The University of Melbourne BHP Billiton School Mathematics Competition 2007 INTERMEDIATE DIVISION SOLUTIONS

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Bruno Logan
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1 The University of Melbourne BHP Billiton School Mathematics Competition 2007 INTERMEDIATE DIVISION SOLUTIONS 1. A rectangle with integer side lengths has its perimeter equal to its area. Find all possible side lengths of such a rectangle. Solution 1: Let the dimensions of the rectangle be a and b and assume without loss of generality that a b. Then the perimeter of the rectangle is 2a + 2b while the area of the rectangle is ab. Therefore, the following equation must be true. ab = 2a + 2b ab 2b = 2a b(a 2) = 2a b = 2a a 2 Since a b, we obtain the inequality a a 2 which implies that a 4. Substituting a = 1, 2, 3, 4 only yields two integer solutions for b namely (a, b) = (3, 6) and (a, b) = (4, 4). Solution 2: With the same setup as the previous solution, we have the following equation. ab = 2a + 2b ab 2a 2b = 0 ab 2a 2b + 4 = 4 (a 2)(b 2) = 4 Since a and b are positive integers, a 2 and b 2 must be integers which are at least 1 and multiply to give 4. The only possible solutions with a b are (a 2, b 2) = (1, 4) and (a 2, b 2) = (2, 2). Therefore, the only possible solutions are (a, b) = (3, 6) and (a, b) = (4, 4). 2. Tom has change in his pocket made up of 5,, 20, 50 cents and $1 coins. (He may have more than one, or none, of a type of coin.) Tom needs exactly $2 but finds he cannot exactly make up $2 with his change. What is the maximum amount in change Tom can have? Solution: We may assume that Tom has exactly one 5c coin. If he had more, then he could exchange them for c coins and still not be able to make up $2 exactly. And if he had no 5c coins, then giving him a 5c coin would not help him to make up $2 exactly, since it is clearly impossible to make up $1.95 without 5c coins. Using a similar idea, we may assume that Tom has at most one c coin. If he had more, then he could exchange them for 20c coins and still not be able to make up $2 exactly. By the same reasoning, we may assume that Tom has no more than four 20c coins and 1 50c coin. Also, it is clear that Tom has at most one $1 coin. Therefore, the maximum amount of change that Tom has in his pocket is 1 5c + 1 c c c + 1 $1 = $2.45 2a 1

2 However, if Tom had exactly $2.45, then he could make up $2 exactly with a $1 coin, a 50c coin, two 20c coins and a c coin. Therefore, Tom can t have exactly $2.45. We have shown that Tom definitely has a 5c coin, so the next highest amount that Tom could possibly have is $2.35. It is clear now that this is the maximum possible, since with one 5c coin, four 20c coins, 1 50c coin and one $1 coin, it is impossible to exactly make up $2. 3. Agatha sold some books and counted the money she received for them. The two most expensive books equalled 1/5 of the money she received, and the six cheapest books equalled 1/2 of the money she received. How many books did she sell? Solution: Since = 7, Agatha must have sold more than the eight books mentioned in the problem. Suppose that Agatha sold k other books. Each of these books costs at least as much as each of the six cheapest books and at most as much as each of the two most expensive books. So each of these books must cost at least as much as the average of the six cheapest books. In other words, each of these k books costs at least = 1 12 of the money she received. Similarly, each of these books costs at most as much as the average of the two most expensive books. In other words, each of the k books costs at most = 1 of the money she received. However, note that all together, these k books must cost 3 of the money Agatha received. If k 4, then the k books would be worth at least = 1 3 of the money she received, which is more than the 3 required. Similarly, if k 2, then the k books would be worth at most 2 1 = 1 5 of the money she received, which is less than the 3 required. So it follows that k = 3 and Agatha sold = 11 books overall. 4. Katrin is in a rectangular pool in her backyard. She is 1m, 7m and 8m from three of the corners of the pool, as in the diagram. How far is she from the fourth corner? Solution: Let x be Katrin s distance in metres from the fourth corner of the rectangular pool. Let a, b, c, d be the lengths labelled in the diagram below. a b c x 8 d 1 7 2

3 Then by Pythagoras Theorem, we have the following four equations. a 2 + c 2 = x 2 b 2 + d 2 = 7 2 a 2 + d 2 = 1 2 b 2 + c 2 = 8 2 Adding the first two equations gives a 2 + b 2 + c 2 + d 2 = x while adding the last two equations gives a 2 + b 2 + c 2 + d 2 = Therefore, x = from which it follows that x 2 = 16 and x = Alex decides to buy an igadget which costs between $1 and $0. He notices that the price, in cents, is a palindrome it reads the same forwards as backwards, such as $12.21 or $8.38. He also notices that the price in cents is a prime number its only factors are 1 and itself. Alex is so happy with this that he decides to buy two igadgets, and even happier when the total price in cents is again a palindrome. What possible prices can the igadget be? Solution: Let X be the value of the igadget in cents and note that X must have three digits or four digits. If X is a four-digit palindrome of the form abba (that is, with the digits a, b, b and a), then we can write X = 00a + 0b + b + a = 01a + 1b = 11(91a + b). However, a four-digit number divisible by 11 cannot possibly be prime, so we deduce that X must be a three-digit number. In fact, we can deduce that 2X cannot be a four-digit number. Otherwise, it would be divisible by 11 and hence, not double a three-digit prime. Therefore, X must be a three digit number less than 500. Let X be of the form aba so that we can write X = 0a + b + a. We have already determined that a = 1, 2, 3 or 4. Clearly, we can rule out a = 2 and a = 4, because they would result in an X which is even. Therefore, we only have twenty numbers to consider. 1, 111, 121, 131, 141, 151, 161, 171, 181, , 313, 323, 333, 343, 353, 363, 373, 383, 393 Of these, we can eliminate those which do not yield palindromes when doubled to create the following list. 1, 111, 121, 131, 141, 303, 313, 323, 333, 343 3

4 And of these, we can eliminate those which are composite to leave the following possibilities. 1, 131, 313 In conclusion, we find that the price of the igadget can only be $1.01, $1.31 or $ At a meeting consisting of 2007 people everyone shook hands with at least one other person. After the meeting, people who shook hands with no more than people were given a red shirt with the words I am unfriendly printed on it. People who shook hands only with those who were given red shirts were given blue shirts with the words All my friends are unfriendly. Some lucky people got both a red and blue shirt. Prove that the number of blue shirts is no more than the number of red shirts. Solution: Let us call a person blue if they received a blue shirt only, red if they received a red shirt only and purple if they received a blue shirt and a red shirt. Let B be the number of blue people, R the number of red people and P the number of purple people. Furthermore, let X be the total number of handshakes which occurred between someone with a blue shirt and someone with a red shirt. Let Y be the total number of handshakes which occurred between two purple people. It follows from these definitions that Y X. Note that each blue person shook hands with at least 11 people otherwise, they would have received a red shirt as well. Furthermore, all of those 11 people must have received a red shirt. It follows that 11B + Y X. Now note that each red person shook hands with at most people, and hence, at most people with a blue shirt. It follows that R + Y X. Putting these two inequalities together, we obtain 11B + Y X R + Y from which it follows that 11B R. Since B and R are non-negative, this gives us the fact that B R and hence, that B + P R + P. But this tells us precisely that the number of blue shirts is no more than the number of red shirts. 7. Let (i, j, k) be an ordered triple of positive integers such that i + j + k = 23. For example (1, 1, 21), (21, 1, 1) and (6, 9, 8). For each triple we can compute the product P of i, j and k. In our three examples this gives P = 21, 21 and 432 respectively. Let S be the sum of the P for all allowed triples (i, j, k) Show that 23 divides S. S = Solution: We will show that S is equal to the binomial coefficient ( 25 5 ). Imagine that we have 25 white boxes lined up in a row. Suppose that we colour 2 of them blue so that the remaining boxes divide into three groups: left boxes, middle boxes and right boxes. In particular, notice that to do this, we cannot colour an 4

5 end box blue, nor can we colour two consecutive boxes blue. If there are L left boxes, M middle boxes and R right boxes, then it is clear that L + M + R = 23. Let us choose three boxes to colour red one from the left boxes, one from the middle boxes and one from the right boxes. Clearly, the number of ways to do this is L M R. This represents one term in the sum S so it follows that S is simply the number of ways to choose two blue boxes from the 25, without choosing an end box or two consecutive boxes, followed by choosing three red boxes, one from each group created by the blue boxes. This somewhat complicated counting problem is actually not as hard as it looks! It is simply equal to the number of ways of choosing five boxes from the 25 available and colouring them alternately red, blue, red, blue, red. Therefore, S is equal to the number of ways of choosing 5 objects from 25 objects, a number which we denote by ( 25) 5. It remains to show that this number is divisible by 23, which we can verify as follows. ( ) = =

BRITISH COLUMBIA SECONDARY SCHOOL MATHEMATICS CONTEST, 2017 Junior Preliminary Problems & Solutions

BRITISH COLUMBIA SECONDARY SCHOOL MATHEMATICS CONTEST, 017 Junior Preliminary Problems & s 1. If x is a number larger than, which of the following expressions is the smallest? (A) /(x 1) (B) /x (C) /(x