2005 Pascal Contest (Grade 9)

Transcription

1 Canadian Mathematics Competition An activity of the Centre for Education in Mathematics and Computing, University of Waterloo, Waterloo, Ontario 005 Pascal Contest (Grade 9) Wednesday, February 3, 005 Solutions c 005 Waterloo Mathematics Foundation

2 005 Pascal Contest Solutions Page 1. Calculating each of the numerator and denominator first, = = 7. Answer: (E). Simplifying the terms in pairs, 6a 5a + 4a 3a + a a = a + a + a = 3a. 3. When we substitute = 3, the product becomes 3()(1)(0)( 1). Since one of the factors is 0, the entire product is equal to 0. Answer: (C) 4. Of the numbers, 3, 4, 5, 6, 7, only the numbers, 3, 5, and 7 are prime. Since 4 out of the 6 numbers are prime, then the probability of choosing a ball with a prime number is 4 6 = 3. Answer: () 5. Calculating from the inside out, = 36 4 = 144 = When half of the water is removed from the glass, the mass decreases by = 300 g. In other words, the mass of half of the original water is 300 g. Therefore, the mass of the empty glass equals the mass of the half-full glass, minus the mass of half of the original water, or = 400 g. Answer: () 7. Solution 1 Since 1 = 1, then = 3 1 = 36, so 1 = 1 (36) = Solution Since 3 1 = 1, then 1 = 3 ( 1) = 3 (1) = Answer: (C) 8. We calculate the square of each number: ( 5) = 5, ( 3 ) = 9, 4 = 4, ( 3 5 ) = 9, and 5 8 = 64. We can quickly tell that each of 5, and 8 is less than its square. Since 3 = 1 1 and 9 = 1, then However, 3 = is also less than its square. 9 is larger than its square. 5 Answer: () 9. Solution 1 Since BA is isosceles, BA = AB =. Since CA is isosceles, CA = AC = y. A B C

3 005 Pascal Contest Solutions Page 3 Therefore, BAC = ( + y). Since the sum of the angles in ABC is 180, then Therefore, + y = y + ( + y) = 180 ( + y) = y = y = 90 Solution Since BA is isosceles, BA = AB =. Therefore, looking at the sum of the angles in AB, we have = 180 or = 180 or = 76 or = 38. Since BA and AC are supplementary, then AC = = 76. Since CA is isosceles, CA = AC = y. A B C Therefore, looking at the sum of the angles in CA, we have y + y + 76 = 180 or y + 76 = 180 or y = 104 or y = 5. Therefore, + y = = 90. Answer: () 10. The third term in the sequence is, by definition, the average of the first two terms, namely 3 and 8, or 1(3 + 8) = 1 (40) = 0. The fourth term is the average of the second and third terms, or 1 (8 + 0) = 14. The fifth term is the average of the third and fourth terms, or 1 (0 + 14) = 17. Therefore, = If a and b are positive integers with a b = 13, then since 13 is a prime number, we must have a and b equal to 1 and 13, or 13 and 1, respectively. If b = 1, then since b c = 5, we must have c = 5. But then we could not have c a = 4. So b cannot be 1. Thus, b = 13, so a = 1, and c = 4 (which we can get either from knowing b = 13 and b c = 5, or from knowing a = 1 and c a = 4. Therefore, a b c = = 5. Answer: (E)

4 005 Pascal Contest Solutions Page 4 1. Solution 1 To get from K to M, we move 6 units to the right and 9 units up. y M(10,11) L(6,w) O K(4,) Since L lies on the line segment KM and to get from K to L we move = 1 6 units to the 3 right, then we must move 1 9 = 3 units up. 3 Thus, w = + 3 = 5. Solution The slope of line segment KM is = 9 6 = 3. Since L lies on line segment KM, then the slope of line segment KL will also be 3. Therefore, w 6 4 = 3 or w = 3, so w = 3 or w = 5. Answer: (B) 13. Solution 1 Each unit cube has three eposed faces on the larger cube, and three faces hidden. Therefore, when the faces of the larger cube are painted, 3 of the 6 faces (or 1 of the surface area) of each small cube are painted, so overall eactly 1 of the surface of area of the small cubes is painted. Solution Since the bigger cube is by by, then it has si by faces, for a total surface area of 6 = 4 square units. All of this surface area will be painted. Each of the unit cubes has a surface area of 6 square units (since each has si 1 by 1 faces), so the total surface area of the 8 unit cubes is 8 6 = 48 square units. Of this 48 square units, 4 square units is red, for a fraction of 4 48 = 1. Answer: (C) 14. Every integer between 005 and 3000 has four digits. Since palindromes are integers whose digits are the same when read forwards and backwards, then a four-digit palindrome must be of the form yy where and y are digits. Since 3000 is not a palindrome, then the first digit of each of these palindromes must be, ie. it must be of the form yy. Since each palindrome is at least 005 and at most 3000, then y can take any value from 1 to 9. Therefore, there are 9 such palindromes: 11,, 33, 44, 55, 66, 77, 88, and 99. Answer: (C)

5 005 Pascal Contest Solutions Page When 14 is divided by n, the remainder is, so n must divide evenly into 14 = 1. Therefore, n could be 1,, 3, 4, 6, or 1. But the remainder has to be less than the number we are dividing by, so n cannot be 1 or. Thus, n can be 3, 4, 6 or 1, so there are 4 possible values of n. Answer: () 16. To get the largest possible number by using the digits 1,, 5, 6, and 9 eactly once, we must choose the largest digit to be the first digit, the net largest for the net digit, and so on. Thus, the largest possible number is Also, using this reasoning, there are only two numbers that use these digits which are at least , namely and Therefore, must be the largest even number that uses each of these digits eactly once. Reversing our logic, the smallest possible number formed using these digits is 1 569, and the only other such number smaller than is 1 596, so this must be the smallest even number formed using these digits. Calculating the difference, we get = Let the side length of each of the squares be. P S Q R Then the perimeter of P QRS equals 8, so 8 = 10 cm or = 15 cm. Since P QRS is made up of three squares of side length 15 cm, then its area is 3(15 cm) or 3(5) cm = 675 cm. Answer: (B) 18. First, 005 = 40005, so the last two digits of 005 are 5. We need to look at 005 5, but since we only need the final two digits, we don t actually have to calculate this number entirely. Consider = = When we carry out this multiplication, the last two digits of the product will only depend on the last two digits of the each of the two numbers being multiplied (try this by hand!), so the last two digits of are the same as the last two digits of 5 5 = 15, ie. are 5. Similarly, to calculate 005 4, we multiply (which ends in 5) by 005, so by the same reasoning ends in 5. Similarly, ends in 5. Therefore, 005 and both end in 5. Also, = 1, so the epression overall is equal to = Therefore, the final two digits are The easiest way to count these numbers is to list them by considering their hundreds digit. Hundreds digit of 1: No such numbers, since tens digit must be 0, which leaves no option for units digit. Hundreds digit of : Only one possible number, namely 10.

6 005 Pascal Contest Solutions Page 6 Hundreds digit of 3: Here the tens digit can be or 1, giving numbers 31, 30, 310. Hundreds digit of 4: Here the tens digit can be 3, or 1, giving numbers 43, 431, 430, 41, 40, 410. Therefore, there are 10 such numbers in total. Answer: (B) 0. We label the five junctions as V, W, X, Y, and Z. S V W X Y Z A B C From the arrows which Harry can follow, we see that in order to get to B, he must go through X (and from X, he has to go to B). So we calculate the probability that he gets to X. To get to X, Harry can go S to V to W to X, or S to V to Y to X, or S to V to X directly. At V, the probability that Harry goes down any of the three paths (that is, towards W, X or Y ) is 1. 3 So the probability that Harry goes directly from V to X to 1. 3 At W, the probability that Harry turns to X is 1, so the probability that he goes from V to W to X is 1 1 = At Y, the probability that Harry turns to X is 1, so the probability that he goes from V to Y 3 to X is 1 1 = Therefore, the probability that Harry gets to X (and thus to B) is = 6+3+ = Answer: (C) 1. We try each of the five possibilities. If m : n = 9 : 1, then we set m = 9 and n =, so 9 + = 300 or 10 = 300 or = 30, so m = 9(30) = 70 and n = 30. If m : n = 17 : 8, then we set m = 17 and n = 8, so = 300 or 5 = 300 or = 1, so m = 17(1) = 04 and n = 8(1) = 96. If m : n = 5 : 3, then we set m = 5 and n = 3, so = 300 or 8 = 300 or = 75, so m = 5( 75) = 375 and n = 3( 75) = 5. If m : n = 4 : 1, then we set m = 4 and n =, so 4 + = 300 or 5 = 300 or = 60, so m = 4(60) = 40 and n = 60. If m : n = 3 :, then we set m = 3 and n =, so 3 + = 300 or 5 = 300 or = 60, so m = 3(60) = 180 and n = (60) = 10. The only one of the possibilities for which m and n are integers, each greater than 100, is m : n = 3 :. Answer: (E)

7 005 Pascal Contest Solutions Page 7. Let AS = and S = y. Since SAP and SR are isosceles, then AP = and R = y. Since there are two pairs of identical triangles, then BP = BQ = y and CQ = CR =. A P y B y Q S y y R C SR is right-angled (since ABC is a square) and isosceles, so its area (and hence the area of BP Q) is 1 y. Similarly, the area of each of SAP and QCR is 1. Therefore, the total area of the four triangles is ( 1 ) + ( 1 y ) = + y, so + y = 00. Now, by the Pythagorean Theorem, used first in P RS, then in SAP and SR, P R = P S + SR = (SA + AP ) + (S + R ) = + y = (00) = 400 so P R = 0 m. Answer: (B) 3. From the centre, there are 8 possible 0s to which we can move initially: 4 side 0s and 4 corner 0s

8 005 Pascal Contest Solutions Page 8 From a side 0, there are 4 possible 0s to which we can move: side 0s and corner 0s From a corner 0, there are possible 0s to which we can move: side 0s From a side 0, there are 3 possible 5s to which we can move From a corner 0, there are 5 possible 5s to which we can move So the three possible combinations of 0s are side-side, side-corner and corner-side. For the combination side-side, there is a total of 4 3 = 4 paths (because there are initially 4 side 0s that can be moved to, and from each of these, there are side 0s that can be moved to, and from each of those, there are 3 5s that can be moved to). For the combination side-corner, there is a total of 4 5 = 40 paths. For the combination corner-side, there is a total of 4 3 = 4 paths. Therefore, in total there are = 88 paths that can be followed to form 005. Answer: (E) 4. Since we have an increasing sequence of integers, the 1000th term will be at least We start by determining the number of perfect powers less than or equal to (This will tell us how many integers less than 1000 are skipped by the sequence.) Let us do this by making a list of all perfect powers less than 1100 (we will need to know the locations of some perfect powers bigger than 1000 anyways):

9 005 Pascal Contest Solutions Page 9 Perfect squares: 1, 4, 9, 16, 5, 36, 49, 64, 81, 100, 11, 144, 169, 196, 5, 56, 89, 34, 361, 400, 441, 484, 59, 576, 65, 676, 79, 784, 841, 900, 961, 104, 1089 Perfect cubes: 1, 8, 7, 64, 15, 16, 343, 51, 79, 1000 Perfect 4th powers: 1, 16, 81, 56, 65 Perfect 5th powers: 1, 3, 43, 104 Perfect 6th powers: 1, 64, 79 Perfect 7th powers: 1, 18 Perfect 8th powers: 1, 56 Perfect 9th powers: 1, 51 Perfect 10th powers: 1, 104 In these lists, there are 41 distinct perfect powers less than or equal to Thus, there are 959 positive integers less than or equal to 1000 which are not perfect powers. Therefore, 999 will be the 959th term in the sequence. (This is because 1000 itself is actually a perfect power; if it wasn t, 1000 would be the 959th term.) Thus, 1001 will be the 960th term. The net perfect power larger than 1000 is 104. Thus, 103 will be the 98nd term and 105 will be the 983rd term. The net perfect power larger than 104 is Therefore, the 104 will be the 1000th term. The sum of the squares of the digits of 104 is = 1. Answer: (E) 5. By the Pythagorean Theorem in AE, A = AE + E = = 565, so A = 75. Since ABC is a rectangle, BC = A = 75. Also, by the Pythagorean Theorem in BF C, F C = BC BF = = 3600, so F C = 60. raw a line through F parallel to AB, meeting A at X and BC at Y. To determine the length of AB, we can find the lengths of F Y and F X. A 1 E 45 F X Step 1: Calculate the length of F Y B Y C The easiest method to do this is to calculate the area of BF C in two different ways. We know that BF C is right-angled at F, so its area is equal to 1 (BF )(F C) or 1 (45)(60) = Also, we can think of F Y as the height of BF C, so its area is equal to 1 (F Y )(BC) (F Y )(75). or 1 F B Y 75 C

10 005 Pascal Contest Solutions Page 10 Therefore, 1 (F Y )(75) = 1350, so F Y = 36. (We could have also approached this by letting F Y = h, BY = and so Y C = 75. We could have then used the Pythagorean Theorem twice in the two little triangles to create two equations in two unknowns.) Since F Y = 36, then by the Pythagorean Theorem, so BY = 7. Thus, Y C = BC BY = 48. Step : Calculate the length of F X BY = BF F Y = = 79 Method 1 Similar triangles Since AE and F X are right-angled at E and X respectively and share a common angle, then they are similar. Since Y C = 48, then X = 48. Since AE and F X are similar, then F X X = AE E or F X 48 = 1 so F X = Method Mimicking Step 1 rop a perpendicular from E to A, meeting A at Z. A Z X 48 E F We can use eactly the same argument from Step 1 to calculate that EZ = 504 and 5 that Z = Since F is a straight line, then the ratio F X : EZ equals the ratio X : Z, ie. F X = or F X 504 = or F X = 14. Method 3 Areas Join A to F. Let F X = and EF = a. Then F = 7 a. A 1 E 7 a X F Since AE = 1 and E = 7, then the area of AE is 1 (1)(7) = 756. Now, the area of AE is equal to the sum of the areas of AEF and AF, or = 1 (1)(a) + 1 (75)() so 1a + 75 = 151 or a = 7. Now in F X, we have F X =, G = 48 and F = 7 a.

11 005 Pascal Contest Solutions Page 11 By the Pythagorean Theorem, + 48 = (7 a) = ( 5 7 ) = 65 Therefore, 48 = , or = 14. Therefore, AB = XY = F X + F Y = =