is 20% of the price. Increasing the price by $20 increases the profit to one third of the price of the dress. What was the original price of the dress? Let the Cost Price of the dress be Y and the original selling price be X. We are given: X Y = X/5 and X + 20 Y = (X + 20)/3 These are solved simultaneously to give X = $100 For Years 10 to 13 Professor Al Jebra only has white or black socks. In his sock draw are a number of socks. He has (correctly) calculated that the probability of his choosing two socks at random and forming a black pair with them is 50%. What is the probability he can form a white pair? The professor can only have 4 socks three black and one white. Let s label the socks: B 1 B 2 B 3 W There are six possible choices of two socks: B 1 B 2 B 1 B 3 B 2 B 3 B 1 W B 2 W B 3 W Only three of these choices (in bold) form a pair. Hence the probability of forming a black pair is 3/6 or 50% None of the choices allow a white pair to be formed. So the probability of forming a white pair is 0%

10 Solution To Maths Challenge #39 For Years 6 to 9 One invention saves 30% on fuel, another saves 45% and a third saves 25%. If all three inventions are used together, what percentage of fuel is saved? Look at the first invention. If it saves us 30% on fuel, it means we now only use 70% of what we used to use. It s a similar picture with the other inventions the one that saves us 45% really means we use 55% of the old amount, and an invention that saves us 25% means we use 75% of the old amount of fuel. With these three inventions together, we now use 70% of 55% of 75% of the old amount of fuel. This is 0.7 x 0.55 x 0.75 or or % This means we save % of our fuel.

11 For Years 10 to 13 Two people arrive at a restaurant independently. Their arrival times are random and uniformly distributed between 5:00 p.m. and 6:00 p.m. What is the probability that the two people arrive within 10 minutes of each other? This probability question is solved by graphing an algebraic system of equations that models the problem. Let x denote one person's arrival time (minutes after 5:00pm) and y the other person's arrival time. We observe that 0 x 60 and 0 y 60. The condition that the arrival times differ by no more than 10 minutes may be represented by the inequality x y 10. The graph of this system provided on the 60x60 square (left) shows the solution region represented by a diagonal stripe across the square. The area of the entire square is The area of the portions outside the stripe is Thus, the area of the stripe is Because of the uniformity of the arrival times, the probability of arriving within 10 minutes is the area of the stripe divided by the area of the square = 11/36.

12 Solutions to Maths Challenge #40 Which is the better fit a square peg in a round hole, or a round peg in a square hole? A round peg in a square hole: A square peg in a round hole: The percentage of space wasted is (2a) 2 πa 2 x 100% (2a) 2 = 4 π x 100% 4 = 21.46% approximately So a round peg in a square hole is a better fit. The radius of the circle must be 2a The percentage of space wasted is 2πa 2 - (2a) 2 x 100% 2πa 2 = 2π 4 x 100% 2π = 36.34% approximately Great effort from: Dinky, Marya and Niharika (all 6M, all Simba)

13 Maths Challenge #41 For All Years The Problem of the Year (2014) Using the digits 2, 0, 1 and 4 exactly once, and any mathematical symbols you know, generate as many of the numbers from 1 to 40 as you can (you won t get them all)! Extra credit is earned if you use the digits in the order given. For example: 7 = gets you 1 point, but 7 = gets you 2 points Target Arithmetic Score Target Arithmetic Score 1 2 x x ( x 1) + 4! ( ) + 4! x x (- 2 x 0 x 1) + 4! x ( ) + 4! x x 1 + 4! ! 2 8 (2 + 0) x 1 x ! ! 2 9 (2 +0!)! (2 + 0!)! 1 + 4! 2 10 (2 +0!)! + 1 x (2 + 0!)! x (1 + 4) 2 11 (2 +0!)! (2 + 0!)! ! (- 2) x ! + 4! ! ! x x x x (2 1) 1

14 Solutions to Maths Challenge #42 More Problems of the Year When 2014 x 2013 x 2012 x 2011 x x 3 x 2 x 1 is evaluated, how many zeroes are at the end? 2. What is the last digit of ? 3. What is the number written as a fraction in its simplest form? 4. When all the natural numbers which start with a 2 are written down in increasing order, which is 2014 th on the list? 1. The number of zeroes is The last digit of is a 6. Each decade that contains numbers ending in 4 and 5 produces a zero, each multiple of 10 produces a zero, each multiple of 100 produces 2 and each multiple of 1000 produces 3. There is a simple pattern in the last digit of 2014 n, which is the same as the pattern in the last digit of 4 n. The digits progress: 4, 6, 4, 6 etc = 2014 = Number(s) How many of these there are Thus, by the time is written down, this is the 1111 th number beginning with 2. The 2014 th number is thus the 903 rd number in the range which is

15 Solutions to Maths Challenge #43 Find three consecutive whole numbers which multiply together to give Let the numbers be (n 1), n and (n + 1) (n 1)n(n + 1) = These three numbers (n, n- 1 and n- 2) are pretty close together, almost giving us the equation n 3 = so the answer for n will be pretty close to the cubed root of (which is almost 74). Also, because ends in a zero, one of the numbers could end in a 4, another in a 5. This leads us to try 73 x 74 x 75 which is indeed

16 Solutions to Maths Challenge #44 For Years 6 to 9 5cm The diagram shows two squares placed side by side. The small square is 2cm on a side, and the large one is 5cm on a side. 2cm What is the value, in cm 2, of the shaded area? Let the height of the shaded region be x. Using similar triangles 5/x = 7/2 x = 10/7 Area of shaded triangle = ½ x 2 x 10/7 = 10/7 cm 2 For Years 10 to 13 Given x, y and z are all positive integers, find all possible solutions for (x, y, z) in the equation xy + 17xz = 53 The equation factorises to x(y + 17z) = 53. Given that x, y, z are positive integers, and given the fact that 53 is prime: It must be that EITHER x = 1 and (y + 17z) = 53 OR x = 53 and (y + 17z) = 1 (which we can ignore) Thus x = 1 and y + 17z = 53 which, in turn, gives y = 2 and z = 3 or y = 19, z = 2, or y = 36, z = 1 Thus the possible solution sets for (x, y, z) are: (1, 2, 3), (1, 19, 2) and (1, 36, 1)

17 Solutions to Maths Challenge #45 For All Years If Germany = 73 Tonga = 47 Fiji = 24 Azerbaijan = 77 Break the code and find out what number JAPAN is! If we assign the simple cipher: A=1, B=2, C=3, D=4.. Y=25, Z=26 then add the numbers in the countries names we get: Germany = which is 83 Tonga = which is 57 Fiji = which is 34 And these numbers are all 10 less than the numbers in the code. Azerbaijan = which is 87 So, Japan must be ( ) 10 which is 32 Correct Answer from Joy (Y10, Pemba)

18 Solutions to Maths Challenge #46 For All Years 1 st envelope has 1p, 2 nd has 2p, 3 rd has 4p and 4 th has 8p Correct answers from Ramona (Y8, Pemba) and Jennifer (Y7, Pemba) and Mary- Jo (Y9, Zanzibar)

19 Solution to Maths Challenge #47 For All Years There were 4 football matches taking place on Sunday afternoon. John thought the winners would be Liverpool, Spurs, Chelsea and Everton. Sue thought that Everton, Man City, Chelsea and Southampton would be the winning teams. Nick said Man Utd, Man City, Spurs and Everton would all win their matches and that Arsenal would score no goals. Who played who? There are 8 teams altogether. Everton can t possibly play Liverpool, Spurs, Chelsea, Man City, Southampton or Man Utd. This means they must play Arsenal. Chelsea can t (obviously) play Everton or Arsenal. Nor can they play Spurs, Liverpool, Southampton or Man City. This means they play Man Utd. Man City can t be playing Spurs, Chelsea, Everton, Arsenal, Southampton or Man Utd. This means they play Liverpool. And that leaves Spurs playing Southampton. Correct Answers from: Jennifer (Y7, Pemba) and Ramona (Y8, Pemba)

20 Solutions to Maths Challenge # All the positive integers are written in the cells of a square grid, as shown. Starting from 1, the numbers spiral anticlockwise. The first part of the spiral is shown in the diagram What number is immediately below 2014? The key here is to notice that, starting from 1, the leading diagonal (going down) contains the odd squares. Similarly, starting from 4, the leading diagonal going up contains the even squares. These are highlighted in the diagram to the left. Now 44 2 = 1936 and 45 2 = 2025 This gives us a clue as to where 2012 will be in the expanded diagram

21 Thus the number beneath 2014 is

22 Solution to Maths Challenge #49 For Years 6 to 9 Find the fraction between 5 9 and 4 7 that has the smallest denominator.!! can be written as and!! can be written as ½ - way between these fractions is!".!. more commonly written as 71!" (Any other fraction in the gap would need to use a bigger denominator) For Years 10 to 13 Q and R are positive whole numbers and Q 2 R 2 = 116. What is Q? Q 2 R 2 = (Q R)(Q + R) = 116 and we know that Q and R are positive integers Either (Q R)(Q + R) = (1 x 116) or (2 x 58) or (4 x 29) The only possibility that works is the middle one, giving Q = 30 and R = 28

23 Solution to Maths Challenge #50 For All Years Patterns are really important in Maths, but sometimes they are difficult to see! Find the next number in the sequence: 0, 1, 2, 4, 5, 6, 8, 10, 40, 46, 60, 61, 64, 80, 84,? These numbers are special. Write them out ZERO, ONE, TWO, FOUR, FIVE, SIX etc What about the missing numbers THREE, SEVEN, NINE etc? You soon see that the numbers given never use the same letter twice! And the next number you can say that about is the number 5000 (FIVE THOUSAND)

24 Solution to Maths Challenge #51 For All Years Can you find the integer x such that: 2x is a square and 3x is a cube? This problem can be tackled using trial and error methods. Technology is a great help, though especially a spreadsheet such as the one below: As you can see, the first integer this works for is 72 i.e. 2x72 = 144 and 144 is a square number (12 2 ), and 3x72 = 216 and 216 is a cubic number (6 3 )

25 Solution to Maths Challenge #52 For Years 6 to 8 When all the numbers from 1 to 1000 are written out, which digit (if any) is used least often? The answer is the 0. By way of explanation, look for example at the 90 numbers from 10 to 99. Although there is an equal distribution of digits in the 2 nd position, no zeroes ever occupy the first position. A similar thing happens with the 900 numbers between 100 and 999. For Years 9 to 13 Find the area of this quadrilateral 7cm 3cm 9cm 7cm y x 3cm Draw the extra line, so splitting the shape into 2 right- angled triangles. Then use Pythagoras Theorem to find the missing sides x and y 9cm y 2 = y = 130 x! = 130 3! x = 11 Area = (½ x 7 x 9) + (½ x 3 x 11)cm 2 = 48cm 2

26 Solution to Maths Challenge #53 For All Years Push- ups! Ahmed exercised each day this week. On Monday he did 8 push- ups. Each day his average number of push- ups increased by 1. How many push- ups did he do on Friday? On Monday, his average is obviously 8. For his average to rise to 9 on Tuesday, he would have to do 10 push- ups that day (check: (8 + 10) 2 = 9) For his average to rise to 10 on Wednesday, he would have to do 12 push- ups that day (check: ( ) 3 = 10) For his average to rise to 11 on Thursday, he would have to do 14 push- ups that day (check: ( ) 4 = 11) For his average to rise to 12 on Friday, he would have to do 16 push- ups that day (check: ( ) 5 = 12) Correct Solution from: Jennifer, 7D, Pemba

27 Solution to Maths Challenge #54 For All Years The diagram shows an equilateral triangle with its corners at the midpoints of the sides of a regular hexagon. What fraction of the hexagon is shaded? The trick is to draw some extra lines like in the diagram below: It is clear from this diagram that 9 of the 24 small equilateral triangles are shaded. Thus 9 / 24 is shaded.

28 Solution to Maths Challenge #55 For All Years The above diagram shows part of a pattern of regular hexagons. Each side of the hexagon is 1cm. Every time a new hexagon is added it shares one side with the previous added hexagon. How many hexagons are needed for the perimeter of the final shape to be 2014cm? When the pattern is complete, the first and last hexagons each contribute 5cm to the perimeter, while all the ones in- between contribute 4cm If we want the total perimeter to be 2014cm, this means that 10cm comes from the end hexagons and 2004cm comes from the middle ones. So there has to be 2004/4 or 501 middle hexagons When we add in the end hexagons, this gives us 503 hexagons in the pattern. Correct Solutions from: Judah and Kashyep, Y10, Mafia

29 Solution to Maths Challenge #56 For Years 6 & 7 4 black cows and 3 brown cows give as much milk in five days as 3 black cows and 5 brown cows give in four days. Which cow gives more milk the black or the brown? Assume the black cow gives x litres/day, and the brown gives y litres/day. Thus, 5(4x + 3y) = 4(3x + 5y) 20x + 15y = 12x + 20y 8x = 5y y > x i.e The brown cow gives more milk For Years 8-10 Four men, one of whom was known to have committed a certain crime, made the following statements when questioned by the police: Archie: Dave did it. Gus: I didn't do it. Dave: Tony did it. Tony: Dave lied when he said I did it. If only one of these four statements is true, who was the guilty man? If the culprit was Then these men s statements would be Archie s Dave s Gus s Tony s Archie False False True True Dave True False True True Gus False False False True Tony False True True False The only combination in which three statements are false and one true happened when Gus committed the crime.

30 Solutions to Maths Challenge #57 All Years A large square is cut into 37 smaller squares. Of these small squares, 36 of them each have an area of 1cm 2. What is the length of the original square? Possibility 1 Possibility 2 In other words, the original square was of side 10cm

31 Solutions to Maths Challenge #58 Years 6-7 Broken Clock A clock broke into two pieces. The numbers on each of the pieces add up to the same total. Draw a diagram to show how the clock cracked. If the clock breaks as shown, one side will have 10, 11, 12, 1, 2 and 3 (sum = 39) and the other side has 4, 5, 6, 7, 8 and 9 (sum = 39 also). Correct answers from Safaniya, Mandeep, Dinky and Reshma (all Y6) For Years 8 to 10 In the calculation , which two adjacent digits should be swapped in order to increase the result by 100? If after division by 18 the number has to be increased by 100, then before the division it needs to be increased by = So we need to find two adjacent digits in the number which when swapped increase it by If is increased by 1800 it becomes = We obtain from by swapping the adjacent digits 2 and 4.

32 Solutions to Maths Challenge #59 Years 6-7 On a (24- hour) digital clock displaying hours, minutes and seconds, how many times in a day to all the digits in the display change? This only happens three times in a day at 09:59:59, 19:59:59 and 23:59:59 For Years 8 to 10 If p, q are distinct primes (both) less than 7, what is the largest possible value of the highest common factor of 2p 2 q and 3pq 2? Given the initial conditions, the possibilities are: (i) p=2, q = 3 (ii) p=3, q=2 (iii) p=2, q=5 (iv) p=5, q =2 (v) p=3, q = 5 and (vi) p=5, q=3 p=2, q = 3 p=3, q=2 p=2, q=5 p=5, q =2 p=3, q = 5 p=5, q=3 2p 2 q pq HCF of 2p 2 q, 3pq The table tells us that the HCF is 45.