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SASMO 2014 Round 1 Secondary 2 Solutions 1. The first day of 2014 was a Wednesday. There are 365 days in 2014. In what day of the week will 2015 begin? Solution 365 days = 52 weeks and 1 day Since the first day of 2014 was a Wednesday, then the first day of 2015 will be Thursday . 2. What is the maximum number of parts that can be obtained from cutting a circular cake using 3 straight cuts? Solution Maximum number of parts = 8 3. Evaluate 2014  2014  2013  2015. Solution Method 1 2014  2014  2013  2015 = (2013 + 1)  2014  2013  (2014 + 1) = 2013  2014 + 2014  2013  2014  2013 = 2014  2013 = 1 Method 2 2014  2014  2013  2015 = 2014  2014  (2014  1)  (2014 + 1) = 2014  2014  (2014  2014  1) = 1 4. Solve = 3 . Solution = 3 = 9 First C ut Second Cut Third Cut Свалено от Klasirane.Com

x + 3 = 9 x = 6 5. Find the exact value of 1  0.99999… Solution Let x = 0.99999… Then 10 x = 9.99999…  9x = 10 x – x = 9 x = = 1  0.99999… = 1 i.e. 1  0.99999… = 0 Note: 0.00000… is not acceptable since it suggests that the participant does not know that the answer is exactly 0. 6. Mersenne primes are prime numbers of the form M p = 2 p  1, where p is a prime. For example, 3 = 2 2  1 is a Mersenne prime. Find the 4 th largest Mersenne prime. Solution M 2 = 2 2  1 = 3 is the smallest Mersenne prime. M 3 = 2 3  1 = 7 is prime. M 5 = 2 5  1 = 31 is prime. M 7 = 2 7  1 = 127 is prime.  the 4 th largest Mersenne prime is 127 . 7. Simplify ( x  a) ( x  b) ( x  c) … ( x  z). Solution Since x  x = 0, then (x  a) ( x  b) ( x  c) … (x  z ) = 0. 8. How many squares are there in a 5  5 square grid? Solution No. of 1  1 squares = 25 = 5 2 No. of 2  2 squares = 16 = 4 2 Свалено от Klasirane.Com

No. of 3  3 squares = 9 = 3 2 No. of 4  4 squares = 4 = 2 2 No. of 5  5 squares = 1 = 1 2  total no. of squares in a 5  5 square grid = 12 + 2 2 + 3 2 + 4 2 + 5 2 = 55 9. If x and y are positive integers, find the values of x and y which satisfy the equation x2  4y2 = 41. Solution x 2  4y2 = 41 ( x + 2 y )(x  2y) = 41 Since 41 is prime, there are only two factors of 41, i.e. 1 and 41. Since x and y are positive integers and x + 2 y > x  2y, then x + 2 y = 41 and x  2y = 1. Solving, x = 2 1 and y = 10 . 10. Find the dimensions of all the rectangles with integral sides whose area and perimeter are numerically equal. Solution Method 1 Let the dimensions of a rectangle be l units by b units. Then area = perimeter Since b  2 > 0, then b > 2. W hen b = 3, l = 6. When b = 4, l = 4. When b = 5, l = is not an integer. When b = 6, l = 3, which is essentially the same as b = 3, l = 6. If b > 6, l < 3. By symmetry, , i.e. l > 2. So there are no other solutions.  the dimensions of all the rectangles with integral sides whose area and perimeter are numerically equal are 3 by 6 and 4 by 4. Me thod 2 Let the dimensions of a rectangle be l units by b units. Then area = perimeter Свалено от Klasirane.Com   bl lb  2 2 2   b b l 2 2   l l b   bl lb  2

Since b  2 > 0, then b > 2. F or l to be an integer, must also be an integer. This means that 4  b – 2, i.e. b  6. So the only possible solutions are when b = 3, 4, 5 and 6. When b = 3, l = 6. When b = 4, l = 4. When b = 5, l = is not an integer. When b = 6, l = 3, which is essentially the same as b = 3, l = 6.  the dimensions of all the rectangles with integral sides whose area and perimeter are numerically equal are 3 by 6 and 4 by 4. 11. A whole number is between 40 and 70. When it is divided by 3, the remainder is 1. When it is divided by 7, the remainder is 2. Find the number. Solution Method 1 Since the number leaves a remainder of 1 when divided by 3, then the possible values of the number are 40, 43, 46, 49, 52, 55, 58, …, 70. Since the number leaves a remainder of 2 when divided by 7, we can either check each of the above possible values one by one, or we can start with 7  6 + 2 = 44, and then add 7 until we get a number in the first list , i.e. 44, 51, 58, …  the number is 58 . Method 2 Let the number be N. Then N  1 (mod 3) and N  2 (mod 7). Since the moduli 3 and 7 are relatively prime, we can use the Chinese Remainder Theorem as followed. 7 N  1 (mod 3)  N  1 (mod 3)  N 1 = 4; 3 N  2 (mod 7)  N 2 = 3. Then N  7  4 + 3  3 = 37 (mod 21). Since N is between 40 and 70, then N = 37 + 21 = 5 8. 12. Find the value of . Свалено от Klasirane.Com 2 2   b b l 2442    b b  2422    b b 2 4 2   b 2 4  b             2 2 12 2 2 12 2 2

Solution Let = x. Then = x, i.e. 12 x2 + 2 x  2 = 0.  6x2 + x  1 = 0 implies (3 x  1) (2 x + 1) = 0, i.e. x = or  (rejected because x > 0)  = . 13. Find the value of . Solution Method 1 1 + 2013 = 2014 3 + 2011 = 2014 5 + 2009 = 2014 1005 + 1009 = 2014 1007 = 1007 So 1 + 3 + 5 + … + 2013 = 503  2014 + 1007  = = = 503 + = 503.5 Method 2 1 + 2013 = 2014 3 + 2011 = 2014 5 + 2009 = 2014 2013 + 1 = 2014 So 1 + 3 + 5 + … + 2013 = = 1007  1007  = = = = 503.5 14. Find the sum of the terms in the n th pair of brackets: There are = 503 pairs of numbers that add up to 2014 There are = 1007 pairs of numbers that add up to 2014 Свалено от Klasirane.Com             2 2 12 2 2 12 2 2 x 12 2 2              2 2 12 2 2 12 2 2

(1, 2, 3, 4), (5, 6, 7, 8), (9, 10, 11, 12), … Solution The sums of the terms in each pair of brackets form the following sequence: 3, 7, 11, 15, … Method 1 Common difference between consecutive terms, m = 4 Term before the first term, c = 3  4 = 1  sum of the terms in the n th pair of brackets = mn + c = 4 n  1 Method 2 Common difference between consecutive terms, d = 4 First term, a = 3  sum of the terms in the n th pair of brackets = a + ( n  1) d = 3 + ( n  1)  4 = 4 n  1 15. In the diagram, PQ and RS are parallel, PA = PB and RA = RC . Find BAC . Solution Let PAB = x. Then PB A = x (base s of isos. ABP) APB = 180  2x ( sum of  ABP) ARC = 180    APB (corr.  s; PQ // RS) = 180   (180   2x) = 2 x RAC = (base s of isos. ARC ) = 90   x  BAC = 180    PAB  RAC (adj. s on a str. ln) = 180   x  (90   x) = 90  16. There are 367 students in the school hall. What is the probability that two of the students have their birthday falling on the same day of the year? P Q R S A B C Свалено от Klasirane.Com

Solution There are at most 366 days in a year. Suppose none of the first 366 students have their birthday on the same day of the year. Then the last student’s birthday must fall on the same day of the year as the birthday of one of the first 366 students.  probability two of the students have their birthday on the same day of the year = 1 17. Let x be a number such that . Find the value of . Solution = 5 2  = 25  = 23 =  =  = = 5  (23  1) = 110 18. Divide the following shape into 4 identical parts. Solution The shape actually consists of three identical squares. But 3 and 4 are relatively prime, so it is not easy to divide the three squares into four identical parts. So we divide the shape into LCM(3, 4) = 12 parts first. From the 12 parts, we then try to regroup into 4 identical parts as shown below: 19. Find all the solutions to the trigonometric equation sin2 x  cos 2 x = 1 for 0  x  360 . Свалено от Klasirane.Com 5 1  x x 3 3 1 x x  2 1        x x 2 2 1 2 x x  2 2 1 x x                x x x x 11 2 2 3 3 11 x x xx  3 3 1 x x               x x x x 11 2 2        x x 1              1 11 2 2 x x x x

Solution Since sin 2 x  1 and cos 2 x  0, then the only possible solution to sin 2 x  cos 2 x = 1 is when sin 2 x = 1 and cos 2 x = 0, i.e. when sin x = 1 and cos x = 0.  x = 90  or 270  20. The diagram shows a square AEFG with an inscribed circle. ABCD is a rectangle such that AB = 2 cm and AD = 4 cm. Find the radius of the circle. Solution Draw CJ and OH parallel to AE, and OI parallel to AG. Then OCJ is a right-angled triangle. Let the radius of the circle be r cm. Then OC = OH = OI = r. So CJ = r  2 and OJ = r  4. By Pythagoras’ Theorem, OC 2 = CJ 2 + OJ 2 , i.e. r2 = ( r  2) 2 + ( r  4) 2 . Simplifying, we have r2  12 r + 20 = 0, i.e. ( r  2) ( r  10) = 0. Since r > 4, then the radius of the circle is 10 cm. A B C D E F G A B D G F E C I H J O Свалено от Klasirane.Com

21. In the following cryptarithm, all the letters stand for different digits. Find the final four-digit product. H E  E H * 8 + * * * * * * Solution Replace the * with letters, which can be the same. H E  E H A 8 + B C D F G I First I = 8. Since the addition of two digits will give a maximum of 18, or a maximum of 19 if there is a carryover (or renaming) of 1, this means that the maximum carryover is 1. So B = 9 and there is a carryover of 1 for B to give 10, i.e. D = 1 and F = 0. For HE  H = A8 to be a two-digit number, H < 4. If H = 1, E = 8; if H = 2, E = 4 or 9; if H = 3, E = 6. For HE  E = BC to be a two-digit number, if H = 1, then E  6, so E  8; if H = 3, then E  2, so E  6; if H = 2, then E  4, so E  9, but E = 4. So H = 2 and E = 4.  the final four-digit product is HE  EH = 24  42 = 1008 Свалено от Klasirane.Com

22. A farmer wants to plant 10 trees in 5 rows such that there are exactly 4 trees in each row. Draw a diagram to show how the trees should be planted. Solution If you start with 10 trees and then try fitting 5 lines onto them, you will realise that the lines must overlap because there are not enough trees, and it’s not easy to fit 5 lines onto 10 trees.  try to draw 5 overlapping lines first, and a common figure with 5 overlapping lines is the following star: Then put in the trees and yes, it works. 23. Find the last three digits of 25 25 . Solution Since the last three digits of a product ab depends only on the last three digits of a and of b, then 25 = 025 025  25 = 625 625  25 = 15 625  the last three digits are always 625, with the exception of the first power.  the last three digits of 25 25 are 625 . 24. Find the remainder when 3 2014 is divided by 5. Solution Method 1 Observe the following pattern: when divided by 5, 3 1 leaves a remainder of 3, 3 2 = 9 leaves a remainder of 4, 3 3 = 27 leaves a remainder of 2, 3 4 = 81 leaves a remainder of 1, 3 5 = 243 leaves a remainder of 3, 3 6 = 729 leaves a remainder of 4, 3 7 = 2187 leaves a remainder of 2, 3 8 = 6561 leaves a remainder of 1, … This means that the remainder will repeat with a period of 4. Since 2014 = 503  4 + 2, then 3 2014 , when divided by 5, will leave a remainder of 4 . Свалено от Klasirane.Com

Method 2 When divided by 5, 3 1 leaves a remainder of 3. If 3 k leaves a remainder of 3 when divided by 5, it means 3 k = 5 p + 3 for some integer p . Then 3 k +1 = 3(5 p + 3) = 15 p + 9 = 15p + 5 + 4 = 5(3 p + 1) + 4, i.e. 3 k +1 will leave a remainder of 4 when divided by 5. Let 3 k +1 = 5 q + 4 for some integer q . Then 3k +2 = 3(5 q + 4) = 15 q + 12 = 15 q + 10 + 2 = 5(3 q + 2) + 2, i.e. 3 k +2 will leave a remainder of 2 when divided by 5. Let 3 k +2 = 5 r + 2 for some integer r. Then 3 k +3 = 3(5 r + 2) = 15 r + 6 = 15 r + 5 + 1 = 5(3 r + 1) + 1, i.e. 3 k +3 will leave a remainder of 1 when divided by 5. Let 3 k +3 = 5 s + 1 for some integer s. Then 3 k +4 = 3(5 s + 1) = 15 s + 3, i.e. 3 k +4 will leave a remainder of 3 when divided by 5. This means that the remainder will repeat with a period of 4. Since 3 1 leaves a remainder of 3 when divided by 4, and 2014 = 503  4 + 2, then 2 2014 , when divided by 4, will leave a remainder of 4 . 25. Find the largest number of composite numbers less than 2014 that are relatively prime to one another, i.e. the highest common factor of any two of them is 1. Solution If 6 = 2  3 is one of the composite numbers, then all the multiples of 2 and all the multiples of 3 will not be relatively prime to 6. Thus, to get the largest number of composite numbers, we have to choose 2 n and 3 m instead of 6 k . In other words, we have to choose pn , where p is prime. Since the composite numbers must be less than 2014, then n must not be too big, esp. when p is big enough. Now, the largest value for p is 43, since 432 = 1849 < 2014 but 47 2 = 2209 > 2014. Without loss of generality, to get the largest number of composite numbers less than 2014 that are relatively prime to one another, choose the following numbers: 2 2 , 3 2 , 5 2 , 7 2 , …, 43 2 (although we can choose, e.g. 2 10 = 1 024 < 2014 instead of 2 2 ).  largest number of composite numbers less than 2014 that are relatively prime to one another = number of primes  43 = 14 Свалено от Klasirane.Com