WELCOME TO THE 2 ND ANNUAL DAMIEN HIGH

WELCOME TO THE 2 ND ANNUAL DAMIEN HIGH SCHOOL MATH COMPETITION 2011 If you have not yet registered and received a name tag, please make your way to the front of the Activity Center.

SUPER QUIZ BOWL • 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

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

PROBLEM #1 : SOLUTION a , b , c , d , a , 2 , -1 , 1 13 , , 21 21 … … -3 1 , , 00 , , 11 , , 22 , , 33 , , 55 , , 88 , , 13 … Since each term is the sum of the previous two terms, d can be found by realizing that d + 0 = 1 so d = 1 c + 1 = 0 so c = -1 b + -1 = 1 so b = 2 a + 2 = -1 so a = -3

PROBLEM #2 : SOLUTION 7 A 2 7 6 2 - B 9 4 8 9 __________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________ C 7 3 3 2 7 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.

PROBLEM #3 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.

PROBLEM #3 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.

PROBLEM #3 : SOLUTION

PROBLEM #4 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. Translated into simpler English, it says: Diophantus’ Epitaph read: “Here lies Diophantus’, the Diophantus‘ youth lasted 1/6 of his life. He wonder behold. Through art algebraic, the stone tells grew a beard for the next 1/12 of his life. At how old: ‘God gave him his boyhood, one-sixth of his the end of the following 1/7 of his life, life; One twelfth more as youth, while whiskers grew Diophantus got married. Five years from then rife; And then yet one-seventh, ere marriage begun; In his son was born. His son lived exactly 1/2 five years there came, a bouncing new son. Alas, the of Diophantus‘ life. Diophantus died 4 years dear child of master and sage. After attaining half the after the death of his son. measure of his father's life, chill fate took him. After consoling his fate by the science of numbers for four

PROBLEM #4 : SOLUTION We could use an equation to reflect the several ages of Diophantus: 1 If we let d = Diophantus’ age at his death, then we get… 1 1 1 d + d + 5 + d + 4 = d 6 12 7 2 By multiplying both sides by the common denominator of 84, we get : 14 d + 7 d Got married + 12 d + 420 Son lived for ½ + 42 d + 336 =All of this adds Youth 84 d of Diophantus’ Lasted after next 1/7 life By combining like terms, we get : 1/6 of life 75 d + 756 = 84 d 756 = 9 d Grew beard for next 1/12 of life Son born 5 yrs later to his age at death Died 4 yrs after son’s death d = 84 So Diophantus was 84 years old when he die

PROBLEM #5 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)… 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) How much more ribbon would you need to buy?

PROBLEM #5 : SOLUTION Ribbon pulled tightly around Earth Ribbon pulled 1 foot above equator Extra Necessary Ribbon r r Earth r + 1 Earth 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 r + 2 π - 2 π r = 2 π feet

PROBLEM #6

PROBLEM #6 : SOLUTION

PROBLEM #7 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).

PROBLEM #7 : SOLUTION k - 5 = 5 (s - 5) k = 5 s - 20 k + 5 = 3 (Her son’s age s + 5) - 8 k = 3 s + 2 Kate’s age 5 years ago By the transitive property, if a = b & a = c , then b = c. So… Was 5 times Was 8 less 5 s - 20 = 3 s + 2 than 3 times Kate’s age 5 years from now 2 s = 22 Her son’s age 5 years from now Son’s Age = 11 k = 5 (11) - 20 Kate’s Age = 35

PROBLEM #8 Here is a conversation between two friends that takes a little math and a lot of logic to decipher the solution: Thomas : Peter, how old are your children? Peter : Well, there are three of them and the product Thomas : That is not enough of their ages is 36. . Peter : Well then, the sum of their ages is exactly th 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. Hint : Every one of Peter’s responses is important and all ages are whole numbers! How old were each of Peter's children?

PROBLEM #8 : SOLUTION Since Peter says that the product of his kids ages is 36, the EIGHT possibilities are: 1*1*36 1*2*18 1*3*12 1*4*9 1*6*6 2*2*9 2*3*6 3*3*4 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 : 1+1+36=38 1+2+18=21 1+3+12=16 1+4+9 =14 1+6+6=13 2+2+9 =13 2+3+6=11 3+3+4=10 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. 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

PROBLEM #9 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.

PROBLEM #9 : SOLUTION

PROBLEM #10 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?

PROBLEM #10 : SOLUTION When the test began, the professor started both hourglasses running. When the 7 min hourglass ran out, he turned it around. 4 minutes later, the 11 min hourglass ran out, and he promptly turned the 7 min hourglass around again, so the 4 min ran back again. 11+4=15, and the test was over.

PROBLEM #11 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.

PROBLEM #11 : SOLUTION

PROBLEM #12 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.

PROBLEM #12 : SOLUTION • 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. • 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). • Since EDCBA begins with 8, B × 4 must be either 1 or 0. But (Dx 4)+3 must be odd, so B must be 1. A B C D E × 4 E D C B A 2 B C D E × 4 E D C B 2 3 2 B C D 8 × 4 8 D C B 2 2 1 C D 8 × 4 8 D C 1 2

PROBLEM #12 : SOLUTION • (Dx 4)+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 4 x 1 to be 7, is if there is a carried 3. Only 4 x 8 or 4 x 9 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 2 1 C 7 8 × 4 8 7 C 1 2 3 3 2 1, 9 7 8 × 4 8 7, 9 1 2