UNIVERSITY OF NORTHERN COLORADO

MATHEMATICS CONTEST

First Round

For all Colorado Students Grades 7-12

November 3-6, 2011

You have 90 minutes- no calculators allowed

• A regular hexagon has six sides with equal length and six angles with equal measure.

• The positive integers are 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, …

1. A 4 x 9 cardboard rectangle is cut up and the pieces rearranged, without gaps or overlap, to

form a square. What is the perimeter of that square?

2. Solve for N: 23 x 54 x 72 = 250 x N

3. In triangle ABC, side AB has length 6, side BC has length 5, side AC

has length 7. Segment CD is perpendicular to AB and point D divides

segment AB into two pieces. What is the length of the longer piece?

4. The ones digit in the number 24 = 16 is 6.

a. What is the ones digit in the number 26 ?

b. What is the ones digit in the number 28 ?

c. What is the ones digit in the number 22011 ?

5. The hexagon ABCDEF has one internal angle greater

than 180 degrees, angle BCD. What is the largest number of

internal angles greater than 180 degrees that any single

hexagon can contain?

6. Find the shortest distance from point A to point B, measured on the curved

surface of the cylinder. Segment PQ is a diameter of the circular base, and

the base has circumference 6 centimeters. Point A is 2 centimeters above

point P. Point B is 6 centimeters above point Q.

Over

B

P

A

Q

BA D

C

BC

DE

F

A

7. A drawer contains 24 utensils: one knife, one fork, and one spoon, each in 8 different colors. If

you pull items at random from the drawer without looking, what is the smallest number of items

you must take to be certain to have pulled out a complete matching table setting, containing a

knife, fork, and spoon of the same color?

8. Find the product of all of the positive integers n that satisfy the following inequality.

n < 12 < n + 17 < 2n + 10 < n2 ! 51

9. Square Meal You want to eat a lump of cookie dough in stages. A cookie

press converts the dough into a square of uniform thickness. On day 1 you

divide the square into 4 equal smaller square pieces, using a 2x2 grid, then

eat one of these 4 pieces. On day 2 you press the remaining dough into a

new square, subdivide it using a 3x3 grid, and eat one of these 9 pieces.

Continue pressing, subdividing, and eating pieces of the remaining dough.

What fraction of the original lump remains immediately after the 100th

meal? Give your answer as a fraction c /d, expressed in lowest terms.

10. Treasure Chest You have a long row of boxes.

The 1st box contains no coin. The next 2 boxes each contain 1 coin. The next 4 boxes each

contain 2 coins, the next 8 boxes each contain 3 coins, the next 16 boxes each contain 4 coins,

and so on. (The number of boxes that contain N coins is twice the number of boxes that contain

(N-1) coins.)

(a) How many coins are in the 100th

box?

(b) How many coins are there when the contents of the first 100 boxes are combined?

11. Hex Consider the sequence of honeycomb-shaped figures below. The first figure has

one cell and is made of 6 line segments. The second figure has 7 cells and is made

of 30 line segments. How many line segments are there in the 20th figure? (The next

page is a sheet of paper tiled in hexagons for your use in considering this problem.)

First figure Second figure Third figure

over

UNIVERSITY OF NORTHERN COLORADO

MATHEMATICS CONTEST

FINAL ROUND

For Colorado Students Grades 7-12

January 21, 2012

You have three hours. No calculators are allowed. Show your work for each problem

on pages behind your answer sheet. Your score will be based on your answers and

your written work, including derivations of formulas you are asked to provide.

• The positive integers are 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, …

• An ordinary die is a cube whose six faces contain 1, 2, 3, 4, 5, and 6 dots.

1. (a) What is the largest factor of 180 that is not a multiple of 15?

(b) If satisfies , then what is the largest perfect square

that is a factor of ?

2. Four ordinary, six-sided, fair dice are tossed. What is the probability that the sum of the

numbers on top is 5?

3. Mrs. Olson begins a journey at the

intersection of Avenue A and First Street in

the upper left on the attached map. She ends

her journey at one of the Starbucks on

Avenue D. There is a Starbucks on Avenue

D at every intersection from First Street

through Sixth Street! If Mrs. Olson walks

only East and South, how many different

paths to a Starbucks on Avenue D can she

take? Note that Mrs. Olson may pass one

Starbucks on her way to another Starbucks

farther to the East.

4. (a) What is the largest integer for which is divisible by ?

(b) For how many positive integer values of is divisible by ?

5. What is the remainder when is divided by seven?

6. How many 5-digit positive integers have the property that the product of their digits is 600?

7. A circle of radius 1 is externally tangent to a circle of radius 3

and both circles are tangent to a line. Find the area of the shaded

region that lies between the two circles and the line.

8. An ordinary fair die is tossed repeatedly until the face with six dots appears on top. On

average, what is the sum of the numbers that appear on top before the six? For example, if

the numbers 3, 5, 2, 2, 6 are the numbers that appear, then the sum of the numbers before the

six appears is . Do not include the 6 in the sum.

9. Treasure Chest . You have a long row of boxes. The 1st box contains no coin. The next

2 boxes each contain 1 coin. The next 4 boxes each contain 2 coins. The next 8 boxes each

contain 3 coins. And so on, so that there are boxes containing exactly coins.

(a) If you combine the coins from all the boxes that contain 1, 2, 3, or 4 coins you get 98

coins. How many coins do you get when you combine the coins from all the boxes that

contain 1, 2, 3, …, or coins? Give a closed formula in terms of . That is, give a

formula that does not use ellipsis (…) or summation notation.

(b) Combine the coins from the first boxes. What is the smallest value of for which

the total number of coins exceeds 20120? (Remember to count the first box.)

10. An integer equiangular hexagon is a six-sided polygon whose side

lengths are all integers and whose internal angles all measure

120 degrees.

(a) How many distinct (i.e., noncongruent) integer equiangular

hexagons have no side length greater than 6? Two such hexagons

are shown.

(b) How many distinct integer equiangular hexagons have no side

greater than ? Give a closed formula in terms of .

(A figure and its mirror image are congruent and are not

considered distinct. Translations and rotations of one another are

also congruent and not distinct.)

11. Construct a 4th

degree polynomial that meets as many of the

following conditions as you can: The sum of the roots is 1, the sum of the squares of the roots

is 2, the sum of the cubes of the roots is 3, and the sum of the 4th

powers of the roots is 4.

UNC MATH CONTEST SOLUTIONS

FINAL ROUND JANUARY 2012

(1.A.) 36

180= 2 x 2 x 3 x 3 x 5. Factors of 180 are produced by selectingsubsets of these prime factors: 2 x 2 = 4 and 2 x 3 x 5 = 30, forinstance. Any factor that is not a multiple of 15 must leave outeither the 5 or both of the 3s. The largest factor is found byleaving out the 5. That factor is 2 x 2 x 3 x 3= 36. This questionand 1.B extend Problem 2 from the First Round.

(1.B) 196 = 142

Factor 99000 into prime factors, so that the equation becomes25 ! 32 ! 54 ! 73 ! 11 = 23 ! 32 ! 53 ! 11! N. Cancel com-mon factors and deduce 22 ! 5! 73 = N. The largest perfectsquare factor of N is constructed by taking each prime factorto the largest available even power. You get 22 ! 72 = 142 =196.

(2.) 1/324

To get a sum of 5 you must get three 1s and one 2. There are fourdifferent ways to get this combination: 2111, 1211, 1121, and1112. Each of these has probability 1/64, so the total probabilityis 4/64 = 1/324.

(3.) 126

Solution (i). Starting in the upper left corner, begin tabulatingthe number of pathways that reach intermediate destinationson the grid. Each path that ends at an intermediate destination

must approach its destination from a neighboring intersectionthat is one block away either north or west. Therefore, the num-ber of paths to any intersection is the sum of the numbers ofpaths to these two nearest neighboring intersections. This is ex-actly the rule for generating Pascal’s triangle, as you can see inthe diagram. (Not all path counting problems will be this regu-lar.) Take the sum of all such path-numbers along the Starbuckson Avenue D: 1 + 4 + 10 + 20 + 35 + 56 = 126.

FIGURE 1. Path-counting, step by step

Solution (ii). (33) + (4

3) + (53) + (6

3) + (73) + (8

3) = 126. Justification.Consider first a typical path, described as a list of eastward andsouthward steps. The path SSEESEE goes to the Starbucks onFifth Street. All paths going to this Starbucks will have sevenletters and exactly four of those letters will be E’s and three willbe S. The number of such paths is (7

3). In general, to reach theintersection of Avenue D and Nth Street, going only east andsouth, a path must go N " 1 blocks east and 3 blocks south, orN + 2 blocks in all. Out of these N + 2 blocks, exactly 3 will besouth. The number of choices will be (N+2

3 ). Compute this forN = First, Second,. . ., Sixth Streets, and add.

1

2 FINAL ROUND JANUARY 2012

Solution (iii) Here is a solution that counts the paths to all theStarbucks on Avenue D at once. Imagine a rope that followsMrs. Olson’s path from the starting point at the intersectionof Avenue A and First Street to her destination Starbucks onAvenue D, and then continues eastward to the lowest right corner ofthe diagram: the intersection of Avenue D and Sixth Street. The ropemust always travel eight blocks. Place a knot at each intersec-tion, including the start and end. No matter which path Mrs.Olson takes, there will be exactly nine knots on the rope. Thereare three special knots on the rope where Mrs. Olson makes thedecision to head south to the next Avenue after passing throughan intersection, and there is a final fourth knot where she de-cides to stop, somewhere on Avenue D. Mark these four specialknots X. For any rope that has four of its nine knots marked X,there will be one and only one path Mrs. Olson can choose thatwill correspond to this marking of the knots. Therefore, Mrs.Olson has exactly (9

4)=126 paths from which to choose.

!"#$%&'(&")*&+,-*.&/012&+,-*.&/31

4

4

4

4

56%&5

56%&7

56%&8

56%&9

.:;&!; +(<&!; -=<&!; >;?&!; @;?&!; A;?&!;

!;BC

FIGURE 2. Nine knots with four X’s

Solution (iv) Here is another solution that counts all the pathsat once. We discuss this solution at length because it will playa role in the solution of Problem 10. Let W, X, Y and Z be the

street numbers of the locations of the four knots in solution (iii)with 1 # W # X # Y # Z # 6. We want to count the numberof choices we have for the ordered list W, X, Y, Z. Imagine arow of rooms 1, 2, 3, 4, 5, and 6 with five walls dividing them:

#1 #2 #3 #4 #5 #6

FIGURE 3. Rooms with walls

Think of the letters as balls that will be dropped into the roomscorresponding to their values. Several balls can go into a singleroom and some rooms may be empty. For the example abovewith W = 1, X = 1, Y = 3, and Z = 5 we get

oo o o

#1 #2 #3 #4 #5 #6

FIGURE 4. Balls with walls

We can use the shorthand o o | | o | | o |. Each solutionW, X, Y, Z corresponds to a string of nine symbols, four of whichare balls o and five of which are walls |. Therefore, there are(9

4)=126 solutions in all. That is, the number of ways to pick anordered list of four numbers in nondecreasing order W # X #Y # Z from the set S = {1, 2, 3, 4, 5, 6} is (9

4). More generally,the number of ways to choose four numbers in nondecreasingorder W # X # Y # Z from the set S = {1, 2, 3, 4, 5, 6, . . . n}is (n+3

4 ). This counting technique is sometimes called ”balls andwalls”, ”stars and bars”, or ”sticks and stones.”A handful of students interpreted the problem to say that Mrs.Olson could pass at most one Starbucks on her way to her fi-nal destination Starbucks. So interpreted, the problem becomessomewhat harder; and full credit was given for correct solu-tions to this variant interpretation, whose answer is 91 paths.

UNC MATH CONTEST SOLUTIONS 3

(4.A) 289

Use long division to get

(n3 + 1631)(n + 11)

= (n2 " 11n + 121) +300

(n + 11).

The polynomial n2 " 11n + 121 is an integer for all integer val-ues of n. The largest n for which 300

(n+11) is integer is the n thatmakes the denominator 300: n = 300" 11 = 289.

(4.B) 11

Each factor m of 300 that satisfies m $ 12 will produce a posi-tive integer solution n = m" 11. Count all the factors of 300,and then discard the seven small factors 1, 2, 3, 4, 5, 6, 10 that aresmaller than 12. To find all the factors of 300, use the prime fac-torization 300 = 22 ! 31 ! 52. Each factor of 300 can be writtenas 2r ! 3s ! 5t with r = 0, 1, 2; s = 0, 1; and t = 0, 1, 2. Thereare 3! 2! 3 = 18 choices for these powers, hence 18 factors of300. After discarding the seven factors that are too small, elevenfactors remain.

(5.) 0

Strategy: Look for cyclic patterns in the powers. This can bedone by computing powers of 11 and 12, or more efficiently byfirst reducing these mod 7 and then computing powers: 12 %5 mod 7, hence122011 % 52011mod 7; and similarly 112012 %42012mod 7.

Now tabulate the powers working mod 7, and look for cyclicpatterns:

k = 1 2 3 4 5 6 7 8 9 10 11 1212k % 5 4 6 2 3 1 5 4 6 2 3 111k % 4 2 1 4 2 1 4 2 1 4 2 1

The table reveals that powers of 12 and 11 repeat with periodsof length 6 and 3 respectively. Thus each power can be reducedby removing whole multiples of its period. Reduce the power2011 % 1 mod 6, and deduce 122011 % 52011 % 51 % 5 mod 7.Similarly, reduce 2012 % 2 mod 3, and deduce 112012 % 42012 %42 % 2 mod 7 . Therefore 122011 + 112012 % 5 + 2 = 7 % 0 mod 7.This problem is an extension of Problem 4 on the First Round.

6. 210

Factoring into primes, 600 = 2! 2! 2! 3! 5! 5. These sixprimes must be placed in five digit places, and some primesmust share a place. Clearly both 5s must be alone, occupy-ing two places. Now use trial and error to place the other fourprimes in three places. The possible unordered lists of five dig-its are found to be

(i) 2,3,4,5,5; (ii) 1,8,3,5,5; (iii) 1,4,6,5,5; and (iv) 2,2,6,5,5.

Now count the possible orderings of the digits in each list. Tocount the orderings of list (i)=2,4,3,5,5, first choose two slots forthe 5s: there are 10 ways. Then rearrange the remaining threedigits 3!=6 different ways. Thus there are 10! 6 = 60 different5 digit numbers whose digits are 2, 3, 4, 5 and 5. Similarly thereare 60 orderings of list (ii) and 60 orderings of list (iii). List (iv)has only 10! 3=30 distinct orderings, because of the repeated2. The total is 210.

7. 4&

3" 11!6

Draw a trapezoid ADFE by dropping perpendicular feet fromthe two centers A and D to the points E and F on the tangentline. The area sought is found by removing two circular sectorsfrom the trapezoid. Note that because AD has length 4, DG haslength 2, and the triangle is right, we see that triangle ADG isa 30-60-90 triangle. Therefore AG has length 2

&3. The area of

the trapezoid is the average of AE and DF times the distancebetween two tangent points on the line, or 4

&3. Angle GAD is

4 FINAL ROUND JANUARY 2012

!

1

1

32

G

E F

A

D

!

FIGURE 5. Trapezoid minus two sectors

30 degrees, so angle EAD is 30+90=120 degrees and the sectorin the smaller circle is one third of that circle. It has area !

3 .Angle GDA is 60 degrees and the sector in the larger circle isone sixth of that circle, hence has area 3!

2 .

Desired area= 4&

3" !3 "

3!2 =4

&3" 11!

6 .

8. 15

On any single roll, 5/6 of the time you win some points. Onaverage this win W adds (1 + 2 + 3 + 4 + 5)/5 = 3. On theother hand, 1/6 of the time you roll a losing value L = 6 thathalts the game. Now let S be the average sum of all possiblestrings of rolls. Below is a table of all possible strings. Each Win a string contributes 3 points to its average value.

String Average Value ProbabilityL 0 1

6WL 3 5

6 !16

WWL 6 ( 56 )2 ! 1

6WWWL 9 ( 5

6 )3 ! 16

. . . . . . . . .

Taking the sum of these values, weighted by the probability ofeach case, gives S = (3 ! 1

6 ) !"k=1 kpk where p = 5

6 . This is

computable by the methods used in Problem 9 below, but thereis also a shortcut method to do this problem:

Note that the table contains a duplicate of itself after the firstroll. This means that 5/6 of the time the first roll is a win thatadds 3, and then we get to start rolling all over again. ThusS = (1/6)(0) + (5/6)(3 + S), which implies S = 15.

(9.a) (N " 1)2N+1 + 2

0

1 1

2 2 2 2

3 3 3 3 3 3 3 3

4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4

FIGURE 6. Coins in boxes

Let TN = 1 · 2 + 2 · 4 + 3 · 8 + 4 · 16 + . . . + N · 2N be the totalnumber of coins in all the boxes in all the rows up to the rowwith N coins per box. Let SN = 1 + 2 + 4 + 8 + 16 + . . . + 2N bethe total number of boxes, including the box at the top with zerocoins in it. The question asks for a formula for TN . First get aformula for SN , which is a geometric series, by using a standardmethod: investigate the effect of doubling the sum.

SN = 1 + 2 + 4 + 8 + 16 + . . . + 2N

2 · SN = 2 + 4 + 8 + 16 . . . + 2N + 2N+1

Subtract the first equation from the second one, and cancel pairsof duplicated terms to deduce that 2SN " SN = 2N+1" 1.

Now try a similar trick on the TN : investigate the effect of dou-bling the sum that defines Tn.

2 · TN = 1 · 4 + 2 · 8 + 3 · 16 + 4 · 32 + . . . + N · 2N+1

TN = 1 · 2 + 2 · 4 + 3 · 8 + 4 · 16 + . . . + N · 2N

UNC MATH CONTEST SOLUTIONS 5

Subtract to get

2TN " TN =

(0" 1)2 +(1" 2)4 +(2" 3)8 + . . . +(N" 1"N)2N + N2N+1

= N2N+1 " (2 + 4 + 8 + . . . + 2N)

= N2N+1 " (SN " 1)

= N2N+1 " (2N+1 " 2) = (N " 1)2N+1 + 2

(9.b) 2201

Note that if (N" 1)2N+1 + 2 ' 20, 000 then (N" 1)2N ' 10, 000.Tabulate powers of 2 and estimate N2N . Guessing N ' 10, wecheck that indeed with this choice

(N " 1)2N+1 + 2 = 9! 2048 + 2 = 18434,

which is ' 20000. So far we have taken all boxes up to all theboxes with 10 coins. We still need to get 20121" 18434 = 1687more coins using boxes that now have 11 coins. Divide 1687/11to get 153 and a remainder that forces us to take one more box,making 154 more in all. (There are 211 boxes with 11 coins each,so there are plenty available.) Thus we need S10 + 154 = 2047 +154 = 2201 boxes. It is possible to do part (b) without gettingthe formula in (a). Compute explicitly how many coins are inthe row with one coin per box, two coins per box, and so on.A number of students did this. This problem is an extension ofProblem 10 from the First Round.

(10.a) 126

One approach is to enumerate the possibilities. In addition tounderstanding the geometric properties of equiangular hexagons,one must have a careful and systematic method to do the enu-meration (see appendix). One student did this correctly. Obvi-ously, another approach is to complete part (b) for the explicitcase n=6.

(10.b) (n+34 )

!"#$%&'(&")*&+*+,-&./01&+*+,-&.20

A

B

C

a

b

a

a

b

cc

c

b

FIGURE 7. Triangles erected exterior to hexagon

Label the sides of the equiangular hexagon so that A is longest,and a is opposite it. Let B be the longer side adjacent to a, andC be the shorter side adjacent to a. Use lower-case letters to la-bel their opposite sides. Erect exterior equilateral triangles tothe lower-case sides. This frames the hexagon inside an equi-lateral triangle. Equate the lengths of the three sides of theequilateral frame: c + A + b = c + B + a = b + C + a; henceA" a = B" b = C" c. Call this common difference d. It satis-fies the inequality 0 # d # C" 1. The side length c = C" d sat-isfies 1 # c # C. Every equiangular hexagon determines a listof four such whole numbers A $ B $ C $ c $ 1. Conversely,by reversing the steps in the construction, we see that for everylist of four such whole numbers there exists exactly one equian-gular hexagon that has these given side lengths. The reversedconstruction starts by computing d = C " c, b = B " d, anda = A" d; then drawing an equilateral triangle with side lengthd (possibly zero); then assembling the equiangular hexagon byadjoining parallelograms as drawn in the figure below.

6 FINAL ROUND JANUARY 2012!"#$%&'(&")*&+*+,-.&/012&+*+,-.&/31

a

b

c

A

B

C

d

d

d

FIGURE 8. Parallelograms around a triangle

We now count all such lists that satisfy n $ A $ B $ C $ c $ 1by recognizing that this is version (iv) of Problem 3 above! Thenumber of such lists is (n+3

4 ).

The appendix shows all 126 of the hexagons for n=6. Theyare ordered and color-coded first by the length of the longestside. Then they are ordered by the lengths of the three othersides.

(11) x4 " x3 " 12 x2 " 1

6 x + 124

The polynomial P(x) = (x " r)(x " s)(x " t)(x " u) has rootsr, s, t, and u. Multiply out:

P(x) = x4" (r + s + t + u)x3 +(rs + rt + ru + st + su + tu)x2

"(rst + rsu + rtu + stu)x + rstu.

The coefficient a of x3 is a = "(r + s + t + u), the coefficientb of x2 is b = (rs + rt + ru + st + su + tu), the coefficient c of

x is c = "(rst + rsu + rtu + stu), and the constant term d isd = rstu.

We are told that r + s + t + u = 1 and r2 + s2 + t2 + u2 = 2 andr3 + s3 + t3 + u3 = 3 and r4 + s4 + t4 + u4 = 4. Therefore a ="1. Next observe that (r + s + t + u)2 " (r2 + s2 + t2 + u2) =2(rs + rt + ru + st + su + tu), which is 2b. Since we know that(r + s + t + u) = 1 and the sum of the squares is 2, 2b = 1" 2 ="1 or b = " 1

2 . That is, the coefficient of x2 is b = " 12 . We use

similar observations to find c and d. It will be convenient toname the sums of the powers of the roots: R1 = r + s + t + u =1, R2 = r2 + s2 + t2 + u2 = 2, R3 = r3 + s3 + t3 + u3 = 3, andR4 = r4 + s4 + t4 + u4 = 4. The coefficient c = "(rst + rsu +rtu + stu). To find c, look at combinations of a, b, and R(s thatproduce terms with three factors of the roots r, s, t, and u; thatis, look at things like R3 or R1b or R2a. Keep in mind that thefirst two useful equalities were "a = R1 and "2b = aR1 + R2.Observe that in fact"3c = bR1 + aR2 + R3. Put in the values weknow for the right side and "3c = (" 1

2 )(1) + ("1)(2) + 3 = 12 .

Thus c = " 16 . Similarly, "4d = cR1 + bR2 + aR3 + R4 = " 1

6 +(" 1

2 )(2) + ("1)(3) + 4 = " 16 and d = 1

24 . We have determinedthe last coefficient that we need.

The identities relating the Rs and the a, b, c and d are very oldidentities called the Newton-Girard formulae. The relationshipsbetween the roots of a polynomial and the coefficients are alsovery old and are known as Vieta’s formulae. The R(s are some-times called power functions and the expressions rs + rt + ru +st + su + tu, rst + rsu + rtu + stu and so on are known as sym-metric functions in the roots r, s, t and u. Students were notexpected to have seen these before! The challenge was to workout the various relationships.The contest writing team this year included Oscar Levin, Rich Mor-row, Richard Grassl, Katie Diaz, contest director Ricardo Diaz, andRocke Verser, who submitted Problems 3, 10, and 11.

UNC MATH CONTEST SOLUTIONS 7