1. If you have not yet registered and received a name tag, please make your way to the front of the Activity Center. Schedule 8:00 – 8:30 Check-In 8:30 – 8:45 Greeting from the Principal 8:45 – 9:15 Math Medley Exam 9:15 – 9:30 Break 9:30 – 10:00 Individual Subject Competition 10:00 – 10:15 Break 10:15 – 11:45 Super Quiz Bowl and Solutions 11:45 – 12:00 Results and Awards

2. This is a group competition in which many problems must be solved through collaboration. Scratch paper is available at your tables, but the final answer for each problem must be written legibly in the box on the provided answer forms. The final solution must be true for all the given clues. Each question is worth 5 points; partial points may be given on some problems.

3. In the famous Fibonacci Sequence, each term is the sum of the two terms to its left. 0, 1, 1, 2, 3, 5, 8, 13, 21 … If we were to continue the sequence TO THE LEFT, what would be the values of a, b, c, & d ? a, b, c, d,

4. a, b, c, d, 0, 1, 1, 2, 3, 5, 8, 13, 21 … Since each term is the sum of the previous two terms, d can be found by realizing that d + 0 = 1 so d = 1 a, b, -1, 1, 0, 1, 1, 2, 3, 5, 8, 13, 21 … a, 2, -1, 1, 0, 1, 1, 2, 3, 5, 8, 13, 21 … -3, 2, -1, 1, 0, 1, 1, 2, 3, 5, 8, 13, 21 … a, b, c, 1, 0, 1, 1, 2, 3, 5, 8, 13, 21 … c + 1 = 0 so c = -1 b + -1 = 1 so b = 2 a + 2 = -1 so a = -3

5. If A, B, and C are digits in the following 3 digit subtraction problem : 7 A 2 - 4 8 B _______________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________ C 7 3 What are the values of A, B, & C ?

6. 7 A 2 - 4 8 B __________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________ C 7 3 7 A 2 - 4 8 9 __________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________ C 7 3 7 6 2 - 4 8 9 __________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________ C 7 3 7 6 2 - 4 8 9 __________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________ 2 7 3 Since we start on the right when subtracting, we can see that nothing subtracts from 2 to get 3. So we must have borrowed 10 from the 10’s column. The only number that subtracts from 12 to get 3 is 9. The only number minus 8 that gives 7 is 15. A can’t be fifteen, but it could be 5, if we borrowed from the 100’s column. But remember we also borrowed from the 10’s in the first step, so A must be 5+1 or 6. Since we just borrowed from the 100’s column in the last step, the 7 became a 6. So 6 – 4 is 2.

7. Mr. Geiger loves the feast of Epiphany. The January 6 th holiday celebrates the 3 wise men giving gifts to baby Jesus. In some countries, Christmas presents are exchanged on this day instead of December 25 th. Mr. Geiger also likes January 6 th because it’s his birthday! Damien would like to give you a gift to celebrate the upcoming Christmas season and to thank you for competing.

8. The 7 pieces in each bag is what’s known as a Tangram, an ancient Chinese puzzle game. Use all the pieces in a bag (one color per puzzle) to build the following shapes: Gaspar, Melchior, Belthasar (the 3 wise men) and a camel. Use the papers provided as a stencil.

10. We know very little about the life of the mathematician Diophantus, often known as the 'father of algebra‘, except that he came from Alexandria and he lived around the year 250 AD. However, there remains a riddle that describes the spans of Diophantus's life. Diophantus‘ youth lasted 1/6 of his life. He grew a beard for the next 1/12 of his life. At the end of the following 1/7 of his life, Diophantus got married. Five years from then his son was born. His son lived exactly 1/2 of Diophantus‘ life. Diophantus died 4 years after the death of his son. How long did Diophantus live? Translated into simpler English, it says: Diophantus’ Epitaph read: “Here lies Diophantus’, the wonder behold. Through art algebraic, the stone tells how old: ‘God gave him his boyhood, one-sixth of his life; One twelfth more as youth, while whiskers grew rife; And then yet one-seventh, ere marriage begun; In five years there came, a bouncing new son. Alas, the dear child of master and sage. After attaining half the measure of his father's life, chill fate took him. After consoling his fate by the science of numbers for four years, he ended his life.’”

11. By combining like terms, we get : By multiplying both sides by the common denominator of 84, we get : We could use an equation to reflect the several ages of Diophantus: If we let d = Diophantus’ age at his death, then we get… So Diophantus was 84 years old when he died. d 4 d 2 1 5 d 7 1 d 12 1 d 6 1  Youth Lasted 1/6 of life Grew beard for next 1/12 of life Got married after next 1/7 of life Son lived for ½ of Diophantus’ life Son born 5 yrs later Died 4 yrs after son’s death All of this adds to his age at death ddddd843364242012714  dd8475675  d9756  84  d

12. A ribbon is tied tightly around the earth at the equator. If you were to pull the ribbon 1 ft. above the equator everywhere around the earth (like a halo)… How much more ribbon would you need to buy? Note : You do not need to know the radius or circumference of the earth to solve this problem. Just recall that Circumference = 2 π (radius)

13. Extra Necessary Ribbon Earth r r Ribbon pulled tightly around EarthRibbon pulled 1 foot above equator r + 1 Length of Ribbon = Circumference of equator = 2 π r Length of entire ribbon pulled away from earth= 2 π (r + 1) Length of extra ribbon needed to connect the ends is … 2 π (r + 1) - 2 π r = 2 π feet = 2 π r + 2 π - 2 π r

16. 5 years ago Kate was 5 times as old as her son. 5 years from now, Kate’s age will be 8 less than three times the corresponding age of her son (5 years from now). (Hint : Let k=Kate’s age & s= Son’s age) Find their ages (in years).

17. By the transitive property, if a = b & a = c, then b = c. So… Kate’s age 5 years ago Was 5 times Her son’s age 5 years ago  555  sk 205  sk  8535  sk Kate’s age 5 years from now Was 8 less than 3 times Her son’s age 5 years from now 23  sk 23205  ss222  s Son’s Age = 11 20)11(5  k Kate’s Age = 35

18. Thomas : Peter, how old are your children? Peter : Well, there are three of them and the product of their ages is 36. Hint : Every one of Peter’s responses is important and all ages are whole numbers! How old were each of Peter's children? Here is a conversation between two friends that takes a little math and a lot of logic to decipher the solution: Thomas : That is not enough... Peter : Well then, the sum of their ages is exactly the number of sodas we have drunk today. Thomas : That is still not enough. Peter : OK; the last thing is that my oldest child wears a red hat.

19. Since Peter says that the product of his kids ages is 36, the EIGHT possibilities are: But as Thomas noticed, he needed more information. So Peter said the sum of their ages is the number of sodas they drank. Weird, because we don’t know how many sodas they drank. The sums are : Now we must assume Thomas knew how many sodas they drank. What’s most important at this point is that Thomas STILL didn’t know the ages, which implies there was more than one group of ages with the same sum. Either 1+6+6 =13 or 2+2+9=13. 1*1*361*2*18 1*3*121*4*9 1*6*62*2*9 2*3*63*3*4 1+1+36=381+2+18=21 1+3+12=161+4+9=14 1+6+6=132+2+9=13 2+3+6=113+3+4=10 So what’s up with the red hat !?!?! Well if Peter has an oldest child, then his oldest MUST NOT BE A TWIN. So his kids are 2, 2, & 9

20. The 7 pieces in each bag is what’s known as a Tangram, an ancient Chinese puzzle game. Use all the pieces in a bag (one color per puzzle) to build the following shapes: angel, 2 different Advent candles, and a Christmas box. Use the papers provided as a stencil.

22. An eccentric professor used a unique way to measure time for a test lasting 15 minutes. He used just two hourglasses. One measured 7 minutes and the other 11 minutes. During the whole time he turned the hourglasses only 3 times. How did he measure the 15 minutes?

23. When the test began, the professor started both hourglasses running. When the 7min hourglass ran out, he turned it around. 4 minutes later, the 11min hourglass ran out, and he promptly turned the 7min hourglass around again, so the 4 min ran back again. 11+4=15, and the test was over.

24. Using the numerals 1, 9, 9 and 6, mathematical symbols +, -, x,, root and brackets, create the following numbers: 29, 32, 35, 38, 70, 73, 76, 77, 100 and 1000. All the numerals must be used in the given order (each just once) and without turning upside down.

26. Assuming A, B, C, D, & E each represent different digits between 0 and 9, find the digits such that : A B C D E × 4 E D C B A Hint : Remember to carry your tens digits.

27. Since EDCBA must be a five digit number, A must be either 1 or 2; but it can’t be 1, since all multiples of 4 are even. 2 B C D E × 4 E D C B 2 3 2 B C D 8 × 4 8 D C B 2 A B C D E × 4 E D C B A So E must be 3 or 8 (which make 12 or 32). But E can’t be 3, since no five digit number times 4 is less than 40,000 (product can’t begin with 3). 3 2 1 C D 8 × 4 8 D C 1 2 Since EDCBA begins with 8, B × 4 must be either 1 or 0. But (Dx4)+3 must be odd, so B must be 1.

28. (Dx4)+3 can only end in 1, if D is 2 or 7. But 2 has already been used as A. So D must be 7. The only way to get 4x1 to be 7, is if there is a carried 3. Only 4x8 or 4x9 begin with 3. Since 8 has already been used as E, C must be 9. 3 2 1 C D 8 × 4 8 D C 1 2 3 3 2 1, 9 7 8 × 4 8 7, 9 1 2 3 2 1 C 7 8 × 4 8 7 C 1 2