Download Answer: 64

State Math Bowl XXXVII – 2015
Brigham Young University – Hawaii
Round #1, Problem A: (4 points/10 minutes)
All answers must be expressed exactly
√
Simplify (1 + i 3)6
Answer:
64
Solution:
(Quick Solution)
√
(1 + i 3) = (2 cis60◦ )6 = 64 cis360◦ = 64
Solution:
(Brute Force Solution)
Multiply it out completely using the Binomial theorem:
√
√
√
√
√
√
√
(1 + i 3)6 = 1 + 6(i 3) + 15(i 3)2 + 20(i 3)3 + 15(i 3)4 + 6(i 3)5 + (i 3)6
√
√
√
= 1 + 6i 3 − 45 − 60i 3 + 135 + 54i 3 − 27
= 64
State Math Bowl XXXVII – 2015
Brigham Young University – Hawaii
Round #1, Problem B: (5 points/10 minutes)
All answers must be expressed exactly
The Foodland market in Laie paid a total of $67 for mangoes and avocados.
The mangoes were sold at a profit of 20%, but the avocados started to spoil
and ultimately were sold at a 2% loss. The market made an overall profit of
$8.56 on all of these fruit. How much did the market pay for each kind of
fruit?
Answer:
Mangoes are $45 and Avocados are $22
Solution:
Let M be the amount paid for mangoes and A be the amount paid for
avocados.
M + A = 67
0.2M − 0.02A = 8.56
Solving the system yields M = $45 and A = $22
State Math Bowl XXXVII – 2015
Brigham Young University – Hawaii
Round #1, Problem C: (6 points/10 minutes)
All answers must be expressed exactly
In a pirate ship off Laie point, seventeen pirates steal between 500 and 700
gold coins from a wealthy BYU-Hawaii math professor. After dividing the
coins equally, they find 3 coins left over. Hence, a fight occurs over the
extra coins and one pirate dies. With one less comrade, they attempt to
divide the coins equally and find 7 coins left over. Another fight and
another death later, they are finally able to divide the coins equally. How
many coins did the pirates steal?
Answer:
615 coins
Solution:
The number of coins C is a multiple of 15, leaves 7 when divided by 16, and
leaves 3 when divided by 17. We are given that C is between 500 and 700.
The first multiple of 15 in that range is 510, which leaves 14 when divided
by 16 and leaves zero when divided by 17. The next multiple of 15 is 525,
which leaves 13 when divided by 16 and leaves 15 when divided by 17. The
next multiple of 15 is 540, which leaves 12 when divided by 16 and leaves 13
when divided by 17. Notice that with each successive multiple of 15 the
remainder when divided by 16 decreases by 1 and the remainder when
divided by 17 decreases by 2. Five more multiples of 15 later is 615, which
leaves 7 when divided by 16 and leaves 3 when divided by 17. The pirates
stole 615 coins.
State Math Bowl XXXVII – 2015
Brigham Young University – Hawaii
Round #2: (5 points/5 minutes)
All answers must be expressed exactly
Subtracting the squares of 6,278 and 1,443 yields a number known to have
only four prime divisors. Find those four prime divisors.
Answer:
5, 7, 967, and 1103
Solution:
Using the difference of squares formula, the number we start with equals
(6278 + 1443)(6278 − 1443) = 7721 · 4835.
The factors of 7721 are 7 and 1103. The factors of 4835 are 5 and 967. We
are given that the number has only four prime factors, so these numbers
must all be prime. Thus the four prime factors are 5, 7, 967, and 1103.
State Math Bowl XXXVII – 2015
Brigham Young University – Hawaii
Round #3: (6 points/6 minutes)
All answers must be expressed exactly
A modernistic painting consists of triangles, rectangles and pentagons. Two
red roses are drawn within each rectangle and five carnations are drawn
within each pentagon. How many triangles, how many rectangles and how
many pentagons appear in the painting if it contains 40 geometric figures,
153 sides of geometric figures and 72 flowers?
Answer:
T = 13, R = 21, and P = 6
Solution:
Let T be the number of triangles, R be the number of rectangles, and P be
the number of pentagons. Then
T + R + P = 40;
3T + 4R + 5P = 153; and
2R + 5P = 72.
Solving the system yields T = 13, R = 21, and P = 6.
State Math Bowl XXXVII – 2015
Brigham Young University – Hawaii
Round #4: (7 points/7 minutes)
All answers must be expressed exactly
The Aliquot Sum of a positive integer n is the sum of all of the proper
positive divisors of n. For example, the Aliquot Sum of 8 is 7, since 1 + 2 +
4 = 7. The Aliquot Sequence of n is the recursive sequence that starts with
n and in which each successive term is the Aliquot sum of the previous
term. The sequence continues until there is a repetition or until the number
0 is reached. For example, the Aliquot Sequence of 8 is {8, 7, 1, 0}. Find
the Aliquot Sequence of 144.
Answer:
{144, 259, 45, 33, 15, 9, 4, 3, 1, 0}
Solution:
(Brute Force)
n
144
259
45
33
15
9
4
3
1
An
1+2+3+4+6+8+9+12+16+18+24+36+48+72 = 259
1 + 3 + 7 = 45
1 + 3 + 5 + 9 + 15 = 33
1 + 3 + 11 = 15
1+3+5=9
1+3=4
1+2=3
1
0
An interesting fact is that no one knows whether there are numbers that have an infinite Aliquot
sequence. There are several numbers, including 276, 306, 396, 552, 564, etc. for which it is not
known if their Aliquot sequence is infinite or finite.
Solution:
(An for large numbers)
Let s(n) be sum of the divisors of n and let An indicate the Aliquot Sum of n. It follows that
An = s(n) − n. For any prime p and integer a, s(pa ) = 1 + p + p2 + · · · + pa . Next,
p · s(pa ) = p + p2 + · · · + pa + pa+1 . It follows that p · s(pa ) − s(pa ) = pa+1 − 1, which gives
s(pa ) =
pa+1 − 1
p−1
.
The sum of the divisors of a number is multiplicative in nature. For example,
s(a · b · c) = s(a) · s(b) · s(c). Thus, if you know the prime number factorization of a number, it
makes it easier to calculate the Aliquot Sum. Since 144 = 24 · 32 , then
A144 = s(144) − 144 = s(24 )s(32 ) − 144
25 − 1 33 − 1
=
− 144 = 31(13) − 144 = 259
2−1
3−1
State Math Bowl XXXVII – 2015
Brigham Young University – Hawaii
Round #5, Problem A: (Algebra-Plane Geometry/6 points/8 minutes)
All answers must be expressed exactly
Solve for x: 16x = 2(9 · 4x − 16)
Answer:
1
2
,2
Solution:
Let b = 4x . Simplifying b2 = 2(9 · b − 16) yields
b2 − 18b + 32 = (b − 2)(b − 16) = 0.
Thus, b = 2, 16. It follows that x = 12 , 2
State Math Bowl XXXVII – 2015
Brigham Young University – Hawaii
Round #5, Problem B: (Trigonometry-Analytical Geometry/6 points/8 minutes)
All answers must be expressed exactly
A lune is a region bounded by two circular arcs of unequal radii as shown. Starting with the isosceles right
triangle ABC, draw a circular arc with center C and
radius a. Let M be the midpoint of the line segment
AB. Then draw the circular arc with center M and
radius AM . What is the area of the lune?
Answer:
B
M
A
a
a
C
a2
2
Solution:
The area of the triangle ABC is 12 a2 . The area of the quarter of a circle
with center C and radius a is 14 πa2 . Next, the area of the half circle from A
2
1
to B with center at M is 2 π √a2 = 14 πa2 . Putting it all together yields
Area of Lune = Half Circle − (Quarter Circle − Triangle)
1 2
1 2
1 2
πa − a
= πa −
4
4
2
1
= a2
2
State Math Bowl XXXVII – 2015
Brigham Young University – Hawaii
Round #5, Problem C: (Probability-Statistics/6 points/8 minutes)
All answers must be expressed exactly
HPD compiled the following probabilities for New Year’s Eve
P (driving drunk or accident) = 0.35
P (driving drunk) = 0.32
P (driving accident) = 0.09
What is the probability that a driver is drunk given that driver had an
accident on New Year’s Eve? Round to the nearest hundredth.
Answer:
0.67
Solution:
Let D = driving drunk and A = driving accident. It follows that
P (D and A) = P (D) + P (A) − P (D or A) = 0.32 + 0.09 − 0.35 = .06
Thus,
P (D given A) =
P (D and A)
P (A)
=
.06
.09
=
2
3
≈ .67
State Math Bowl XXXVII – 2015
Brigham Young University – Hawaii
Round #6: (9 points/9 minutes)
All answers must be expressed exactly
A box has two square sides opposite of each other, while the rest are
rectangular. It is open on one of the rectangular sides. The sum of the
length of all the edges is 40. The square of its longest spatial diagonal is
equal to the sum of 7 times the length of the box and 2 times the square
side. Find the total surface area of the box.
Answer:
54,
350
9
Solution:
Let the square side = x and the length of the box = y.
8x + 4y = 40
2x + y = 10
y
= 10 − 2x
We also have x2 + y 2 + x2 = 7y + 2x Substitute y = 10 − 2x
x2 + (10 − 2x)2 + x2
x2 + 100 − 40x + 4x2 + x2
6x2 − 40x + 100
6x2 − 28x + 30
3x2 − 14x + 15
x1,2
x1 =
14−4
6
y1 = 10 − 2
=
5
3
=
5
,
3
20
,
3
=
=
=
=
=
=
=
=
x2 =
7(10 − 2x) + 2x
70 − 14x + 2x
70 − 12x
0
0 √
14± 196−180
√6
14± 16
6
14±4
= 53 , 3
6
14+4
6
= 3
y2 = 10 − 2(3) = 4
So the total surface area is
3(3 × 4) + 2(3 × 3) = 54 and
5
5
350
20 5
3
×
+2
×
=
3
3
3
3
9
State Math Bowl XXXVII – 2015
Brigham Young University – Hawaii
Round #7, Problem A: (4 points/15 minutes)
All answers must be expressed exactly
Solve this system of equations. Answer as an ordered pair.
log10 x2 = y + 3
log10 x = y − 1
Answer:
(10000, 5)
Solution:
Since log10 x2 = 2 log10 = y + 3, then y + 3 = 2(y − 1). This leaves y = 5
Thus, log10 x = 4, which gives x = 10000.
State Math Bowl XXXVII – 2015
Brigham Young University – Hawaii
Round #7, Problem B: (6 points/15 minutes)
All answers must be expressed exactly
A train leaves point A two hours behind its scheduled time and runs from
point A to point B at 50% more than its usual rate, arriving on time. If it
had run from A to B at 25 miles per hour (and started on time), it would
have been 48 minutes late. What is the distance from A to B?
Answer:
170 miles
Solution:
Converting 48 minutes to hours yields 48/60 = .8 hours, so a table that
relates the d, r,and t, is
d r
t
d 25 t + 0.8
d 1.5r t − 2
It follows that
rt = 1.5r(t − 2) =⇒ t = 1.5(t − 2) =⇒ t = 6.
We also know that d = rt = 25(t + .8) = 25(6) + 20 = 150 + 20 = 170 miles.
State Math Bowl XXXVII – 2015
Brigham Young University – Hawaii
Round #7, Problem C: (10 points/15 minutes)
All answers must be expressed exactly
If
2
x
−
2
y
= 1 and y − x = 2, evaluate (x + y)2
Answer:
20
Solution:
First, note that 1 =
2
x
−
2
y
=
2y − 2x
xy
(x + y)2 =
=
=
=
=
4
xy
. This means that xy = 4. Thus,
x2 + 2xy + y 2
x2 − 2xy + y 2 + 4xy
(y − x)2 + 4xy
4 + 16 = 20
State Math Bowl XXXVII – 2015
Brigham Young University – Hawaii
Round #8: (10 points/10 minutes)
All answers must be expressed exactly
The ancient Babylonians used the base 60 (sexagesimal) number system, rather than our base 10
r
system. A number written p, q; r in base 60 is equivalent to 60p + q + 60
in base 10. Note that the
semicolon is like a decimal point while the comma separates powers of 60. the following solution
to a quadratic equation of the form x2 + ax = b is explained on an ancient Cuneiform Tablet as
follows:
“Take half of 1, which is 0; 30, and multiply 0; 30 by 0; 30, which is 0; 15. Add this to
14, 30 to get 14, 30; 15. This is the square of 29; 30. Now add 0; 30 to 29; 30, and the
result is 30.”
Given that a and b are integers, what are they? Give your answers in base 10. It might help to
compare this ancient solution with the modern day quadratic formula.
Answer:
a = −1, b = 870
Solution:
The solution to the quadratic given (x2 + ax − b = 0) using the quadratic formula is
√
−a ± a2 + 4b
x=
2
Decimal
We need to translate the key numbers written in sexagesi- Sexagesimal
1
1
mal into decimal form. They are given to the right. Next,
0; 30
30/60 = 1/2
If you write down the equation that yields x = 30 as the
0; 15
15/60 = 1/4 or (1/2)2
solution and compare to the quadratic solution above, we
14, 30
60 · 14 + 30 = 870
can find the value of a and b:
s √
√
1 2
1
1 + 1 + 4 · 870
−a ± a2 + 4b
x = 30 =
+ 870 + =
=
2
2
2
2
The only values that fit are a = −1 and b = 870.
Note also that since the quoted text is the solution only, then any number that seems to appear
out of nowhere has to be either a, −a, b, or −b. 1 is such a number and so is
14, 30 = 60 · 14 + 30 = 870.
Note: Remnants of the ancient Babylonian base 60 number system remain in how we tell time
and in angle degree measure. For the text of the original problem, see the book, “Mathematical
Cuneiform Texts”, by O. Neugebauer and Abraham J. Sachs,
http://www.jstor.org/stable/3608807?seq=2
State Math Bowl XXXVII – 2015
Brigham Young University – Hawaii
Tie-Breaker #1
All answers must be expressed exactly
Kimo went to the Waimanalo Nursery and bought 4 plants; A, B, C and D.
The clerk told him that all 4 had been watered that day. She then
instructed Kimo to have 1 day between watering (water it every 2 days) for
plant A, 2 days between watering for plant B, 3 days between watering for
plant C and 4 days between watering for plant D. Assume that the day of
purchase is day 0. Assuming Kimo follows the instructions exactly, on what
day will Kimo first water all 4 plants?
Answer:
12
Solution:
Let x be the number of days after the purchase of the plants. Plant A will
be watered every multiple of 2 days, plant B every multiple of 3 days, and
plant C every multiple of 4 days. We want to find the lowest common
multiple of 2,3, and 4. That is 12.
State Math Bowl XXXVII – 2015
Brigham Young University – Hawaii
Tie-Breaker #2
All answers must be expressed exactly
The price of a new car was reduced by 40%. When the car still did not sell,
it was reduced by 40% of the reduced price. If the price of the car after
both reductions was $9,360, what was the original price of the car?
Answer:
26,000
Solution:
The first time, the price is reduced to P − 0.4P = 0.6P . The next time the
price is further reduced 40%. So .60P − 0.4(.60P ) = .36P . Thus,
.36P = 9360 =⇒ P = 26,000
State Math Bowl XXXVII – 2015
Brigham Young University – Hawaii
Tie-Breaker #3
All answers must be expressed exactly
If there are 14 standbys waiting for a Hawaiian Airlines flight and only 6
seats available, in how many ways can those 6 people be selected?
Answer:
3003
Solution:
The answer is
14 C6 =
=
14 · 13 · 12 · 11 · 10 · 9
6·5·4·3·2·1
7 · 2 · 13 · 12 · 11 · 10 · 3 · 3
6·5·4·3·2·1
= 7 · 13 · 11 · 3
= 3003