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M2C1 M2C1 ! "# $ % ! " # & % % % '! % ! % (% % * ) +, *- ! ( % ( &! .+ % % % */ % % , 01 (% 2 % %! #+ - % %* 0 %! ! ! % %& ! ( %& ! % ( ! $+ -& *3 ( 4% ! ! 4! 4 %&4% 5 ! 5 %& ! !! % ! ! ! 0 ! ! % %! ( % *""444+ + &" ! "4% " 6 %! 7 && %! 7 ! % ( !8& %+ 2 M2C1 - Meet #2, Mystery Category 1 Mystery Meet #2, December 2005 1. Some parents are needed to volunteer at a school store. There are 180 days in the school year and three parents are needed each day. If each parent works a total of 15 days, how many parent volunteers are needed? Hint: Imagine dividing the 180 days into 12 chunks of 15 days each. 2. The numbers on each line segment indicate the sums of the numbers in the two circles at the ends of the segment. What’s the sum of the three numbers in the circles? 22 25 31 Hint: You could figure out what is in each circle, but that’s the hard way. 3. When farmer Tom got tired of his chickens, he sold half (1 ) of them to his 2 neighbor Fred, he gave one fourth (4 ) of them to his uncle Bob, he sold one eighth 1 (1 ) of them to the local butcher, and he gave one ninth (1 ) of them to the local 8 9 food pantry. The remaining eight (8) chickens made a fabulous feast for all his friends and relatives. How many chickens did farmer Tom have before he got rid of them? Answers 1. _______________ 2. _______________ 3. _______________ 3 M2C1 - Meet #2, Mystery Solutions to Category 1 Mystery Meet #2, December 2005 Answers 1. If three parents are needed for each of the 180 days of the school year, then 3 × 180 = 540 parent-days are 1. 36 needed. If each parent works a total of 15 days, then 540 ÷ 15 = 36 parents are needed. 2. 39 3. 576 2. Since 25 is 3 more than 22, we might realize that the number in the bottom right circle must be 3 more than the number in the bottom left circle. If we imagine taking 3 away from this number, then we would have two equal numbers whose sum is 28 instead of 31. The number in the bottom left circle must be 28 ÷ 2 = 14. Now we can figure out that the top number must be 22 – 14 = 8 and the bottom right number must be 31 – 14 = 17. The 8 desired sum is 8 + 14 + 17 = 39. Alternatively, we might note that each number in the circles is used twice in the sums on the line segment. This means that the sum of the 22 25 numbers in the circles is exactly half the sum of the numbers on the line segments. (22 + 25 + 31) ÷ 2 = 78 ÷ 14 31 17 2 = 39. 3. Adding the fractional amounts of chickens that farmer Tom sold or gave away, we get 1 1 1 1 36 + 18 + 9 + 8 71 + + + = = . 2 4 8 9 72 72 The 8 chickens that became the feast must represent the 1 missing of the chickens. Thus the total number of 72 chickens must have been 8 × 72 = 576. 4 M2C1 - Meet #2, Mystery Category 1 Mystery Meet #2, November 2004 1. A classroom has three light switches, one for each row of lights on the ceiling. Including the possibility of all three switches off, how many different ways can the lights be set in this classroom? 2. At stage 0, the triangle below has an area equal to 1 square unit. At stage 1, a similar triangle has been removed from the middle, and the area is now equal to 3/4 of a square unit. At each new stage, the middles of the remaining triangles are removed, and the area is 3/4 of what it was in the previous stage. How many square units are in the area of the remaining shaded regions at stage 4? Express your answer as a common fraction in lowest terms. Stage 0 Stage 1 Stage 2 3. Let’s say that a “comprimesite” number is a product of exactly two primes. The number 6, for example, is a comprimesite number, since it can be written as 2 × 3. The number 4 is also comprimesite, since it can be written as 2 × 2. The numbers 7 and 8 are not comprimesite, since they cannot be written as a product of two primes. Find the greatest comprimesite number less than 200. Answers 1. _______________ 2. _______________ 3. _______________ 5 M2C1 - Meet #2, Mystery Solutions to Category 1 Mystery Meet #2, November 2004 Answers 1. Each of the three light switches can be either on or off. Thus there are 2 × 2 × 2 = 8 ways the lights can set 1. 8 in the classroom. If we use 1 for “on” and 0 for “off”, we can represent the 8 different ways as three-digit binary 81 numbers as follows: 000, 001, 010, 011, 100, 101, 110, 2. 111. 256 3. 194 2. At stage 4, the remaining area will be 4 3 3 3 3 3 81 = ⋅ ⋅ ⋅ = . 4 4 4 4 4 256 3. The greatest “comprimesite” number less than 200 is 194, which is 2 × 97. 199 - prime 198 = 2x99, and 99 is not prime 197 - prime 196 = 2x98 and 98 is not prime 195 = 5x39 and 39 is not prime 194 = 2x97 and 97 is prime! 6 M2C1 - Meet #2, Mystery Category 1 Mystery Meet #2, November 2003 1. In the following shape sentences, same shapes have the same numbers and different shapes have different numbers. Find the value of the circle. + + = 14 + – = 13 + = 19 Hint: What is the 3rd equation minus the 2nd equation? 2. Together Albert and Brendan weigh 121 pounds. Brendan and Camille weigh 113 pounds together. Camille and Dana together weigh 83 pounds. How many pounds do Albert and Dana weigh together? 3. A regular hexagon lives in the first quadrant of a Cartesian Coordinate System as shown below. How many units are in the perimeter of the hexagon? y (3,4) (9,4) Answers 1. _______________ 2. _______________ x 3. _______________ Hint: Divide up the hexagon like a pie into 6 pieces Are they equilateral triangles? 7 M2C1 - Meet #2, Mystery Solutions to Category 1 Mystery Meet #2, November 2003 Answers 1. The third shape sentence shows that a triangle and a circle together have a value of 19. The second shape 1. 17 sentence has the same triangle and circle, but the value of the square has been subtracted, reducing the 19 to 13. 2. 91 The value of the square must be 6. Writing the value 6 in both squares of the first shape sentence, it becomes clear 3. 18 that the triangle must have a value of 2. Finally, returning to the third shape sentence, we see that the circle must have a value of 17. 2. If we combine the first and the third sentences, we will know that Albert and Brendan with Camille and Dana must weigh 121 plus 83 or 204 pounds. We know from the second sentence that the combined weight of Brendan and Camille is 113 pounds. If we subtract this 113 from 204, we get 91 pounds, which must be the combined weight of Albert and Dana. y 3. A regular hexagon can be subdivided into 6 equilateral triangles as shown in the figure below. Since the horizontal distance (3,4) (9,4) between (3,4) and (9,4) is 9 – 3 = 6 units, the side length of the equilateral triangles must be 6 ÷ 2 = 3 units. Thus the perimeter of the hexagon is 3 × 6 = 18 units. x 8 M2C1 - Meet #2, Mystery Category 1 Mystery Meet #2, November, 2002 1. I am a two-digit prime number. The product of my digits is 15. What number am I? 2. Misha bought a candy bar and paid for it with a $50 bill. For change, the cashier gave her three each of three different kinds of bills and three each of four different kinds of coins. What is the price of the candy bar if it is less than five dollars? Express your answer in dollars or cents with the appropriate symbol. 3. Regions A, B, and C in the figure below each contain a different number. The number in each circle is the sum of the numbers in the two adjacent regions. What is the value of the region with the greatest value? A 28 B 31 33 C Answers 1. _______________ 2. _______________ Hint: Make 3 equations such as A+B =28 A+ C=31 3. _______________ Easier: A and B average to 14 because they add to 28. But which is greater and by how much? 9 M2C1 - Meet #2, Mystery Solutions to Category 1 Mystery Meet #2, November, 2002 Answers 1. If the product of the two digits is 15, the digits can only be 3 and 5. The number 35 is composite, but 53 is 1. 53 prime. 2. 77¢ or $0.77 *Or 2¢ or $0.02 2. The cashier will not give Misha three $20 bills in change, so the largest bills could be $10 bills. Also, if we 3. 18 use tens, fives, and two dollar bills, it’s too much and if we don’t use the tens, it’s too little. So the bills must be: $10 + $10 + $10 + $5 + $5 + $5 + $1 + $1 + $1 = $48 That leaves $2.00, or 200¢, for the change in coins and the price of the candy bar. Again, we can eliminate the use of $1 dollar coins and half dollar* coins. The four sets of three coins can only be quarters, dimes, nickles, and pennies, with a value of: 25¢ + 25¢ + 25¢ + 10¢ + 10¢ + 10¢ + 5¢ + 5¢ + 5¢ + 1¢ + 1¢ + 1¢ = 123¢ or $1.23. The candy bar must have cost 200¢ – 123¢ = 77¢ or $2.00 – $1.23 = $0.77. *11/2006 Note: 2¢ also works: 3 half-dollars, 3 dimes, 3 nickels, plus 3 pennies is $1.98. 3. Since the sum of the values in region B and C is two more than the sum of the values in regions A and C, we know that B must be two more than A. If we took two away from B, then A and B would be the same and the sum of their values would be 26 instead of 28. The value of A must be half of 26 or 13. That means B is 15, two more than 13, and C must be 18, since 33 – 15 = 18. To be sure, we should verify that A + C = 31 and indeed 13 + 18 = 31. The largest value is 18. 10 M2C1 - Meet #2, Mystery Category 1 Mystery Meet #2, November, 2001 N 1. A soldier was standing at attention facing North when he was given the W E following sequence of commands in quick succession: Left Right Left face! Left face! Left face! Right face! About face! S Left face! About face! Left face! Left face! If “left face” means turn 90 degrees to the left, “right face” means turn 90 degrees to the right, and “about face” means turn 180 degrees around, which direction was the soldier facing after executing all these commands? 2. Jane, Tara, Peter, and Marina live in different houses and go North to the same school. Use the clues and the graph to find out which point represents Marina’s house. D a. Tara and Peter both live seven blocks east of the school. A b. Peter and Jane walk the same number of blocks to school. B c. Marina likes strawberry ice cream. School C East 3. If the total length of the line is 1 inch at stage 0 and 1 1 inches at stage 1, how many inches 3 long will the line be at stage 4? Express your answer as a mixed number in lowest terms. Stage 0 Stage 1 Stage 2 Answers 1. _____________ 2. _____________ 3. _____________ 11 M2C1 - Meet #2, Mystery Solutions to Category 1 Mystery Meet #2, November, 2001 Answers 1. The soldier faces west, then south, then east, then south, then north, then west, then east, then north, and 1. west finally west. 2. C 2. From clue a, we know that points D and B represent 13 Tara and and Peter’s houses. From clue b, we know that 3. 3 points A and B represent Peter and Jane’s houses. 81 Combining these clues, we can determine that Peter must live at point B. This means Tara must live at D and Jane must live at A. Marina must live at point C regardless of ice cream preference. 3. The length of the line is growing at each stage by a ( 43 ) 2 factor of 4 3 . At stage 2, the line will be = 16 = 1 9 9 7 ( 4) 3 inches long. At stage 3, it will be 3 = 64 27 = 2 10 27 inches long. And at stage 4, it will be ( 4 ) 4 = 256 = 3 13 3 81 81 inches long. 12 M2C1 - Meet #2, Mystery Category 1 Mystery Meet #2, December 2000 1. John has just purchased five 12-foot planks from which he will cut a total of twenty 34-inch boards for a ramp. If we disregard the thickness of the cut, how many total inches of board will be left over when John is done cutting? (Reminder: 1 foot = 12 inches.) 2. Pick a four-digit number and write it down. Now rearrange the digits to form a second four- digit number. Subtract the smaller number from the larger number and divide the result by 9. What is the remainder? 3. Just for today, the symbol ( N ) signifies the value obtained by alternately adding and + − subtracting, from least to greatest, each of the positive factors of N, not including N. For example: + (20)− = +1 − 2 + 4 − 5 + 10 = 8 . Find the value of + ( 36) − . Answers 1. _____________ 2. _____________ 3. _____________ 13 M2C1 - Meet #2, Mystery Solutions to Category 1 Mystery Meet #2, December 2000 Answers 1. John will be able to cut four 34-inch boards from each of the five 12-foot planks. Since 12 feet is 144 inches and 1. 40 4 × 34 = 136 inches, John will have five 8-inch boards left over for a total of 40 inches left over. 2. 0 2. The remainder will be zero regardless of the number − used. For example, start with the four-digit number 3285. 3. 11 Rearrange the digits to form 8523. Now subtracting the smaller number from the larger number we obtain: 8523 − 3285 = 5238 . Now dividing by 9, we get: 5238 ÷ 9 = 582 , remainder 0. In general, any particular digit may move to a different place value and the difference between those place values is itself a multiple of nine. Say the digit d started in the thousands place and moved to the tens place, then we will have the difference 1000d − 10d , which can be rewritten as (1000 − 10 )d = 990d . 990d is clearly a multiple of 9. A similar argument can be made for each of the four digits used. Another explanation involves the divisibility test for nine. Since the sum of the digits will be the same regardless of their order, we can say that the original number and the rearranged number will have the same remainder when divided by nine. When we subtract one number from the other, that remainder will be lost, leaving a number that is divisible by nine. 3. Alternately adding and subtracting the proper factors of 36 from least to greatest, we get: + (36)− = +1 − 2 + 3 − 4 + 6 − 9 + 12 − 18= − 11 14 M2C1 - Meet #2, Mystery

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