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1 005 MATHCOUNTS CHAPTER SPRINT ROUND. We are given the following chart: Cape Bangkok Honolulu London Town Bangkok Cape 6300, Town Honolulu 6609, London To find the distance between Honolulu and Cape Town, go to the row labeled Honolulu and look for the intersection with the column labeled Cape Town. The value there is,535. (Or go to the column labeled Honolulu and look for the row labeled Cape Town. The answer is still the same.),535.. We are asked to find the percent of Canadian non-users for whom cost is not the primary barrier in the pie chart shown below (redrawn in Excel). Other 7% No need 8% Lack of skills Not enough time Cost 7% 7% 3% 8% Access to computer or internet According to the chart, 8% said cost was the barrier to using the internet. Therefore 00-8 = 7% did not say cost was a factor It is :00:00 midnight. In hours, 39 minutes and 44 seconds the time will be A:B:C on a -hour digital clock. hours is 5 days + hours (4 5 = 0). Thus the clock will show :39:44. Therefore, A + B + C = = Shooting hoops for 30 minutes burns 50 calories. Do this for 7 days you will burn 7 * 50 = 050 calories Olga purchases a rectangular mirror that fits exactly inside a frame. The shaded region represents the mirror itself. The outer perimeter of the frame measures 60 cm by 80 cm and the width of each side of the frame is 0 cm. To determine the width and height of the mirror, subtract 0 (0 for each side of the border) from the width of the frame and 0 from the height of the frame = = 40 Thus, the mirror itself is 60 cm wide and 40 cm high = The product of two positive whole numbers is 005. To determine what those numbers are, find the factors of 005. Of course is a factor, but we re told we can t use it. Since the number ends in 5, it is clear that 5 is a factor. 005 = = Three friends each order a large pizza. Shauntee eats 3 of her pizza. Carlos eats 9 8 of his and Rocco eats 7 6 of his pizza. To find out how much pizza is left, add up all the remaining pieces. Shauntee

2 has 3 left. Carlos has 9 left and Rocco has left It takes 4 minutes for Jana to walk one mile. So at this rate how far will she walk in 0 minutes? If Jana walks mile in 4 minutes she must walk of a mile in 4 minute. In ten minutes she must walk Roslyn has 0 boxes. 5 contain pencils, 4 contain pens and contain both pens and pencils. If boxes contain both pens and pencils and 5 contain pencils then 5 = 3 boxes contain only pencils. If boxes contain both pens and pencils and 4 contain pens, then 4 = boxes contain only pens. Thus, we have containing only pens, 3 containing only pencils and containing both = 7. Yet there are 0 boxes total, so 0 7 = 3 boxes do not have anything in them How many combinations of pennies, nickels and/or dimes are there with a total value of 5? Start with dimes. D + 5 P # D + N # We ve exhausted dimes. D + 3N #3 D + N + 5P #4 D + N +0P #5 D + 5P #6 We ve exhausted dime. So forget about dimes. Now only nickels and pennies. 5N #7 4N + 5P #8 3N + 0P #9 N + 5P #0 N + 0P # No more nickels. Now only pennies. 5P #. Let s insert parentheses into the expression in the following manner: ( 3) + (5 7) + (9 ) + (4 43) + (45 47) + 49 Note: The value of the expression does not change. We realize that there are 5 odd numbers in the integers through 50 and every two of them in this expression sum up to -. There are then pairs that each sum to and the 49 is left over. (- ) + 49 = = 5. A rectangular tile measures 3 inches by 4 inches. What is the fewest number of these tiles that are needed to completely cover a rectangular region that is feet by 5 feet? Well, you could draw it like this: But it s easier to see that it takes six 4-inch lengths to make feet, and it takes twenty 3-inch lengths to make 5 feet. So we have 6 of these tiles going down and 0 of these tiles going across. 6 0 = 0 3. Trapezoid ABCD has vertices A(, -), B(, ), C(5, 7) and D(5, ). To find the area of the trapezoid, draw it and draw the line BD. Two right triangles have been formed. The area of triangle BDC is ½ 6 4 =. The area of triangle BAD is ½ 4 3 = = 8

3 4. Trey receives a 5% commission. If he sells a coat for $60 his commission is $60.05 or $3. If, however, the coat were discounted by 0%, its cost would be $ or $48. Taking the commission on $48 we get $48.05 or $.40. $3 - $.40 = $.60, which is 60 cents. 5. Five distinct points A, B, C, D and E lie on a line. E is the midpoint of segment AB or AE = EB. A E B D is the midpoint of segment AE. AD = DE = ½ EB A---D---E B Both C and E are the same distance from B. A---D---E B C EB = BC The distance from D to B is 9 units. DE + EB = DB and DE = ½ EB EB + EB = DB = 9 3 EB = 9 3EB = 8 EB = 6 DC = DE + EB + BC = = 5 ANS farthings = pence pence = shilling 0 shillings = pound pound + 5 pence = 0 shillings + 5 pence = 0 pence + 5 pence = 40 pence + 5 pence = 45 pence = 45 4 farthings = 980 farthings 7. What is the sum of all the distinct positive two-digit factors of 44? First, start by figuring out just what the factors of 44 are 44 = = 7 44 = = = = = = That s all the factors since if you go any further than the square root of 44 (i.e., ), you ll only repeat the factors we have, but switched (i.e., 6 9, 8 8, etc.) Looking at all the factors we can see that only 7 are two-digit numbers = 6 8. The points B(, ), I(, 4) and G(5, ) are plotted to form triangle BIG. The triangle is translated five units to the left and two units upward to form triangle B I G as in the diagram below: Translating to the left means subtracting 5 from the x-coordinate. Translating two units upward means adding to the y-coordinate. If B is (, ), then B is (-5, +) or (-4, 3). If G is (5, ) then G is (5-5, +) = (0, 3). Thus the midpoint between (-4, 3) and (0, 3) is 4 0,3 = (-,3). 9. The positive difference of the cube of an integer and the square of the same integer is 00. Let x be the integer. Then: x³ - x² = 00 x²(x ) = 00 Factor 00 looking for squares = 0 0 and 0 = = 900; doesn t work. Continuing: 50, = 5 5 and 5 = If you take the sheet of paper and fold it twice, then unfold and mark the quarters through 4, it looks like this:

4 + + = 5 Turn the paper over and number the side behind as 5, the side behind and 6, the side behind 4 as 7 and the side behind 3 as 8. Now refold and cut. The final image you get is: Thus side 8 is directly behind quarter #3. The bottom right hand corner is actually all 4 corners of the big page. Cutting the wedge in it cuts wedges in all 4 corners. This means we are down to choices A, D, and E. B has the corner cut from the bottom, not from the side. C and F have the edge triangle cut, not a wedge. The top right corner encompasses the center edge between and 3, and and 4. Cutting wedges in there provides two cuts symmetrical around the horizontal fold. This removes choice D (no cuts at all in the middle) and finally choice A because the type of cut is not a wedge. This leaves us with only E as the choice. E. Henry took five tests and got an average of 57. This means a total of 57 5 = 85 points. He scored at least 50 points (out of a possible 00) on each test. To find the highest score, we must minimize the scores on the other four tests. Suppose Henry scored 50 on each of the 4 tests. That sums to 00 points = 85. How many ordered pairs (x, y) satisfy BOTH conditions? Condition I: x = or y = 0 or y = Condition II: x = 0 or x = or y = From condition I, choose x =. Then from condition II we are forced to choose y = or (,). Now from condition I, choose y = 0. From condition II, x must be either 0 or or (0,0) or (,0) for two more choices. Finally choose y = from condition I. Again, x must be either 0 or for (0, ) and (,) for two more choices. 3. If a number is 5 or less, double the number. If a number is more than 5, subtract from it. If the third number in Zan s sequence is 36, we need to find the four distinct numbers that could have been the first number in her sequence. Suppose the second number in the sequence is less than 5. Then the second number is doubled to give 36. Therefore, the second number is 8. What would the first number be? If it is less than 5, it would have been doubled, so the first number could be 9. But if the first number is greater than 5, is subtracted from it to get = 30, so the first number could also be 30. Now go back to the third number and assume the second number is more than 5 so is subtracted from it to get = 48 So the second number is 48. Now assume, the first number is less then 5 so it is doubled to get 48. Thus, the first number could also be 4. Now assume the first number is more than 5 so 48 is less. Thus, the final possibility is 48 + = 60. The four possibilities are 9, 30, 4 and 60. Their sum is: = 3 4. My age has two digits (x and y). If I reverse them, divide by three and add 0, I get my age back again. I can represent my age as 0x + y. Reverse that and it becomes 0y + x. Therefore, 0y x 0 0x y 3 0y + x + 60 = 30x + 3y 7y + 60 = 9x Since x is a single digit, it must be large enough that 9x is greater than 60. Thus x 3. If x = 3, then 9x = 87 and 7y = 7. That won t work. If x = 4, then 9x = 6 and 7y = 56 so y = 8. So I must be 48 years old The sequence of integers in the row of squares and in each of the two columns of squares form three

5 distinct arithmetic sequences. and then a 0-foot plank from D to C, where D is the midpoint of AB. The first information we have is 4 and 8. This says we have an arithmetic sequence where values in the vertical column in the middle differ by 4. This leads us to: Now we have,?,?, 6, where each value in the sequence differs by the same number. If x is the difference between terms in the sequence, we get:, x, x, 3x and 3x = 6 5 = 3x and x = 5 Therefore, the difference between terms in this sequence is 5 and we can fill in some more of the squares. Since D is the midpoint AD = 8. DC = 0. CD is perpendicular to AB and AC is the radius of the circle. Due to the Pythagorean Theorem, AC² = CD² + AD² = 8² + 0² = 64 Area of the circle = π r² = π AC² = 64π 7. f(n) = f(n ) + f(n ) f() = 3 f(3) = 0 f(3) = f() + f() = f() + 3 = 0 f() = 7 f(4) = f(3) + f() = = 7 f(5) = f(4) + f(3) = = 7 f(6) = f(5) + f(4) = = a b 3 b, 4 c 4a = 3b 9b = 8c 3c = d c = 3 d 8, 9 c d 3 6 9b = 8c = 8 d = d d = b 6 We have the final sequence where we let y be the difference between terms in the sequence. So we have: N, N + y, N + y, N + 3y, N + 4y, N + 5y From these terms we know N + 5y = -7 and N + y = -9 4y = -8 and so y = - N - = -9 N = To be able to walk to the center C of a circular fountain, a repair crew places a 6-foot plank from A to B a = 4 3 b b b b ad b b b A play has two male roles, two female roles and two roles that can be either gender. Men must be assigned to male roles and women to female roles. There are 5 men and 6 women. The different

6 situations are: A women play the either/or roles B men play the either/or roles C man and women play the roles Situation A: 4 women get roles and men get roles = 30 0 = 700 ways. Situation B: women get roles and 4 men get roles = = 3600 Situation C: 3 women get roles and 3 men get roles = 30 0 = 700 ways. This is for when the male takes the first role and the female takes the second. We also have to do this one more time when the female takes the first role and the male takes the second for another 700 ways = What is the arithmetic mean of all of the positive two-digit integers with the property that the integer is equal to the sum of its first digit plus its second digit plus the product of its two digits? Let x be the tens digit and y be the ones digit. Then: 0x + y = x + y + xy 9x = xy y = 9 Does this mean that x can be anything? 9 = = Apparently so. 9, 9, 39,, 89, 99 is an arithmetic sequence. Since there are an odd number of values the mean is just the median or 59. TARGET ROUND. Francisco starts with the number 5, doubles it, adds, doubles the result, adds, doubles the result and continues this pattern of two alternating calculations. Francisco does 8 total calculations. Calc : 5 = 0 Calc : 0 + = Calc 3: = Calc 4: + = 3 Calc 5: 3 = 46 Calc 6: 46 + = 47 Calc 7: 47 = 94 Calc 8: 94 + = 95 Phong starts with 5, adds, doubles the result, adds, doubles the result, adds, doubles the result, and continues this pattern of two alternating calculations. Phong also does 8 calculations. Calc : 5 + = 6 Calc : 6 = Calc 3: + = 3 Calc 4: 3 = 6 Calc 5: 6 + = 7 Calc 6: 7 = 54 Calc 7: 54 + = 55 Calc 8: 55 = 0 The positive difference of their results is: 0 95 = 5. All vertices of the cube are to be colored such that no two vertices on the same edge of the cube are the same color. Clearly, one color won t work. But two does. 3. A store purchases a television from a factory for $ The store normally charges 5% of this. $ = The coupon gives 5% off = $ Four different symbols represent unknowns ( ya know, this is just making it take a couple of seconds longer!!!). Let a represent the triangle. Let b represent the square. Let c represent the hexagonallyshaped symbol. Let d represent the upside down trapezoid. Then the 3 equations are: a + b = d

7 a + a = c + c + c + c + c a + a = d + c Let s rewrite that: a + b = d a = 5c a = d + c Since a = 5c and a and c must both be integers less than 0, we can see that this equation is only true if a = 5 and c =. Let s go back to a = d + c. We now have (5) = d +, so d = 8. Using the equation a + b = d, we now know 5 + b = 8, so b = What is the greatest whole number that MUST be a factor of the sum of any four consecutive positive odd numbers? Let x be the first odd number. Then: x + x + + x x + 6 = 4x + = 4(x + 3) x is an odd number. Adding 3 to it will make it even. Thus, x + 3 is divisible by. 4 = 8 6. In the figure below, the smaller circle has a radius of two feet and the larger circle has a radius of four feet. We are asked to find the total area of the shaded regions. First, one can create rectangles by connecting the tangent lines with the diameters of each circle. Each rectangle has a length equal to the radius and a width equal to the diameter. The area of the smaller rectangle is 4 = 8. The area of the larger rectangle is 8 4 = = 40. Now this area includes the shaded area so subtract the areas of the semicircles. The area of the smaller semicircle is: ½πr² = π The area of the larger semicircle is 8π. So the total area to be subtracted is π + 8π = 0π 40-0π Jamie has a jar of coins. The jar contains the same number of nickels, dimes and quarters. The total value of the coins in the jar is $3.0. Let x be the number of coins of each type. Then: 5x + 0x + 5x = 30 (i.e., move to cents; it s easier) 40x = 30 x = John, Mike and Chantel divide a pile of pennies amongst themselves using the following process: - If the number of pennies in the pile is even, Mike will get half of the pile. - If the number of pennies in the pile is odd, one penny will be given to Chantel, and John will get half the pennies in the pile. This process repeats until the pile is empty. They start with 005 pennies. -- The number of pennies is odd so Chantel gets one and John gets half or 00. J: 00, M: 0, C: There are = 00 pennies left. -- The number of pennies is even so Mike gets half or 50. J: 00, M: 50, C: There are 50 pennies left. -3- The number of pennies is odd so Chantel gets one and John gets half of 500 or 50. J: 5, M: 50, C: There are 50 pennies left. -4- The number of pennies is even so Mike gets half or 5. J: 5, M: 66, C: There are 5 left. -5- Chantel gets and John gets half of 4 or 6. J: 34, M: 66, C: 3 There are 6 left. -6-: Mike gets half of 6 or 3. J: 34, M: 657, C: 4 There are 3 left. -7-: Chantel gets, John gets 5 and 5 are left. J: 39, M: 657, C: 5

8 -8-: Chantel gets, John gets 7 and 7 are left. J: 336, M: 657, C: 6-9-: Chantel gets, John gets 3 and 3 are left. J: 339, M: 657, C: 7-0-: Chantel gets, John gets and is left. J: 340, M: 657, C: 8 -: Chantel gets. J: 340: M: 657, C: TEAM ROUND. A competition problem requires one hour to develop. 30,000 students work on the problem for an average of 4 seconds each. The total time spent by the students to solve the problem is 30,000 4 = seconds. One hour is 60 minutes/hour 60 minutes/sec or 3600 seconds. The ratio of the development time to the total time spent by the students is: Select a three-digit multiple of 3. Calculate the sum of the cubes of the digits of that number. Calculate the sum of the cubes of the new number and continue doing so until you arrive at a number that is equal to the sum of the cubes of its digits. I.e., x³ + y³ + z³ = xyz Start with something easy. is a multiple of 3. ³ + ³ + ³ = 3 3³ = 7 ³ + 7³ = = 35 3³ + 5³ + ³ = = 53 ³ + 5³ + 3³ = = A room contains 30 assembled Tworks. There are enough pieces to assemble 00 more Tworks. A Twork takes 8 minutes to assemble and 0 minutes to disassemble. Emma starts assembling Tworks as Ed begins disassembling the ones already made. They both continue to work until there are exactly 35 assembled Tworks in the room (and no partially assembled or disassembled Tworks). We have to find how long it took. The least common multiplier of 0 and 8 is 40, so in 40 minutes 5 Tworks have been assembled and 4 Tworks disassembled. Every 40 minutes one more Twork gets assembled. We need to have 35 so 5 40 = 00 minutes to get 5 more Tworks How many triangles are in the figure? Start with the obvious ones. There are 6 triangles that contain no smaller triangles within them. Now look for triangles that are made out of smaller triangles. There are 0 of those. There cannot be any triangles made out of 3 smaller ones but there are triangles made out of 4 smaller ones. and for a total of 8 triangles there. The only other triangular shape is made out of 8 smaller triangles. and

9 for a total of more = x x For a power of (x ) to equal, x must be or x must be - and 5 - x² must be even. Start with x =. Then x = 3 and 5 - x² = 5-3² = 6. So x can be 3. Now suppose x = -. x = and that will work as well since 5 = 4 and is then the even power we would need for a base of -. But there s one other thing: Any value to the 0 th power is so 5 - x² = 0 will give us other values, x = 5 and -5 does that. This gives us 4 values The minute hand of a clock (that should be a -hour clock!) measures 0 cm from its tip to the center of the clock face. The hour hand measures 5 cm from its tip to the center of the clock face. We are asked to find the sum of the distances traveled by the tips of both hands in one 4-hour period. The minute hand has a length of 0 cm so as it goes round it creates a circle with radius 0 cm. Its circumference is: πr = 0π. The minute hand goes round the circumference time per hour or 4 times per day. 4 0π = 480π The hour hand measures 5 cm so as it goes round it creates a circle with radius 5 cm. Its circumference is πr = 0π The hour hand also goes round the circumference times per day. 0π = 0π The total distance traveled is: 480π + 0π = 500π cm but we were asked for meters which means that the total distance traveled is 5π meters. 5π There are two shoelace patterns for two identical shoes with 4 holes each. Assume that the holes form a rectangular grid and that each hole is cm from its nearest horizontal and vertical neighbor-holes. Determine the ratio of the total length of the shoelace shown in the first pattern to the total length of the shoelace shown in the second pattern. The first pattern looks like this. The straight lines are of length. The diagonal line is the hypotenuse of a cm square so its length is. There are 7 straight lines for a length of 7 and there are 6 diagonal lines for a length of 6. So the length of the shoelace to lace up pattern is The second pattern looks like this: It has one straight line and diagonal lines for length of A four-digit perfect square number is created by placing two positive twodigit perfect square numbers next to each other. We must find the number. So what are the two-digit perfect square numbers? 6, 5, 36, 49, 64, 8 What are the four-digit square numbers that start around 600? 600 and looks promising. How about around 500? 500, then 60. That does it. Each of the other squares can t start this number since the next square will be at least 00 greater Ella rolls a standard six-sided die until she rolls the same number on consecutive rolls. What is the probability that her 0 th roll is her last roll? What s the probability that the first two rolls don t get her the same number twice?

10 Ella can roll any of 6 values the first time and she can roll any of 5 values the second time. So the probability is: The probability that Ella doesn t get it on the third try is the probability that she doesn t get it on the second try and she rolls one of the 5 permissible values on the third try. So similarly to go to the 0 th try, the probability is: The last time she must roll the same number as her 9 th roll and, therefore, there is only one possible roll that will work.) A deck of playing cards has 6 red and 6 black cards. The deck is split into two piles, each having at least one card. There are 6 times as many black cards as red cards in the first pile. The number of red cards is a multiple of the number of black cards in the second pile. Let x be the number of cards in the first pile and y the number of cards in the second pile. x + y = 5 Let a = the number of red cards in the first pile and b = the number of black cards in the first pile. Then: a + b = x and 6a = b Let c = the number of red cards in the second pile and d = the number of black cards in the second pile. Then: c + d = y = 5 x and c = Nd where we do not know N. a + c = 6 and b + d = 6. Going back to the first pile: 7a = x So the number of cards in the first pile is a multiple of 7. The multiples of 7 less than 5 are: 7, 4,, 8, 35, 4, 49 Remember that the number of cards in the second pile must also be a multiple of something so the value better not be prime. 5 - each multiple of 7 is: 45, 38, 3, 4, 7, 0, 3 3, 7 and 3 are primes, ruling out, 35 and 49, leaving 7, 4, 8 and 4 as possibilities. Consider 7 for the number of cards in the first pile. Then a = and b = 6. c = 6 = 5 d = 6 6 = 0 5 and 0 are not multiples of each other. Consider 4: a = and b = c = 6 = 4 d = 6 = 4 4 and 4 are not multiples of each other. Consider 8: a = 4 and b = 4 c = 6 4 = d = 6 4 = c is definitely a multiple of d. How about 4? Then a = 6 and b = 36. c = 6 6 = 0 d = 6 36 = -0 I don t think so! The number of red cards in the second pile is c or.

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