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BHP BILLITON — UNIVERSITY OF MELBOURNE SCHOOL MATHEMATICS COMPETITION, 2003: SENIOR DIVISION 1. Two numbers are called mirror numbers if one is obtained from the other by reversing the order of the digits. For example, 123 and 321 are mirror numbers. Find two mirror numbers whose product is 92565. Solution: The two mirror numbers whose product is 92565 must both have the same number of digits. If they both have 2 or fewer digits, then the product would be at most 99 × 99 = 9801, which is too small. If they both have 4 or more digits, then the product would be at least 1000×1000 = 1, 000, 000, which is too large. Thus, the mirror numbers each have 3 digits. So let the two mirror numbers be abc and cba where each of the letters a, b, c represents a digit. Notice that abc − cba = (100a + 10b + c) − (100c + 10b + a) = 99(a − c). So the difference between the two mirror numbers is a multiple of 99. Thus, the two mirror numbers must leave the same remainders when divided by 99. In particular, they must leave the same remainders when divided by 3 or when divided by 11. However, we require the product of the two mirror numbers to be abc × cba = 92565 = 32 × 5 × 112 × 17. Since their product is divisible by 3, at least one of the mirror numbers must be divisible by 3. But since they both leave the same remainders when divided by 3, we conclude that they must both be divisible by 3. Similarly, since their product is divisible by 11, at least one of the mirror numbers must be divisible by 11. But since they both leave the same remainders when divided by 11, we conclude that they must both 1 be divisible by 11. So we have come to the conclusion that the mirror numbers are of the form abc = 3 × 11 × x and cba = 3 × 11 × y for some positive integers x, y > 3. However, since we require at least one of the mirror numbers to be divisible by 5 and at least one of the mirror numbers to be divisible by 17, we may take abc = 3 × 11 × 5 = 165 and cba = 3 × 11 × 17 = 561. As we can see, the numbers 165 and 561 are indeed mirror numbers and their product is 92565. 2. Some positive integers may be written as sums involving all the digits 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 exactly once. For example 180 = 0 + 16 + 32 + 45 + 78 + 9, 54 = 10 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9. Is it possible to write 100 in this manner? Solution: Recall that the remainder after a number is divided by 9 is equal to the remainder after its sum of digits is divided by 9. So consider a sum which uses all possible digits from 0 to 9 exactly once. Since 0 + 1 + 2 + · · · + 9 = 45 is divisible by 9, then such a sum must also be divisible by 9. But 100 is not divisible by 9, so it is not possible to write it in this manner. 3. Is it possible to draw a ray from the origin such that it avoids all points in the Cartesian plane with integer coordinates except the origin? (Your answer must include valid mathematical reasoning in order to gain credit.) Solution: Consider the ray whose equation is described by y = cx, for x ≥ 0 where c is a positive irrational number. Suppose that this ray passes through the point (m, n) where m and n are non-zero integers. Then we n would have n = cm ⇒ c = m , which contradicts the fact that we have chosen c to be irrational. Therefore, we conclude that it is possible to draw a ray from the origin such that it avoids all points in the Cartesian plane with integer coordinates except the origin. Examples of positive √ irrational numbers are 2 and π. 2 4. Consider an equilateral triangle, in which the two lines AE and BD intersecting at the point F , divide the triangle into four pieces. Given that the triangle AF B and that of the quadrilateral CEF D are equal, find the angle EF D. Solution: Since AB = CA and 6 BAD = 6 ACE = 60◦ , we have the following area(AF B) ⇒ area(AF B) + area(AF D) ⇒ area(ABD) 1 ⇒ AB.AD. sin 6 BAD 2 ⇒ AD = area(CEF D) = area(CEF D) + area(AF D) = area(CAE) 1 CA.CE. sin 6 ACE = 2 = CE Since 4ABD and 4CAE share two common side lengths AB = CA and AD = CE and the included angle 6 BAD = 6 ACE = 60◦ , they are congruent. In particular, we know that 6 CAE = 6 ABD. Thus 6 ABF + 6 BAF = 6 ABD + 6 BAE = 6 CAE + 6 BAE = 60◦ . Considering the sum of the angles in 4AF B yields 6 AF B = 120◦ . So we can determine the opposite angle 6 EF D = 120◦ , as required. 5. We arrange 2003 bins in a circle, and number them 1 to 2003 in clockwise order. Then proceeding clockwise from bin 1, we skip the first bin, place a ball into the second bin, and continue so that, after each placement of a ball, we skip the next bin that is still empty, and place a ball into the very next empty bin. After the placement of 2002 balls, which bin remains empty? Solution: We generalise by considering n bins arranged in a circle and performing the same procedure. Let the number of the last bin that remains empty be denoted by f (n). For small values of n, we have f (1) = 1 f (2) = 1 f (3) = 3 f (4) = 1 f (5) = 3 f (6) = 5 f (7) = 7 f (8) = 1 f (9) = 3 f (10) = 5 f (11) = 7 f (12) = 9 f (13) = 11 f (14) = 13 f (15) = 15 f (16) = 1 f (17) = 3 f (18) = 5 3 A striking pattern seems to be occurring with powers of 2 involved. This motivates us to consider the numbers in their binary representations. Doing so gives us the following f (12 ) = 12 f (1002 ) = 12 f (1112 ) = 1112 f (10102 ) = 1012 f (11012 ) = 10112 f (100002 ) = 12 f (102 ) = 12 f (1012 ) = 112 f (10002 ) = 12 f (10112 ) = 1112 f (11102 ) = 11012 f (100012 ) = 112 f (112 ) = 112 f (1102 ) = 1012 f (10012 ) = 112 f (11002 ) = 10012 f (11112 ) = 11112 f (100102 ) = 1012 Note that in binary, f (n) seems to be obtained from n by removing the leading 1, appending a 1 to the end of the number, and then ignoring any leading 0’s. Let g(n) be the number obtained from n in this manner. We will prove that f (n) = g(n) for all positive integers n by strong induction. From the table above, we know that f (n) = g(n) for all values of n from 1 up to 18. Let us now assume that the equality holds for all values of n less than some positive integer K. We will now show that f (K) = g(K) as well. Case 1: Suppose that K = 2m is even, so that its binary representation has the form 1A0 where A is an arbitrary string of 0’s and 1’s. Then after passing once around the circle, the only numbers left are 1, 3, 5, . . . , 2m − 1. Now the procedure is repeated for these m bins, so by the induction step, the bin in the f (m) = g(m) place is the last one remaining empty. This bin is the one labelled 2g(m) − 1. Thus, f (K) = f (2m) = f (1A0) = 2g(m) − 1 = 2g(1A) − 1 = 2 × A1 − 1 = A10 − 1 = A01 = g(1A0) = g(2m) = g(K). Case 2: Suppose that K = 2m + 1 is odd, so that its binary representation has the form 1A1 where A is an arbitrary string of 0’s and 1’s. Then after passing once around the circle, the only numbers left are 1, 3, 5, . . . , 2m + 1. The next bin to be filled is the one labelled 1 and after that, the only numbers left are 3, 5, 7, . . . , 2m + 1. Now the procedure is repeated for these m bins, so by the induction step, the bin in the f (m) = g(m) place is the last one remaining empty. This bin is the one labelled 2g(m) + 1. Thus, f (K) = f (2m + 1) = f (1A1) = 2g(m) + 1 = 2g(1A) + 1 = 2 × A1 + 1 = A10 + 1 = A11 = g(1A1) = g(2m + 1) = g(K). 4 In either case, it is true that f (K) = g(K). Therefore, by the principle of mathematical induction, f (n) = g(n) for all positive integers n. It is now a simple matter to calculate f (2003) = g(2003) = g(111110100112) = 111101001112 = 1959. 6. Show that for x, y > 0, (x + y)1/3 < x1/3 + y 1/3 . Find the smallest constant K satisfying x1/3 + y 1/3 ≤ K(x + y)1/3 , for all x, y > 0. Solution: For simplification, consider the substitutions a = x1/3 and b = y 1/3 . Thus, we wish to prove the following (a3 + b3 )1/3 ⇔ a3 + b3 ⇔ a3 + b3 ⇔0 a+b (a + b)3 a3 + 3a2 b + 3ab2 + b3 3a2 b + 3ab2 < < < < However, this last inequality is obviously true, since a and b are both positive. We now wish to find the smallest constant K satisfying a + b ≤ K(a3 + b3 )1/3 . When a = b, we have 2a ≤ K(2a3 )1/3 ⇒ K ≥ 41/3 . We will now try to prove that K = 41/3 is indeed the smallest constant K satisfying the inequality. Thus, we wish to prove a+b ⇔ (a + b)3 ⇔ a3 + 3a2 b + 3ab2 + b3 ⇔ 3a2 b + 3ab2 ⇔ a2 b + ab2 ⇔0 ⇔0 ⇔0 ⇔0 ⇔0 ≤ ≤ ≤ ≤ ≤ ≤ ≤ ≤ ≤ ≤ 41/3 (a3 + b3 )1/3 4(a3 + b3 ) 4a3 + 4b3 3a3 + 3b3 a3 + b3 a3 − a2 b + b3 − ab2 a2 (a − b) + b2 (b − a) (a2 − b2 )(a − b) (a + b)(a − b)2 (a − b)2 However, this last inequality is obviously true, since the square of any real number is non-negative. Thus, K = 41/3 and it can be seen that equality is preserved when x = y. 5 7. Euler proved that 232 + 1 641 is an integer. Prove that N can be expressed as the sum of two squares. N= Solution: Suppose that we can express N as a2 + b2 , where a and b are integers. Then, after noticing that 641 = 252 + 42 can also be expressed as the sum of two squares, the given equation for N becomes (a2 + b2 )(252 + 42 ) = (216 )2 + 12 . The left hand side is a product of two numbers, each of which is the sum of two squares while the right hand side is the sum of two squares. This suggests using the famous algebraic identity (a2 + b2 )(c2 + d2 ) = (ad + bc)2 + (ac − bd)2 which may be verified by expanding both sides. Thus, we should use the values c = 25, d = 4, ad + bc = 216 and ac − bd = 1. This then yields the following simultaneous equations 4a + 25b = 216 25a − 4b = 1 Solving these equations gives the values a = 409 and b = 2556. So we can express N as the sum of two squares as follows N = 4092 + 25562 . 6