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					M2C1                                                                 M2C1




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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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