Transcript Welcome to the 1st Annual Damien High School Math Competition

WELCOME TO THE 2ND ANNUAL
DAMIEN HIGH SCHOOL
MATH COMPETITION
Schedule
8:00 – 8:30
8:30 – 8:45
8:45 – 9:15
9:15 – 9:30
9:30 – 10:00
10:00 – 10:15
10:15 – 11:45
11:45 – 12:00
Check-In
Greeting from the Principal
Math Medley Exam
Break
Individual Subject Competition
Break
Super Quiz Bowl and Solutions
Results and Awards
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 points may
be given on some problems.
PROBLEM #1
In the famous Fibonacci Sequence, each term is the
sum of the two terms to its left.
a , b , c , d , 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 ?
PROBLEM #1 : SOLUTION
aa ,, b2b ,, -1
c ,, d11 ,, 00 ,, 11 ,, 11 ,, 22 ,, 33 ,, 55 ,, 88 ,, 13
13 ,, 21
21 …
…
-3
Since each term is the sum of the previous two terms,
d can be found by realizing that
d+ 0= 1
c+ 1= 0
so
so
d= 1
c = -1
b + -1 = 1
so
b= 2
a + 2 = -1
so
a = -3
PROBLEM #2
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 ?
PROBLEM #2 : SOLUTION
7 A
6 2
-
B
4 8 9
C 7 3
2
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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 6th
holiday celebrates the 3 wise men giving gifts to baby
Jesus. In some countries, Christmas presents are
exchanged on this day instead of December 25th.
Mr. Geiger also likes January 6th 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 grew a
wonder behold. Through art algebraic, the stone tells how
beard for the next 1/12 of his life. At the end of
old: ‘God gave him his boyhood, one-sixth of his life; One
the following 1/7 of his life, Diophantus got
twelfth more as youth, while whiskers grew rife; And then
married. Five years from then his son was born.
yet one-seventh, ere marriage begun; In five years there
His son lived exactly 1/2 of Diophantus‘ life.
came, a bouncing new son. Alas, the dear child of master
Diophantus
diedattaining
4 yearshalf
after
the deathof of
son.
and sage. After
the measure
hishis
father's
life, chill fate took him. After consoling his fate by the
longfordid
live?
science of How
numbers
fourDiophantus
years, he ended
his life.’”
PROBLEM #4 : SOLUTION
We could use an equation to reflect the several ages of Diophantus:
If we let d = Diophantus’ age at his death, then we get…
1
1
1
1
d +
d +
d + 5 +
d + 4 = d
6
12
7
2
By multiplying both sides by the common denominator of 84, we get :
of this adds
liveddfor +
½ 336 =All84
+ married
+ 42
14 d + 7 d Got
12 d + 420 Son
d
to
his
age
of Diophantus’
Youth
Lasted
Byofcombining
1/6
life
after next 1/7
like
terms, we
of life
75d + 756 = 84d
756 = 9d
Grew beard
for next 1/12
of life
get :
Son born
5 yrs later
life
at death
Died 4 yrs
after son’s
death
d = 84
So Diophantus was 84 years old when he died.
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
5 = 5 - (5  5) + (5  5)
16 = 5 + 5 + 5 + (5  5)
19 = (5 5) - 5 - (5  5)
35 = 5  (5 + (5  5)) + 5
49 = 5  (5 + 5) - (5  5)
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
-8
k + 5 = 3 (Her
s +son’s
5)age
k = 3s + 2
Kate’s age
5 years ago
5 years ago
By the transitive
property, if a = b & a = c , then b = c . So…
Was
Was
5 times
8 less
5s - 20 = 3s + 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
of their ages is 36.
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.
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 2 , 2 , & 9
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 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.
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 :
ABCDE
×4
EDCBA
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 (Dx4)+3 must be odd, so B must be 1.
3
ABCDE
×4
EDCBA
2BCDE
×4
EDCB2
2BCD8
×4
8DCB2
3
21CD8
×4
8DC12
PROBLEM #12 : SOLUTION
• (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
21CD8
×4
8DC12
3
21C78
×4
87C12
3 3
2 1, 9 7 8
×4
8 7, 9 1 2