The Nautilus Journal of Mathematics Issue I 2012b.pdf

7/27/2019 The Nautilus Journal of Mathematics Issue I 2012b.pdf

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Editors-in-ChiefRikuro Fukusato, Editor of Mathematics

Sophia Oak, Editor of Arts and Operations

Advisor

Mr. James Wong

Staff

Shane BassetDeion Law

Priscilla Luu

Rise MorisatoJerry SuSara Yogi

Joyce Zhang

The Nautilus: Maryknoll Journal of Mathematics

Special thanks to Maryknoll Math Department

Front/Back/Flag Page Art: Sophia Oak

Desktop Publishing Programs: Microsoft Word, The Geometers SketchpadPrinter: Hewlett Parkard Laser Jet 4V

http://www.maryknollschool.org

The nautilus, also known as the living fossil, is a creature that spends most of its life exploringthe great depths of the ocean. It resembles the logarithmic spiral, one that grows without altering itsshape, and was recognized by the great mathematician Jacques Bernoulli as a symbol offortitude andconstancy in adversityas well as growth and renewal. Like the nautilus, Maryknoll students brokeout of their shells and explored the depths of mathematics through challenging monthly problems.

The Nautilus is created by and for students of Maryknoll High School in order to explore therealm of mathematics and to recognize the expression of individuality through problem solving. It isdesigned to showcase student creativity and problem solving ability by publishing the most interestingand elegant solutions. Submissions are accepted monthly from the beginning to end of the month, andfrom September to March.

All original work will not be returned.


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1

1

1

1

11

1

1

1

a = 2

b = 3

c = 4

d = 5

e = 6

f = 7

x = 8

25

20

15

The Nautilus: Maryknoll Journal of Mathematics is a student publication that showcases thebest solutions by Maryknoll students to a set of challenging and non-routine math problems posedby the National Council Teacher of Mathematics and published in the professional journalMathematics Teacher.

Problem 1: Determine the value of x in the spiral diagram.(March 21, 2011)

Solved byJoyce Zhang:

By Pythagoras Theorem, a2 + b2 = c2.12 + 12 = a2 12 + a2 = b2 12 + b2 = c2

2 = a2 12 + ( 2)2 = b2 12 + ( 3)2 = c2

2 = a 3 = b2 4 = c2

3 = b 4 = c

2 = c

12 + c2 = d2 12 + f2 = x2

12 + 2 2 = d2 etc. 12 + ( 7 )2 = x2

5 = d2 8 = x2

5 = d 8 = x

Answer: x = 8

Problem 2: Simplify . (May 29, 2011)

Solved byRikuro Fukusato:

Answer: 32

Problem 3: The rectangle shown in the diagram has width 15 and length 20. Findx, the length ofthe segment perpendicular to the diagonal. (October 28, 2011)

Solved byNathan Goo: In triangles, base1 (height1) = base2 (height2).15(20) = 25x

55

5453

=x

12 = x

Answer: 12


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2

l

a

b

Problem 4: The surface area of a cube is 450. Find the length of an interior diagonal of the cube.[December 19, 2011]

Solved byChloe Fortuna: Let b = length of the interior diagonal,

a = length of the base diagonal, and l = length of the edge.

Answer: 15 units

Problem 5: A light-year is to an astronomical unit (AU) as a mile is to an inch. This rule of thumb

can help us approximate that vast distances that astronomers refer to. (An AU is slightly less thanthe average distance from the Earth to the Sun.) If a light year is 63,240 AU, find the percent errorin the estimate of light-year given by the mile : inch analogy. (September 6, 2011)

Solved byNathan Goo:

1mi 5280ft 12in1mi 1ft

Answer: 0.19%

Problem 6: The midpoints of the sides of the largest square are joined to form a new square. Theprocess is repeated using the new square, and a circle is inscribed in this smallest square. If the areaof the largest square is16/,findtheareaofthecircle.(May2,2011)


Let As1 be the area for the largest square.Let As2 be the area for the middle square.Let As3 be the area for the smallest square.Let ACbe the area for the circle.

As2 = As1 and AS3 = AS2 AS3 = AS1=(16/)

AS3=4/A side of a square is the square root of its area, so the side of S3 and

diameter of the circle is

24or . Therefore, Area of circle AC=(diameter)2

AC=2)

2

2

1(

= 1.

Answer: 1

l = 7 5a = 7 5 + 7 5a = 1 5 0b = 7 5 + 1 5 0b = 15 units

Surface Area SA = 6 l = 450

1 mile = 5280* 12 = 63360 inches


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3

115

115

115

Problem 7: Two thousand patriotic millionaires each decided to contribute 15% of their networth to reduce the national debt. An equally patriotic billionaire decided to match the gifts. Ifeachmillionairesnetworthisexactly1milliondollarsandthebillionairesworthisexactly1billion dollars, what percentage of her net worth did the billionaire need to give to match the gifts?(February 23, 2012)

Solved byFlora Wang:

1 millionaire = 1,000,000

1 billionaire = 1,000,000,000

Answer: 0.3 or 30%

Question 8: Kiri folded a paper square twice. Then she made a single straight cut through allthickness. When she unfolded the square, she saw that she had cut out a square. How might Kirihave folded and cut the paper? (September 8, 2011)


1) Square piece of paper: 2) Fold in half horizontally:

3) Fold in half vertically: 4) Cut across the crease:

Answer:

Problem 9: When Jason arranges his toy soldiers in rows of 6, he has one left over. When hearranges them in rows of 8, he has 3 left over. When he arranges them in rows of 10, he has 5 leftover. What is the smallest possible number of toy soldiers that Jason plays with? (March 2, 2012)

Solved byZacharyKahookele:

Rows of 6 r 1: 1, 7, 13, 19, 25, 31, 37, 43, 49, 55, 61, 67, 73, 79, 85, 91, 97, 103, 109,

Rows of 8 r 3: 3, 11, 19, 27, 35, 43, 51, 59, 67, 75, 83, 91, 99, 107,

Rows of 10 r 5: 5,15, 25, 35 45, 55, 65, 75, 85, 95, 105,

Answer: 115

1,000,000 * 0.15 = 150,0002000(150000) = 300,000,000

= 0.3 or 30%


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4

5

7

8

4

9

1112

136

10

Problem 10: Use each integer from 4 to 13, Solved byJoyce Zhang:inclusive, exactly once to fill the circles sothat each side of the kite has a sum of 29. (February 1, 2012)

Problem 11: The harmonic mean of two integers is 1 less than there arithmetic mean. The largerinteger is twice the smaller integer. What are the two numbers? (January 31, 2009)

Solved byMegan August:

By definition, the harmonic mean is the reciprocal of the average of the reciprocals. Let a, b be the twointegers, where a > b. and a = 2b.The harmonic mean of a and b =

baba

11

2

2

11

1 , which is 1 less than the arithmetic mean

Harmonic Mean:

Arithmetic Mean:

8 = 9 - 1, satisfying the conditions. Therefore, the numbers are 6 & 12.Answer: 6 and 12

.2

ba

12

62

2

a

a

ba

1

2

2

1

2

1

2

bb

bb

1

211

2

ba

ba

22

23

1

2

1

2)2(

b

bb

2

2

2

3

121

2

b

bb

23

)2

3(

4 b

b

233

24 b

b

3233

8)3(

b

b

6

6

698

b

b

bb

.83122

12

32

6

1

12

12

.92

18

2

612

Answer:


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5

1

32 75 2 75 3

751

l

a

b

b - y

y y

a - xx

Y

X

Q(x, y)

B(0, b)

A(a, 0)O

Problem 12: The surface area of a cube is 450. Find the length of an interior diagonal of the cube.(December 19, 2011)

Solution byDevin Rettke:

6squaresarea=450

1squaresarea= = 751 square side = 75

In a cube with side length = 1,

the ratio of side: face diagonal: space diagonal is 1: 2 : 3 .

In a cube with side length = 75,

the ratio of side: face diagonal: space diagonal is 75 1: 75 2 : 75 3 .

Therefore, the space diagonal in a cube with surface area= 450 is 75 3 = 225 = 15

Answer: 15

Problem 13: Line lintersects the positivex- and - andy-axes at A( a, 0) and B (0, b), respectively.Lines parallel to thex- andy-axes are drawn through a pointQ on l, intersecting thex-axis atR andthey-axis atP. Rectangle OPQR and the triangle QAR have the same area. Find the coordinates ofQin terms ofa and b. (PointO is the origin.) (December 2, 2011)

Solution byRikuro Fukusato:

Let (x, y) be the coordinate for Q.

~

Answer: Q: ( a , b)

Area BO A = Area OPQRR + Area QAR + Area BQP

ab = xy + y (a x) + x (b y)ab = 2xy + ay xy + by xyab = ay + bx

Area OPQR = (x) y

AQAR = (a - x) y

xy = (a x) y

x = (a x)

2x = a x3x = a

x = a


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6

70%

30%

Problem 14: The treadmill display indicated that Meg had completed 30% of her workout with 35minutes remaining. How much time had she already spent on the treadmill? (January 13, 2009)

Solved byWensi Paull:

Answer: 15 minutes

Also solved byPriscilla Luu:

Answer: 15 minutes

Problem 15: September 8Problem offered the challenge of cutting out a square from the center ofa folded paper square with a single straight cut. Modify the folding procedure so that you can cut

out a rectangle that is not a square with a single straight cut. (September 16, 2011)


Stage 1: Stage 2: Fold as shown. Stage 3 and 4: Two folds.

Stage 5: Cut as shown. Stage 6: Then unfold the two corners. Stage 7: Unfold again.

Answer:

Problem 16: The sum of two prime numbers is the prime number 89,563. Find the smaller of thetwo addends. (May 3, 2011)

Solved byRikuro Fukusato: Since 89,563 is odd, it is the sum of an even number and an oddnumber. Since 2 is the only even prime number, one of the addend must be a 2, leaving the otheraddend to be (89,563 - 2) = 89,561. The smaller addend is 2.

Answer: 2

100% - 30% = 70% to go = 35 minutes

10% = 5 minutes

100% = 50 minutes50 minutes 35 minutes = 15 minutes completed

7.0

35

3.0

x 0.7x = 10.5x = 15


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7

b c

d

a

54

3

21

Since , , and a=6

Let the length of the mid-size rectangle be c, and its width d.Let the length of the smallest rectangle be b, and its width e.

. Since and b and c are positive integers,

b = 1 and c = 2.

The perimeter is

The sum of the digits is 2+6=8.

Problem 17: On the square geoboard shown below, the distance between adjacent horizontal andvertical pegs is 1 in. What is the area of the 16-gon? (December 16, 2011)

Solution byDawson Zimmermann:

Area of one square = bh = 1 1 = 1 in

Area of five squares = 5 inArea of one triangle = (b*h)2 = (11)/2 = 0.5 in2Area of four triangles = 0.5+0.5+0.5+0.5 = 2 in2

SoArea of total = Area of square+Area of triangle= 7 in

Answer: 7 in

Problem 18: Fill in the cells of the 4X4 magic square with the digits in the number 2011 so thateach row, column, and diagonal contains one 2, one 0, and two 1s. The sum of each row, column,and diagonal will be 4. (March 24,2011)

Solution byJoyce Zhang:

Answer:

Problem 19: Three similar rectangles with integral widths and lengths in the ratio 1: 2 form a non-convex hexagon, as shown. The width of the largest rectangle is 3. Find the sum of the digits of thehexagonsperimeter.(October12th, 2011)


Answer: 8

0 1 1 2

1 2 0 1

2 1 1 0

1 0 2 1


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8

C

D

B

M

N

3

2 1Y

X

N

M

B

D

C

Problem 20: The area of the rectangle shown is 48. The midpoint of is M, and N is 1/3 thedistance from C to B. Compute the area ofDNM. (October 20, 2011)

Solved byBrent Hironaga:

Let x = AB and y = CB .

xy = 48

AreaofRectangle=Area1+Area2+Area3+AreaDMN

So, xy =2

1(2

1x

3

2y) +

2

1(3

1yx)+

2

1(y

2

1x) +DMN

xy =

6

1xy +

6

1xy +

4

1xy +DMN

xy =12

7xy +DMN

12

5xy =DMN

12

5(48) =DMN

Therefore,DMN = 20 Answer: DMN has an area of 20 units2

Problem 21: A 1964 algebra book included the following problem, whichhad appeared in anAmericanalgebratextbookinusenearlyacenturyago:

A pole is one third in the mud, one half in the water, and three feet out ofthe water. What is the length of the pole? (January 7, 2009)

Solved by Nathan Goo:

6

1x = 3

x = 18

Answer: 18 feet

3 feet

x

x


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9

Problem 22: Two pegs on a 2X2 geoboard are connected by a geoband, as shown. Two differentpegs are selected at random and connected by another geoband. If the second band is parallel to thefirst, what is the probability that the 4 pegs are vertices of a parallelogram? (December 12, 2011)

Solution byDeion Law:

4 bands are parallel.

bands will not be aparallelogram

Therefore, will be a parallelogram. Answer: that 4 pegs are vertices of parallelogram

Problem 23: Suppose that f(x) = 1/ (1 +x). Determine all values ofxthat are notin the domain off(f(f(x))). (December 20, 2011)

Solution byDevin Rettke: f(f( f(x))) =(

) Find the domain of f(f( f(x)).

x-2: 11 + 11 + 11 + 2

= 11 + 11 + 11

= 11 + 11 1

= () = . Therefore, 2 . :(

) =

=

.Therefore, 1

.:11 + 11 + 11+ 1.5

= 11 + 11 + 10.5

= 11 + 11+ 2 =1

1 + 11 =11 1 = 10 =

Therefore, 1.5Answer: ,.,

Problem 24: If it takes 12 minutes to cut a log into 4 pieces, how many minutes will it take to cut a

log into 3 pieces? (May 22, 2011) 3 Cuts


Since cutting a log into 4 pieces take 3 cuts, it takes 43

12 minutes for each

cut. Cutting a log into 3 pieces requires 2 cuts, so it would take 42=8minutes.

Answer: 8 minutes


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10

AN

RINI TA

MAHU

x

x

x

x

x x

X

Z

Y

W B

D C

A

Problem 25: The six squares in the diagram are to be folded intoa cube. A humanitarian wants to know which side will be opposite theside labeled AN. Can you help him determine the answer?(October 1, 2011)

Solved byNikka Sonido:

Across of AN is MA when actually applying it and making the box.

It would be MA.

Answer: MA

Problem 26: Compute the digit in the thousands place when calculating the following sum:6 + 66 + 666 + 6666 ++6,666,666,666. (October 10, 2011)


610 = 60

69 + 6 = 60

68 + 6 = 54

67 + 5 = 47

Answer: 7 is in the thousands place.

Problem 27: SquareABCD has points W,X, Y, andZthat are trisection points of the sides of thesquare. W,X, Y, andZare connected to form a rectangle. Find the ratio of the perimeter ofWXYZtothe perimeter ofABCD. (February 26, 2012)

Solution byNathan Goo:

666

6666666

66666666666

666666666666666

666666666+ 6666666666

7400

=

Answer:


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11

d

C = circumference

C

2= 8

5

10

exterior

glass interior

13-3

3

5

13 13

Problem 28: An ant on the outside of a glass, 3 in.from the bottom, sees a drop of honey inside theglass, 5 in. from the top and exactly half way around the glass. The glass is 13 in. high and has acircumference of 16 in. Compute the length of the shortest path for the ant to reach honey.(October 25, 2011)


If we unroll the cylinder and reflect the inside against the rim, the shortest distance between ant andhoney is the hypotenuse d of a triangle with legs C/2=16/2=8 and 13-3+5=15.By Pythagoras Theorem, d2 = 82 + 152. The shortest distance is 17 in.

Answer: 17 in

Problem 29: A bicycle store owner asks one of his employees to count the number of bicycles andtricycles in his store. The employee knows that the owner likes puzzles, so he tells him that thereare a total of 169 wheels but only 152 pedals. How many more bicycles than tricycles are in thestore? (October 30, 2011)

Solved byKristin Yamasaki:152 pedals / 2 pedals per vehicle = 76 vehicles. Solve for a system of equations:Let a = number of bicycles and b = number of tricycles

a + b = 76 -3(a+b = 76) -3a 3b = -228

2a + 3b = 169 2a + 3b = 169 2a + 3b = 169-1a = -59

a = 59Given a=59, substitute it back into the original equation: 2(59) + 3b = 169118 + 3b = 169

3b = 51b = 17

There are 59 bicycles, 17 tricycles. Thus, there are 59-17 = 42 more bicycles than tricycles.

Answer: 42 more bicycles


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12

X-axis

Y-axis

y=0

y=mc

x=c x=9x=3

Problem 30: Letm> 0. Find the value ofc such that the area bounded byy=mx,x=3,x=9, andy=0 isbisected byx=c. (December 5, 2011)

Solution byDeion Law:

Using Calculus,

= = 2 93= = 812 92 = (40.54.5) =36.

Bisected by x = c, so2

1Area = 18m.

= ( 9) =18 9 = 3 6C = 45 = 3 5

Answer: 35Problem 31: A teacher bought 100 puzzles for $82.90 to give out to the members of her mathteam. She bought three different types of puzzles. Type A cost $0.40 each, type B cost $0.70 each,and type C cost $1 each. How many more type C puzzles than type A puzzles did she buy?(October 24, 2011)

Solved byRikuro Fukusato:Let A = Type A puzzle @ $ 0.40; B = Type B puzzle @ $ 0.70; and C = Type C puzzle @ $ 1.00.

A + B + C = 1000.4A + 0.7B + C = 82.90

0.4A + 0.7 (100 C A) + C = 82.90

- 0.3A + .03C = 82.90 70- 0.3A + 0.3C = 12.9

C = 12.9 () + A

C = 43 + A

Problem 32: How many terms are in the expansion of (a+b+c)5 after like terms have beencombined? (May 30, 2011)


Degree Number of Terms0 11 32 6

n2

1x2 +

2

1x, for x = n + 1

52

1(6)2 +

2

16) =

2

42

2

6

2

36 = 21 terms

Answer: 21

Answer: 43 more Type C than As.

Notes1 term since (a + b + c)0 = 13 terms since (a + b + c)1 = a + b + c6 terms since (a + b + c)2 = (a + b + c)(a + b + c), which has

six like terms as a2 + ab + ac + b2 + bc + c2+

x2 + x, where x = n+1,isfortriangularnumbers:1,3,6,


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Problem 33: In a group of 50 girls, each girl is either blonde or brunette, and each has either blueor brown eyes. Find the number of brown-eyed blondes if 14 girls are blue-eyed blondes, 31 arebrunettes, and 18 have brown eyes. (October 9, 2011)

Solved byMitchell Garcia:

50 girls 31 brunettes = 19 blondes19 blondes 14 blue -eyed blondes = 5 brown-eyed blondes

Answer: 5

Also, solved byRikuro Fukusato:

Since a + 31 =50

a = 9

Also, 14 + x = 19

x = 5

Answer: 5

Problem 34: The equation

x2 pxq 0,

q 0, has two unequal roots such that the squaresof the roots are the same as the two roots. Calculate the product

pq. (October 13, 2011)


Let r1 and r2 be the two roots. r12

and r22

are also equivalent.

Since

x2 pxq (x r1)(x r2) 0, foiling gives us:

x2 r1x r2x r1r2 0, or

x2 x(r1 r2) r1r2 0.

r1r2= q, and by substituting r12

and r22, we have:

r1

r2

r1

2r2

2, and

r1

r2

r1

2r2

2 0.

r1

r2

(1 r1

r2

) 0.

Thus,

r1

r2

0 or

r1

r2

1.

Since q=

r1r2 0, r1 r2 must equal 1.

p (r1 r2) (r12 r2

2).

r1 r2 r12 r2

2

.

r1 r2 2r1r2 r12 2r1r2 r2

2

(r1 r2)2(1) (r1 r2)(r1 r2)

(r1 r2)2 (r1 r2)2

0 (r1 r2)2 (r1 r2) 2

0 ((r1 r2) 2)((r1 r2)1) .

(r1 r2) p 2 or

(r1 r2) p 1.

p = 1 or -2.

If p = -2,

x2 pxq x2 2x 1 0 has only one root, so p =1. P = 1 and q = 1.

So, pq = 1(1) = 1.

Answer: 1

Blue Brown Total

Blonde 14 x a

Brunette b c 31

Total d 18 50


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Problem 35: If 121! were written as the product of primes, how many times would 11 appear as afactor? (September 11, 2011)

Solved byJoyce Zhang: By definition, 121!=1234119120121.Amongthefactorsare:

11, 22, 33, 44, 55, 66, 77, 88, 99, 110, 121. Decompose these factors:1 x11, 2 x11, 3 x11, 4 x11, 5 x11, 6 x11, 7 x11, 8 x11, 9 x11, 10 x11, 11 x11Therefore, 11 would appear 12 times in 121!

Answer: 12 times

Problem 36: Three metal spheres with radii in the ratio 3:4:5 are melted and recast into a singlesphere. What is the change (as a percentage) in surface area? (January 24, 2009)


LetA, B, Cbe spheres with radii 3 ,4 ,5 respectively. Let V = Volume and S = Surface Area.

It follows that the new volume Vn = VA + VB + VC= .2883

864

72.0200

144

CBA

n

SSS

S

Vn = 34 (rn)3 Sn=4rn2

2884

3=(rn)3 =4(36) Therefore, the Surface Area

3 216 rn =144 decreased (1-0.72) = 28%.

6 = rn

Answer: 28% decrease

B

C

n

A

VA 4

3rA

3

VA 4

3(27)

VA 36

VB 4

3rB

3

VB 4

3(64)

VB 256

3

VC 4

3rC

3

VC 4

3(125)

VC 500

3

200 CBA SSS


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15

h

10

4

56

10

45

VolumeofWater=456=120

410h=120

h =

Roundtointegers:4y6y = 4 or 5 or 6

If y = 4, x = 2.If y = 5, x = 1.If y = 6, x is not an integer.

So, the positive difference between thetwo numbers is = 9

Answer: 9

Problem 37: A container in the shape of a rectangular solid with dimensions 4 x 5 x 10 is placedso that its height is 10. Water fills the container to a height of 6. The container is then turned so thatthe base dimensions are 4 x 10. Determine the height of the water in the container. (May 7, 2011)


Answer: 3

Problem 38: Find all values ofxsuch that the mean of 77, 137, andxis 1 more than their median.(December 27, 2011)

Solved byChloe Fortuna:

Suppose the median is 77. Then,3

13777 x= 78 x = 20

Suppose the median is 137. Then,3

13777 x= 138 x = 200

Suppose the median is x. Then,3

13777 x= x + 1. 7 7 + 1 3 7 + = 3 + 3 =105.5

Answer: x = 20, 105.5, or 200

Problem 39: Only 2 two-digit integers are equal to three times the product of their digits. Find theabsolute value of the difference of these 2 integers. (October 5th, 2011)


For a two-digit number xy, for positive integers x and y

10x + y = 3(x)yx(10 - 3y) = -y

x y

3y 10 13y109113y101911

3 y

19

3


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16

Problem 40: The product of three positive, consecutive integers is 15 times their sum. Find thethree integers. (March 19, 2012)

Solved byZacharyKahookele:

Product Sum

135=15 1 + 3 + 5 = 991113=12879+11+13=33357=105 3 + 5 + 7 = 15579=315 5 + 7 + 9 = 21 315/21 = 15

Answer: 5, 7, 9

Also solved byCalvin Chang:

Odd integers can always be expressed in the form 2n + 1, where n is any integer.Let 2n + 1, 2n + 3, and 2n + 5 be the three positive, consecutive integer.

(2n + 1)(2n + 3)(2n + 5) = 15((2n + 1) + (2n + 3) + (2n + 5))(4n2 + 8n + 3)(2n + 5) 15(6n + 9)

8n3 + 16n2 + 6n + 20n2 + 40n + 15 90n + 1358n3 + 36n2 + 46n + 15 = 90n + 1358n3 + 36n2 44n120=0 2n3 + 9n2 11n30=0

By Rational Root Theorem, the possible zeros are:

1, 2

1, 2, 3,

2

3, 5,

2

5, 6, 10, 15,

2

15, 30

Testing n = 2 by synthetic division, 2 | 2 9 -11 -304 26 30

2 13 15 | 0 n = 2 is a zero, and a solution.

2n + 1 = 2(2) + 1 = 5; 2n + 3 = 2(2) + 3 = 7; 2n + 5 = 2(2) + 5 = 9.

Checking: (579)=315 15(5 + 7 + 9)= 315 Thus,(579)=15(5+7+9).

Answer: 5, 7, 9

Problem 41: Fivefossilhunterscountedtheirindividualfindsattheendofadayspursuit.The

best mathematician among them noted that the mean number of fossils found was 13, the mode

was 9, and the range was 14. What are all possibilities for the greatest number of fossils found by

one of the five hunters? (September 1, 2011)


Total number of fossils = 13 x 5 = 6565 (n + 2(9) + (n + 14)) = Remainder65 (7 + 18 + 21) = 1965 (8 + 18 + 22) = 1765 (9 + 18 + 23) = 15

Answer: 21, 22, 23


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18

2

2

2

2

2

Problem 45: A puppy weighs of his weight plus lb. How many pounds does the puppy weigh?(October 8, 2011)

Solved byLayne Fujimoto: 1 = of the puppy weighs lb.Let x = weight of puppy.

x = 4(x) = ()4x = 3 pounds

Answer: 3 pounds

Problem 46: A sound barrier along a highway consists of 148 panels. Every 5th panel beginningwith the 4th panel is sand colored. Every 3rd panel beginning with the 5th panel has an oleandershrub plated in front of it. How many of the 148 panels are sand colored but do not have anoleander bush in front? (September 5, 2011)


From the pattern, count the number of sand colored panels

but exclude those with oleander bush in front, there are a

total of twenty panels that fit that description.

Answer: 20

Problem 47: The quadratic equation 2ax2 4ax + a + 1 =0 has two rational roots. If one root isthree times the second root, what is the value of a? (February 24, 2012)

Solved by Rikuro Fukusato:2 4 + + 1 = 02 2 + + 12 = 0

Let the 2 roots be d and e.d = 3e(x - d)(x - e) = 0 + = 0 + ( ) + = 2 + + 12

Answer: 2

+ 12 = = 34-d e = -2

-3e e = -2 = = a=2


21/42


19

4

2

2

4

5 52

1

5

-4

-3

5

4

3

b 2a 28 = 0; a + 2 = 0a = -2

b 2(-2) 28 = 0b = 28 4b = 24

a + b = -2 + 24= 22

Problem 48: How many integers satisfy the following inequality? (December 18, 2011)

3

1

Documents

The Nautilus Journal of Mathematics Issue I 2012b.pdf