MATHCOUNTS 2004-200522

Warm-Up 1Answers

1. 87 (C, F, M)

2. 3184 (C, T)

3. 11 (M, P, T)

4. 66 (C, F)

5. 8 (E, M, S, T)

6. 1.25 (C, M)

7. 108 (C, P, S)

8. 31 (C, M, T)

9. 11 (G, M, P, T)

10. 20 (P, S, T)

Solution - Problem #7Knowing our multiples of 7 may quickly lead us to the fact that our first multiple of seven

within the desired range is 7´5 = 35. We could continue identifying and counting multiples of 7 untilwe reach 790, but that would take a very long time! Dividing 790 by 7 gives us a quotient of 112 witha remainder of 6, so we know that 7´112 < 790 and 7´113 > 790. We�ve discovered that multiplying 7by any integer from 5 through 112 will give us a multiple of 7 within the desired range. How manyintegers are there from 5 through 112? Be careful! Don�t make the mistake of just using thedifference between 112 and 5... you�ll be one integer short! Think of it this way: we want to includeall of the first 112 positive integers except for the first four positive integers. This means there are112 - 4 = 108 integers from 5 through 112, and therefore, there are 108 multiples of 7 between 30 and790.

Solution/Multiple Representations - Problem #3Drawing a picture and watching the ball as it travels around

the circle is a great way to tackle this problem! We need to be surethat there are 11 girls, including Ami, and we can map out the pathof the ball until it reaches Ami again. Notice that by the time itreaches Ami again, it also has landed at each of the other 10positions exactly once (since every position has an arrow pointing toit). Similarly, since each position has exactly one arrow going awayfrom it, each of the 11 girls threw the ball exactly once before itlanded back with Ami.

Another way to approach this problem is to think of every integer in terms of the integer 11.In other words, rather than there being a position 12 on the figure above, we would really be back atposition 1. Similarly, if we were to look for position 35, we would go around the circle three timesand eventually land on position 2. Notice that any integer we choose will correspond to one of the11 positions in our figure, and we can determine the corresponding position by identifying theremainder when the integer is divided by 11. This is called modular arithmetic. Notice that positions14, 25, 36 and 48 would all be equivalent to position 3 since they all have a remainder of 3 whendivided by 11; they are all equivalent to 3 mod 11. Notice that every time a girl throws the ball, theposition of the receiver of the ball is four more than the position of the thrower. Therefore, the ballstarts at position 1 and will land at the positions 5, 9, 13 (or 2 mod 11), 17 (or 6 mod 11),21 (or 10 mod 11), 25 (or 3 mod 11), 29 (or 7 mod 11), 33 (or 0 mod 11), 37 (or 4 mod 11),41 (or 8 mod 11), 45 (or 1 mod 11), etc. We�re looking for when the ball first returns to Ami, which isposition 1. We see that position 1 (which is also position 45) is the 11 th landing point of the sequence,and therefore, there were 11 throws.

MATHCOUNTS 2004-200524

Warm-Up 2Answers

1. 5 (C, F, G, M)

2. 460 (C, T)

3. 5 (C, E, F, G, T)

4.�� (C, F, T)

5. 80 (C, T)

6. 16 (C, F, M, S, T)

7. 12 (C, F, M)

8. 11.4 (C)

9. (1, 3) (E, F, G, T)

10. 43.20 (C)

Solution - Problem #6The solution for this problem uses the Counting Principle, which will be an extremely valuable

tool in many other problems, too. Notice that we must first choose one entrée, and we have fourpossible choices. Then each of these four entrées can be matched with one of the two drinks.Perhaps you can mentally picture these eight possibilities. Finally, each of these eight possibilities canbe matched with one of two dessert options, which brings us to a total of 16 possible mealcombinations. The figure below, known as a tree diagram, illustrates the Counting Principle whichbasically states that when there are m ways to do the first task, n ways to do the second task andp ways to do the third task, there are then m ´n ´p ways to do all three tasks. Knowing the CountingPrinciple, rather than drawing tree diagrams, will save you a lot of time! For this tree diagram, the16 meal combinations are in bold in the middle two columns.

Solution/Multiple Representations - Problem #7There are many ways to find the area of this triangle. Perhaps the easiest way is to graph the

triangle and see that there is a horizontal side of length 8 units from (3, 1) to (11, 1), and then theheight from this horizontal base to the opposite vertex (1, 4) is 3 units. Employing the formulaArea = �

�(base)(height), we would then see that the area of this triangle is ( �

�)(8)(3) = 12 square

units. However, it is also good to know that there are other ways to determine the area of a trianglewhen given the ordered pairs of its three vertices.

One fun method requires us to write the three ordered pairs in a verticallist. We can write the ordered pairs in any order we like; however, the firstordered pair is written again as a fourth ordered pair at the end of the list.This is shown in the shaded region. Next, we will follow the arrows anddetermine the products of these pairs of numbers and write the products off tothe sides. Adding the numbers in each of the two outermost columns we justcreated and finding half of the positive difference of these two sums will giveus the area of the triangle. (This works for convex polygons when the orderedpairs are listed in the order that they appear around the polygon... but don�tforget to add the first ordered pair to the end of the list!)

P/L/Yogurt CD/L/YogurtP/Lemonade CD/Lemonade

P/L/Cookie CD/L/CookiePizza Corn Dog

P/RB/Yogurt CD/RB/YogurtP/Root Beer CD/Root Beer

P/RB/Cookie CD/RB/Cookie

T/L/Yogurt FC/L/YogurtT/Lemonade FC/Lemonade

T/L/Cookie FC/L/CookieTeriyaki Fish & Chips

T/RB/Yogurt FC/RB/YogurtT/Root Beer FC/Root Beer

T/RB/Cookie FC/RB/Cookie

MATHCOUNTS 2004-200528

Warm-Up 3Answers

1. 64.2 (C)

2. 12 (C, G, M P, S, T)

3. 66 (F, M, P, S)

4. 23 (C, F, G, M)

5. 31 (C, F)

6. 41 (C, G, S)

7. 84 (C, F, G)

8. 62 (C, M)

9. 4 (C, E, G, M)

10. 22 (C, E, F, M)

Solution - Problem #9Let�s see if we can set up an algebraic solution for this problem. First, we�ll assign the

variables T = Tyler�s age now and M = Mary�s age now. In four years, Tyler�s age will be T + 4 andMary�s age will be M + 4. According to the first sentence in the problem, T = �

�M. The second

sentence tells us (T + 4) = ��

(M + 4). Let�s substitute ��

M for T in the second equation since we knowthey are equal. We now have ( �

�M + 4) = �

�(M + 4). Multiplying both sides of the equation by 6 gets

us to a nicer equation to work with:

If Mary is now 8 years old, and Tyler is half Mary�s age, then Tyler is 4 years old.

Solution/Representation - Problem #7Understanding the concept of �the mean� is just as important as remembering the formula for

calculating it. Looking at the values (let�s assume that they are test scores), we can see that there area couple in the 70s, a couple in the 80s and a couple in the 90s. Therefore, it would be reasonable toestimate that the mean is close to 85. If the mean is 85, then the total number of points for the sixdifferent scores can be re-distributed to make six scores of 85. One score is already 85, so we won�ttouch that one. The score of 97 is 12 points too many, and the score of 73 is 12 points too few. Let�stransfer those 12 points.

The score of 90 has five extra points. Let�s give four of these points to the score of 81 andthe remaining point to the score of 78. Now we have:

Unfortunately, we don�t have enough points for a mean of 85. (There are no more points totransfer to the 79.) What if we take one point fromeach of the five 85s and give them to the 79?

Now we can see that the original scores have a mean of 84. Though this is not the mostefficient solution, it is a good illustration of what the arithmetic mean represents.

� � � �

� �� � �� �� � ��

�

�

�

�

�� � � � ��

� �

0 0

0 00 0

0

+ = +

+ = ++ = +

=

MATHCOUNTS 2004-200530

Warm-Up 4Answers

1. 22 (C, E, M, P, S)

2. 7 (C, F, T)

3. 2 (C, F, G, M)

4. 10 (C, F, P, T)

5. 50 (C, F)

6. 6 (C, P)

7.�� (C, P, T)

8. 75 (C, F)

9.���� (C, F)

10. 13 (C, F, M, S)

Solution - Problem #4Determining the number of factors of an integer is a fun area of number theory. We could

certainly list out the factors of 48 and count them, but if the question asked us to find the number ofpositive factors of 48,000, we would certainly want a more efficient way to count the factors!

Let�s first start with the prime factorization of 48, which is 24 ´ 31. We know that any factorof 48 could then have no factors of 2, or one, two, three or four factors of 2. That�s five differentoptions for the 2. Any factor of 48 could also have no factors of 3 or one factor of 3. That is twodifferent options for the 3. Going back to the Counting Principle discussed with Warm-Up 2, we seethat there are five ways to incorporate the 2 and two ways to incorporate the 3 into any positivefactor of 48, so there are 5´2 = 10 positive factors. The tree diagram below illustrates how thisworks.

In general, to determine the number of positive factors of an integer, increase each exponentof the prime factorization by 1, and then find the product of these new values.

Solution/Multiple Representations - Problem #7Since rolling a pair of dice is a common scenario for probability problems, it wouldn�t hurt to

be familiar with the 36 possible outcomes. If we list the 36 outcomes, 12 of them result in a sum thatis a multiple of 3. This is a probability of ��

��

�

�= .

We also could consider the outcomes by going through the possible values for the roll of thefirst die, and then determining what values for the second die will get us to a sum that is a multiple ofthree. If we roll a 1 on the first die, a roll of 2 or 5 on the second results in a multiple of three. Ifwe roll a 2 on the first die, then a roll of 1 or 4 on the second die results in a multiple of three. If wecontinue in this manner, we will find our 12 different successful scenarios out of the 36 possibleoutcomes.

Connection to... Pascal�s Triangle (Problem #9)As you look at this problem, you may notice that the first factor of each of the five terms

(1, 4, 6, 4, 1) are the numbers that can be found in Row 4 of Pascal�s Triangle (assuming Pascal�sTriangle starts with Row 0). This is a huge hint that might lead you to see that the expression inProblem 9 is really the expansion of a binomial raised to the fourth power:

� � � � � � � � � � � � � � � ��

�

� � �

�

� � �

�

� � �

�

� � �

�

� � �

�

��

�

�

�

��

�

��

�

�+ = + + + + = + + + +4 9 1 61 6 4 9 1 61 6 4 9 1 61 6 4 9 1 61 6 4 9 1 61 6 4 9 4 9 4 9 4 9 4 9 .

Simplifying this final expression looks daunting, but notice � �

�

��

�

���

��

��

��+ = = =4 9 4 9 .

48 = 24 ´ 31

20 21 22 23 24

20´30 20´31 21´30 21´31 22´30 22´31 23´30 23´31 24´30 24´31

1 3 2 6 4 12 8 24 16 48

MATHCOUNTS 2004-200534

Warm-Up 5Answers

1. 88 (C, F, M P, S, T)

2. 40 (C, G)

3. May 1 (C, P, T)

4.��� (C, F, M)

5. 8.8 (C, F)

6. 2 (C, M, P, S, T)

7. 120 (C, T)

8. 5 (C, E, P, T)

9. 2s (C, F)

10. 1 (M)

Solution - Problem #5

Let�s agree first that (1) multiplying an expression by 1 does not change the value of the

expression and (2) a ratio is equal to 1 if the value in the numerator is equal to the value in the

denominator. Keeping these two facts in mind, consider 6 mph or the ratio ��

PLOHV

KRXU. We want to change

this to "

" VHF

IHHW

RQG. Take a look at the product �

�

�

��

�

��

����

�

PLOHV

KRXU

KRXU IW

PLOH× × ×

PLQ

PLQ

VHF. The first ratio is our

6 mph, and each of the next three ratios is equivalent to 1 since the value in the numerator of each

ratio is equal to the value in its denominator. This ensures that the expression is still equivalent to

6 mph. �Canceling� factors in the numerators with factors in the denominators is a common practice

when we are multiplying fractions, and after performing the same type of procedure with the units in

the expression ��

�

��

�

��

����

�

PLOHV

KRXU

KRXU IW

PLOH× × ×

PLQ

PLQ

VHF, we are left with � ����

�� ��

××

IW

VHF. Dividing out factors of 6,

10 and another 6 from the numerator and denominator, we have ����

� �= � feet/second.

Solution/Multiple Representations - Problem #4We�re told that the three larger sectors of the spinner are congruent, and since each one has a

central angle of 90°, they are each a quarter of the spinner. The smaller sectors (including WIN) thatmake up the remaining quarter of the spinner are congruent, and therefore, they are each an eighth ofthe spinner. Since we are spinning the spinner twice (two events) and determining the probability ofspinning WIN/WIN, we will need to find the probability of the first event (spinning WIN) andmultiply that by the probability of the second event (spinning WIN). Since the sector WIN is aneighth of the spinner, the probability of spinning WIN/WIN is �

�

�

�

�

��4 94 9 = .

We also can rely on a visual representation of the situation. Lettingthe largest rectangle represent our entire sample space, the top figureshows the possible outcomes after one spin. (Remember that WIN is aneighth of the spinner.) The shaded regions represent spinning LOSE, inwhich case we don�t need to spin again because WIN/WIN is no longer apossibility. So already we are limited to an eighth of the largest rectangle.Assuming we�ve already spun WIN, we�re going to spin the spinner again.The possibilities for this second spin are shown in the middle figure. Again,we have shaded the areas corresponding with spinning LOSE. We can seethat there is a very small region now that is not shaded (no LOSE spins).This region is an eighth of the small square, which was an eighth of thelargest rectangle. So the unshaded region is �

�4 9 of �

�4 9 , which is�

�

�

�

�

��4 94 9 = of the largest rectangle. Do you see how this visualrepresentation corresponds to the solution in the paragraph above? (Thebottom figure shows all of the possible outcomes of two spins, with black representing LOSE/LOSE,gray representing LOSE/WIN and WIN/LOSE and white still showing WIN/WIN.)

MATHCOUNTS 2004-200536

Warm-Up 6Answers

1.�� (C, F, M, T)

2. 40 (C, F)

3. 8 (C, F, G, M, T)

4. 5 (C, F, M)

5. 25 (C, P)

6. 25 (C, P, S)

7. 108p (C, F)

8. 68 (C, F, M)

9. 40 (C, F, P, T)

10. 41 (C, F, M)

Solution - Problem #6We�re starting with the expression 810 ́ 522. It seems that a calculator would be nice at this

point, but since this is a Warm-Up, we should be able to do this without a calculator. Let�s take thisdown to its prime factorization: 810 ´ 522 = (23)10 ´ 522 = 230 ´ 522. Notice that we are multiplying a lotof 2s by a lot of 5s. What happens when we multiply a 2 by a 5? We get 10. What if we multiply two2s with two 5s? We get the product of two 10s, which is 100. There is a pattern here that takes us tothis next step: 230 ´ 522 = 28 ́ 222 ́ 522 = 28 ́ 1000...00, such that there are 22 zeros in the secondfactor. Now 28 can be calculated without a calculator and is equal to 256. So now we have256 ́ 1000...00 which is 2,560,00...,000 with a total of 22 zeros. This is a grand total of 25 digits.

Solution/Multiple Representations - Problem #3Algebraically, we can set up the equations T = 3M and M + 1 = �

�T. We are looking for the

value of M + T. If we substitute 3M for T in the second equation, we have M + 1 = �

�(3M) or

M + 1 = ��

M. Multiplying both sides of the equation by 2 gets us to 2M + 2 = 3M. Finally, M = 2. SinceT = 3M, we also know T = 3(2) = 6. Together they have 2 + 6 = 8 coins.

Using the same equations, we can use the following figure to solve theproblem. We know that Tim has three times as many coins as Mike (Figure 1).This means that the total number of coins M + T is also the same as 4M, whichcan be seen in the figure. We�re told that if we add one coin to Mike�s coins,the amount of coins will be equal to half of the coins Tim started with, whichhas been incorporated into Figure 2. Remembering that the entire largerectangle represents 4M, we can now see that (M + 1) + (M + 1) + M = 4M;3M + 2 = 4M; and M = 2. Together, then, they have 8 coins.

Additionally, this problem can be solved with Guess, Check & Revise. The first two rows ofthis table show two initial guesses for the number of coins Tim and Mike have. (If we choose Mike�snumber of coins first, we can then easily multiply this numberby three to determine Tim�s number of coins.) The goal is toget the values of the third and fourth columns equal to eachother. Notice that when Tim has an odd number of coins, thenhalf his number of coins (column D) will not be an integer, andtherefore can�t be equal to the number of Mike�s coins whenhe adds one to his collection (column C). From our twoguesses we can see that the number of coins Tim starts with should be an even multiple of 3. Also, thedifference between columns C and D grew when Mike and Tim�s coin amounts grew. Therefore, let�spick a number of coins smaller than 15 for Tim to start with (remembering that it should be an evenmultiple of 3). Try starting Tim out with 6 coins. Then Mike has 2 coins, and we can see that thecorresponding values in columns C and D are equivalent. This scenario is represented in the third rowof values in the table.

Mike Tim M + 1 T ¸ 2 Compare5 15 6 7.5 6 < 7.510 30 11 15 11 < 152 6 3 3 3 = 3

Figure 1

Figure 2

MATHCOUNTS 2004-200540

Warm-Up 7Answers

1. 2.98 (C, G, S)

2. 10 (G, M)

3. 0 (C, P, S, T)

4. 404 (C, E, G, T)

5. 18 (C, F, M)

6. 92 (C, M)

7. 210 (C, F, M, P, S, T)

8. 3 (C, E, G, M, P, T)

9. 7 (M, P)

10. 21 (F, P, T)

Solution - Problem #3Trying to calculate 72005 on a calculator isn�t going to do us much good since the screen won�t

be able to hold all of the digits we need. So perhaps calculating the final value is not the way to go.Let�s look at the first few powers of 7: 71 = 7, 72 = 49, 73 = 343, 74 = 2401, ... . What you might noticeis that the units digit of 75 will be 7 and a pattern for the units digit emerges: 7, 9, 3, 1, 7, 9, 3, 1, ... .Perhaps there is a pattern for the tens digit, too. Notice that since we are concerned only about thetens digit, it�s necessary to calculate up to only the tens digit of the successive powers of 7. Since74 = 2401, we can calculate 75 = __07, 76 = __49, 77 = __43, ... . We can see that there is also apattern emerging for the last two digits of every power of 7: 7, 49, 43, 01, 07, 49, 43, 01, ... . Ourfinal step is to determine which one of these four two-digit combinations is the one found at the endof 72005. Notice that if the exponent is a multiple of four and the base is 7, then the simplified valueends in 01. From this we know that 72004 = __01, so 72005 = __07. The tens digit is 0.

Solution/Multiple Representations - Problem #9From the given information, we know that the triangle is an isosceles

right triangle in the first quadrant as shown here. We are also told that thereare 15 lattice points in the interior of the triangle (not on the triangle). Theslope of the hypotenuse must be 1 since it goes up N units as it moves to theright N units. Therefore, we can include the coordinates of some of thepoints on the triangle, as shown in this second figure. Notice, too, that theempty circle in this figure represents the point (N - 1, N - 2) and is our firstidentified lattice point in the interior. If we were toidentify the lattice points along the line y = N - 3, wewould see two lattice points inside the triangle. Eachtime we move down a row of lattice points, there isone more lattice point in the interior of the trianglein that row than the number of lattice points in theinterior of the triangle in the row above. Continuingthis pattern until we have 15 lattice points in theinterior leads us to the third figure. From here wecan see that N = 7.

Using Pick�s Theorem is another way to approach solving this problem. The theorem states thatthe area of the triangle is equal to one less than the sum of half the number of lattice points on thetriangle (boundary points) plus the number of lattice points inside the triangle (Area = %

� + I - 1).

From some of our initial deductions in the first solution above, we can see that the area of the triangleis �

�N2 using Area = �

�bh , and the number of lattice points on the triangle is 3N (since there are N

lattice points on each side of the triangle). We now have the equation ��

N2 = ��

1 + 15 - 1. Using somealgebra leads us to N2 = 3N + 30 - 2; N2 - 3N - 28 = 0; (N - 7)(N + 4) = 0; N = 7 or -4; and sinceN > 0, we see N = 7.

MATHCOUNTS 2004-200542

P N Q Total Value (¢)1 - - 12 - - 23 - - 3- 1 - 5- 2 - 10- - 1 25

1 1 - 61 2 - 112 1 - 72 2 - 123 1 - 83 2 - 131 - 1 262 - 1 273 - 1 28- 1 1 30- 2 1 35

1 1 1 311 2 1 362 1 1 322 2 1 373 1 1 333 2 1 38

Warm-Up 8Answers

1.��� (C, E, F, T)

2. 20 (C, F)

3. -32 (C, F)

4. 2 (C, E, G)

5. 132 (C, F, M)

6. 23 (C, F, P, T)

7.��π

(C, F, M)

8. 216 (F, G, P, T)

9.��� (C, F)

10. 160 (C, P, T)

Solution - Problem #8(Before getting into this solution, be sure that you first follow the logic of the solution given

for Warm-Up 4, Problem 4.) We know that 72 = 23 ́ 32, which means that there are 4 ́ 3 =12 factors. If the multiple of 72 included just one other prime factor different from 2 and 3, itwould follow that this multiple m could be represented by m = (23 ´ 32) ´ p, and it would have4 ́ 3 ́ 2 = 24 factors. Bringing a new prime factor into the picture creates too many factors for themultiple. Therefore, let�s try the multiple (23 ́ 32) ́ 2 = 24 ́ 32, which has 5 ́ 3 = 15 factors. Thisdoesn�t satisfy our condition, so we�re left with trying (23 ´ 32) ´ 3 = 23 ´ 33, which has 4 ´ 4 = 16factors. Our multiple is 23 ´ 33 = 216.

Solution/Multiple Representations - Problem #6One solution to this problem involves making a

very organized list. We know that Steve can use onlypennies, only nickels or only the quarter. This is shown inthe six rows of entries in the first section of the list. Wealso know that he can use any combination of two typesof coins (pennies/nickels, pennies/quarters ornickels/quarters). These 11 options are shown in themiddle portion of the list. Finally, he can usecombinations of all three types of coins. This lastpossibility may seem like the most difficult portion of thelist to create, but notice that these six entries match thefirst six entries of the middle portion (pennies/nickels),but with the quarter added to each one. We see thatthere are 6 + 11 + 6 = 23 different values that can bemade from three pennies, two nickels and one quarter.

Because of the number of different types ofcoins, we also can attempt to solve this problem by usingthe Counting Principle. (If there were five pennies, thisapproach would not work since a combination with fivepennies and no nickels would have a value equal to adifferent combination with one nickel and no pennies.)Because of the number of coins available for each typeof coin, we can see that no two distinct combinations ofthese coins will result in the same value, so rather thancounting the different value-amounts that can be formed, we can instead count the number ofcombinations of coins we can create. Notice that we can use 0, 1, 2 or 3 pennies; 0, 1 or 2 nickels;and 0 or 1 quarter. This means that there are four ways to pick the number of pennies, three ways topick the number of nickels and two ways to pick the number of quarters to include, for a total of4 ´ 3 ´ 2 = 24 ways to pick the coins. Our answer is 23, though, since the combination of no pennies,no nickels and no quarter (or 0¢) is one of the 24 combinations we counted, but it is not an option wewould want to include.

MATHCOUNTS 2004-200546

Warm-Up 9Answers

1. 9 (C, G, P)

2. 77 (C, F, G)

3. 62 (C, F, G, M)

4. (10, 0) (C, F, M, P)

5. 3 (E, T)

6. 14 (C, P, T)

7. 20 (C, F, M)

8. 45 (C, F, M)

9.�

��� (C,F, G, S)

10. 100 (C, S)

Solution/Multiple Representations - Problem #1In this arithmetic sequence we know that (12) - (y + 6) = (y ) - (12) since the difference of the

second term and first term must be equal to the difference of the third term and second term. Usingalgebra to solve this equation, we have 12 - y - 6 = y - 12; 18 = 2y ; and y = 9.

You also may see that the difference of the first term and third term is 6, and the values ofthe terms are decreasing. We can deduce, then, that the difference of each pair of consecutive termsis half of 6, which is 3, and that y is less than 12. We see, then, that y = 12 - 3 = 9.

Solution/Multiple Representations - Problem #4The line containing points (9, 1) and (5, 5) has a slope of � �

� �

�

��−

− −= = − . Therefore, theequation of the line is y = -1(x ) + b. Since (5, 5) is on the line, we know 5 = -1(5) + b so b = 10.Knowing that the equation of our line is now y = -1(x ) + 10 and that there is an x-intercept (t , 0), wecan solve 0 = -1(t ) + 10 to see that t = 10. The x-intercept is (10, 0).

If we graph the two points we can see that to go from (5, 5) to(9, 1) we go down four units and right four units. Therefore, we godown one unit for every one unit we move to the right. We can also seeon the graph that to get to the x-axis from the point (9, 1) we just haveto go down one more unit. This means we also must move one unit to theright, which leads us to the point (10, 0).

Connection to... Calendars (Problem #6)This problem assumes that we are using the Gregorian calendars of modern times. Take some

time to investigate how calendars have significantly changed throughout history!

Connection to.... Interior Angles of Regular Polygons (Problem #8)Though this problem asks for the measure of just a portion of the interior angle of the regular

octagon, it is helpful to know the measure of the entire interior angle. To find the measure of aninterior angle of a regular n -gon, we can use the expression � �Q

Q

−� ��� . It may be easier, however, toremember that the sum of an interior angle and its corresponding exterior angle is always 180 degrees.Additionally, the measure of an exterior angle of a regular n -gon is always 360 ¸ n . So for thisoctagon, an exterior angle measures 360 ̧ 8 = 45 degrees, and an interior angle measures180 - 45 = 135 degrees.

MATHCOUNTS 2004-200548

Warm-Up 10Answers

1. 8 (C, P, T)

2. 6 (C, G)

3. 28 (C, E, M, T)

4. 19 (C, P)

5. 10,236 (E, G, S)

6. 5 (C, F, G, T)

7. 6 (C, F)

8. 3 (E, G, M, T)

9. E (M, P, S, T)

10. 3 (C)

Solution - Problem #3A pyramid has a polygon as its base, and then its lateral faces are triangular regions all joining

at a single point. We are to use one of the faces of the prism shown on the previous page as the baseof our new pyramid. Notice that each face of the prism is either a triangle or a rectangle. Since wewant our new pyramid to add on the maximum possible number of faces, vertices and edges, we shouldaffix it to one of the rectangular faces of the prism as shown in Figure 1. This results in a solid witheight faces, seven vertices and 13 edges, for a total value of 28. (We do not count the face that isshared by the prism and pyramid since this would not be a face of the resulting solid when the prismand pyramid are fused together.) Notice that joining the pyramid to the prism at a triangular face(Figure 2) results in only seven faces, seven vertices and 12 edges, for a total value of 26.

Solution - Problem #5We know that the five-digit number must be divisible by each of its non-zero digits. Including

a digit of zero will not be a problem since the zero will not have to serve as a possible divisor.Knowing this, the least possible five-digit number that we could try is 10,234. Any number we choosewill be divisible by one. We also see that it is even, and therefore, divisible by two. However, thetwo-digit number formed by its last two digits (34) is not divisible by four, and therefore, neither isthe five-digit number. We also see that the sum of the five digits is 10, and since 10 is not divisible bythree, neither is the five-digit number. But notice that by increasing the five-digit number by two toform the number 10,236 we create another even number and increase the digit-sum to 12 (which takescare of the number being divisible by three). We�ve now eliminated the digit of four and added thedigit of six, which is fine because 10,236 being divisible by both two and three means that it isdivisible by six. Our five-digit number is 10,236.

Connect to... Order of Operations and Forms of Exponents (Problem #10)

The expression on the previous page is equivalent to � � >� � @� � �� � � ��

�

�

�× + ÷ × − if we use

grouping symbols to show the correct order of operations. (Notice that in the brackets, multiplication

does not come before division. They are �equal� operations and are performed in the order that they

appear from left to right.) Additionally, knowing that � ��

� = and � ��

� �= , we can rewrite the

expression as � � >� � @ � � � � � � �� � �� � � � � � � ��× + ÷ × − = + − =

Figure 2Figure 1

MATHCOUNTS 2004-200552

Warm-Up 11Answers

1.��� (C, M)

2. 24 (E, G, M, T)

3. 995 (C, G, P)

4. 120 (C, M, P)

5. 9748 (E, G, P)

6. 20 (E, F, G, P, T)

7. 600 (C, F, M)

8. 30 (C, F, G, P)

9. 100 (C, F, M)

10. 120 (F, M, P, S, T)

Solution - Problem #10The position-assignment shown for the problem (1, 3, 5, 7) is one example of the spaces that

can be occupied. However, this one position-assignment has 4 ´ 3 ́ 2 ́ 1 = 24 different variationswhen we start considering the colors of the cars that can be in those spaces. For example, thearrangement (R1, W3, B5, G7) shown on the previous page is a different arrangement from (R1, W3,G5, B7). Let�s first find the number of total position-assignments, and then we will multiply thisnumber by 24 to account for the variations of each position-assignment. An organized way of countingis necessary. Let�s start with position-assignments with the first space occupied. They are (1, 3, 5, 7),(1, 3, 5, 8), (1, 3, 6, 8) and (1, 4, 6, 8). Remember that there must be at least one space betweenconsecutive cars, and only one gap can have as many as two spaces, so our arrangements are actuallymore limited than we may have first thought. Now, assuming there is no car in the first space, theremust be a car in the second space. The only possible position-assignment is (2, 4, 6, 8). This is a totalof five position-assignments. Each one has 24 different color-variations, so there is a total of5 ́ 24 = 120 arrangements.

Solution/Multiple Representations - Problem #8Assuming x is an integer, we can let the three consecutive integers be x, x + 1 and x + 2. We

also know x (x + 1)(x + 2) = 33(3x + 3); x (x + 1)(x + 2) = 99(x + 1); and then x (x + 2) = 99. Byobservation we can probably guess two positive integers with a difference of two and a product of99. They are 9 and 11, so x = 9. The sum of the three integers is then 9 + 10 + 11 = 30.

We also could represent the three consecutive integers as x - 1, x and x + 1. Here we wouldhave the equation (x - 1)(x )(x + 1) = 33(3x ). We can further simplify to (x - 1)(x + 1) = 99. Thisequation lends itself to the same reasoning as the equation x (x + 2) = 99 in our last representation.However, we also could take the following steps: (x - 1)(x + 1) = 99; (x 2 - 1) = 99; x 2 = 100; andthen x = 10 or -10. Since we�re limited to a positive integer, x = 10 and our three integers are again 9,10 and 11 with a sum of 30.

Connection to... Side Lengths of Triangles (Problem #2)If we are told only that a triangle has sides measuring 8 units and 6 units, it is impossible for

us to know the length of the third side. However, it is possible for us to know that the third side can�tbe a length like 1 unit or 15 units. The relationship between the third side of a triangle (c ) and theother two sides (a and b ) is a + b > c > a - b. In the progression shown below, we can see that if c istoo large, then the other two sides won�t be long enough to reach each other (far left figure). If c istoo small, then it won�t be long enough to reach the middle-lengthed side (far right figure). Noticethat at the two extremes (8 + 6 = 14 and 8 - 6 = 2), the sides �collapse� and no longer form a triangle.It�s important to keep this rule in mind when solving Problem #2 on the previous page.

MATHCOUNTS 2004-200554

Warm-Up 12Answers

1. 256p (C, F)

2. 49 (C, F)

3. 20 (C)

4. 100 (C, M)

5. 52 (C, F, G, M)

6. 76 (C, E, F, M, P, T)

7. 4 (C, F)

8.����� (C, T)

9. 6 (C, E, G)

10.�� (C, F, G, M)

Solution - Problem #4This problem is not as easy as it may initially appear. Though

we know that at 40 minutes after 4 o�clock, the minute hand is pointingto the eight, the hour hand is no longer pointing to the four. It hasstarted to make its way to the five. In fact, because more than30 minutes have elapsed since 4 o�clock, the hour hand is more thanhalf-way to the five. Since ��

��

�

�= of the 4 o�clock hour has gone by,

the hour hand is ��

of the way from the four to the five. The12 numbers of the clock are equally spaced, so they are placed every360 ̧ 12 = 30 degrees. If the hour hand has moved �

� of the

30 degrees from the four to the five, it has moved ( ��

)30 = 20 degrees.We now know the information seen in the clock shown here. The angleformed by the two hands is then 10 + 30 + 30 + 30 = 100 degrees.

Solution/Multiple Representations - Problem #7The figures illustrate the information we have. We

know that the area of the non-shaded region is three timesthe area of the shaded region. Letting the area of theshaded region be x square inches, the area of the non-shaded region is 3x square inches and the area of the entiredartboard is 4x square inches. (The ratio of the areas of�shaded� to �non-shaded� to �whole� is 1:3:4. This may behelpful in a later solution.) We also know the area of thewhole dartboard is 64p square inches based on the radius.Now we have 4x = 64p, and dividing both sides by four yields x = 16p, which is the area of theshaded region. Letting rs represent the radius of the shaded region, we now have 16p = p(rs )2 and

rs = 4 inches.

We also could approach this problem using ratios. We know that the area of the entiredartboard is 64p square inches. If the area of the shaded region is x square inches, then the area ofthe non-shaded region is 64p - x square inches and �� �

�

π − =[

[. From here we see 3x = 64p - x and

then 4x = 64p. Finally, x = 16p = p(rs )2 and again rs = 4 inches.

Let�s examine another solution that also incorporates ratios. Since the ratio of the area of theentire dartboard to the area of the shaded region is 4:1, we know that the ratio of the linearmeasurements of any corresponding parts of the entire dartboard and shaded region is � �� or 2:1.(This holds true because the two circles are similar.) Since the radius of this entire dartboard is8 inches, the ratio tells us the radius of the shaded circle is 4 inches.

x

3x

MATHCOUNTS 2004-200558

Warm-Up 13Answers

1. 1 (C, F, T)

2. 27 (C, E, P, T)

3.���� (C, F, P)

4. ��

(C, F, M)

5. -3 (C, F)

6. �� � (C, F, M, S)

7. 15 (E, G, P, S, T)

8. 5 (E, G, P, S, T)

9.��� (C, F, P, T)

10. 2003 (E, G, P, T)

Solution/Multiple Representations - Problem #9The first spin has six options, and the second spin has six options, for a total of 6 ́ 6 =

36 spin combinations. The following spin combinations satisfy the condition: (1, 2), (2, 1), (2, 3),(3, 2), (3, 4), (4, 3), (4, 5), (5, 4), (5, 6) and (6, 5). This is 10 of the 36 possible combinations, whichis a probability of ��

��

�

��= .

We also could look at the problem in a different way. Spinning any number from 1 through 6is equally likely for the first spin. We can see that each of the two �extreme values� for the first spin(1 and 6) have only one �good� second spin (2 and 5, respectively). So the probability of meeting thecondition for each of these two first-spin options is �

�. Looking at spinning a �non-extreme value� for

the first spin (either a 2, 3, 4 or 5), we see that each of these four first-spin options has two �good�second spins (a value one less or one greater than the first spin). For each of these four first-spinoptions, the probability of meeting the condition is �

� = �

�. We said before that each of the six first-

spin possibilities is equally likely, so each of their probabilities of success should be equally weighted.If we average the probabilities of success for each of the six first-spin options, we have [2( �

�) +

4( �

�)] ¸ 6 = �

��, which is the probability of Ann spinning integers with a positive difference of 1 on

her first two spins.

Solution/Multiple Representations - Problem #10If 7a + 12b = 1 and a and b are both integers, then a and b will have opposite signs. We also

can guess that D > E since 7a is going to have to be almost equivalent to the opposite of 12b. If wewant a + b to be positive, and we understand the information in the last two sentences, then a must bepositive, and we will need a multiple of 7 that is one greater than a multiple of 12. If we think of oneordered pair (a, b) that satisfies the condition, then perhaps others will follow. Notice that 49 and 48have a difference of one, with the multiple of seven being greater than the multiple of 12, so7(7) + 12(-4) = 1 and (7, -4) is a solution. However, the value of a + b is only 7 + -4 = 3. If we wantto keep this difference of 1 between 7a and ��E , and also keep 7a > ��E , then we must increase7a and decrease 12b by the same amount. This amount must be a multiple of 7 and a multiple of 12.The least multiple of both is 7 ´ 12 = 84. This does work: (49 + 84) + (-48 - 84) = 1 and7(7 + 12) + 12 (-4 - 7) = 7(19) + 12(-11) = 1. Our new a + b is 19 + -11 = 8. We can continue to add 84and subtract 84, but we�re really just increasing the value of a by 12 and decreasing the value of bby seven, which ultimately increases a + b by five each time. Our values for a + b will be 3, 8, 13,18, ... . These are all two less than the positive multiples of five, so the greatest value of a + b that isless than 2005 is 2005 - 2 = 2003.

We know a and b are integers, so a + b must be an integer. We also know a + b < 2005. Let�stry a Guess, Check & Revise approach. Assume a + b = 2004. Let a = x and b = 2004 - x. We thenjust need to see if there exists an integer x such that 7x + 12(2004 - x ) = 1. This simplifies to7x + 24,048 - 12x = 1 or -5x = -24,047. Dividing both sides by five will not yield an integer solutionfor x. Let�s try a + b = 2003, so a = x and b = 2003 - x. Consider 7x + 12(2003 - x ) = 1. We simplifyto 7x + 24,036 - 12x = 1 and finally -5x = -24,035. We see that this yields an integer value of x, andregardless of what that value is, 2003 is the largest possible value of a + b that is less than 2005.

MATHCOUNTS 2004-200560

Warm-Up 14Answers

1. 4 (E, M, P, T)

2. 85 (C, F, T)

3. 22 (P, T)

4. 49 (C, E, G, T)

5. 5 (C, G, M, P, S)

6.����� (P, S, T)

7. -6 (C, F, G, P, S, T)

8. 2 (C, F, M, P)

9.��� (C, F, P, T)

10. 10 (C, F, P)

Solution - Problem #9We know that P(matching pair/any color) = P(matching pair/black) + P(matching pair/brown) +

P(matching pair/gray). Let�s first consider P(matching pair/black). To select a matching pair of blackshoes, we must pick a black shoe on our first selection. The probability of this is ��

�� since there are

12 black shoes and 22 total shoes. Once that black shoe is chosen, we are suddenly limited to sixblack shoes that would be successful for our second selection, so the probability of success on oursecond selection is �

��. (There are only 21 shoes remaining from which to choose.) The probability of

this entire two-step process happening is ����

�

��

��

���× = . The process is similar for P(matching pair/

brown), but we have fewer brown shoes. We see that P(matching pair/brown) = �

��

�

��

��

���× = .

Similarly, we have P(matching pair/gray) = �

��

�

��

�

���× = . Finally P(matching pair/any color) =

��

���

��

���

�

���

��

���

�

��+ + = = .

Solution/Multiple Representations - Problem #8We are given (or can deduce) a lot of information. The three

portions of circles all come from congruent circles, and so the radius ofeach circle is 1 unit. Consider Fig. 1, which is basically the figure given withthe original problem. If G is the center of the upper circle, then we can seethat the area of the shaded portion of circle G that we are trying to find isthe area of the entire circle minus the area of the two �football�-shapedregions outlined with the dotted portion of the circle. If we can determinethe area of these football-shaped regions, we can find the area of theshaded region. In Fig. 2, the shaded circle has been brought forward,ÐDGB is a right angle and it forms a sector that is a quarter of circle G.Triangle DGB is an isosceles right triangle with DG = BG = 1 unit. The areaof circle G is p(1)2 = p square units. The area of the quarter-circle is thenp ̧ 4 square units. The area of triangle DGB is ( �

�)(1)(1) = �

� square units,

so the area of the portion of the quarter-circle not in triangle DGB is(p ̧ 4) - �

�. This is half of the football region. If we reflect this shaded

region over the segment DB, we get the complete football region with an area of 2[(p ̧ 4) - ��

] = π�

�− .Therefore, remembering that the area of the shaded region back in the original problem (Fig. 1) is thecomplete circle minus two of these football regions, we see that our final answer is p - 2( π

��− ) =

p - p + 2 = 2 square units.

Another way to approach this problem is to draw in square BEFD.Notice that the area of square BEFD is the same as the area of the originalshaded region since the two half-football regions not shaded in the squareare congruent to the two half-football regions that are shaded outside of thesquare. Since the diagonal BF of square BEFD is 2 units, the area of thesquare (as well as the shaded region�s area) is �

� � ��

�� �� � � �� �G G = = 2 square units.

Fig. 2

Fig. 1

MATHCOUNTS 2004-200564

Warm-Up 15Answers

1. 4 (C, G, M, P, T)

2. 88 (C, E, G, T)

3. 42 (C, M, P, S, T)

4. 540 (C, F, M)

5.���� (C, F)

6.�� (C, E, F, G, M, T)

7.�� (C, G, P, T)

8.��� (C, M, T)

9. 7 (C, F, G, M, P, S, T )

10. (4, 2) (C, F, M, P)

Solution - Problem #9Being able to determine the number of subsets in a set is critical for this problem. Remember

that a subset of a set can include all, none or any other combination of the members of the set.Notice that if a set has four members, then each of those members can either be included orexcluded; in other words, each member of the set has two options. By the Counting Principle, thereare then 2 ́ 2 ́ 2 ́ 2 = 24 = 16 ways to determine the inclusion/exclusion pattern of the fourmembers, and so there are 16 subsets. (If each member is excluded, that results in the empty set,which is a subset of every set.) From this example, we can make the jump that for any set withn members, the set has 2n subsets. In this problem, we see that 2n + 2 - 2n = 96. Rewriting this slightly,we have (2n ´ 2 ´ 2) - 2n = 96 or 2n (4 - 1) = 96. Dividing both sides by three yields 2n = 32 and we cansee that n = 5. Now we know set B had 5 elements and set A had 7 elements.

Solution - Problem #6We�re not supposed to use a calculator for this problem (since it�s a

Warm-Up), so we can�t simply type the equation into a graphing calculator andsee where on the graph the y -value is always positive. However, knowledgeabout equations in this form can get us our result fairly quickly. We havey = (2x � 1)(4x 2 + 4x + 1), which can be rewritten as y = (2x � 1)(2x + 1)2 ory = (2x � 1)(2x + 1)(2x + 1). We now have three binomial factors for ourequation, and we can deduce that if they were all multiplied out, the highestpower of x would be 3, and the x 3-term would be positive. This tells us thegeneral shape of the graph. It will enter from the bottom left and go upward(since it�s a positive leading term), then probably turn downward and then backupward (since it�s a cubic equation), and exit the graph going out the top right in an upward direction(Figure 1). From the equation we can determine the three x-intercepts (or roots of the equation).Setting each of the factors equal to zero and solving for x will give us the x-intercepts.

Two of the three binomial factors are identical. This is important. Rather than having thethree x-intercepts we expected (as in Figure 1), we have only two distinctx-intercepts. Because − �

� shows up twice, and is therefore a �double-root,�

we know that it is not only an x-intercept, but it is also the tip of one of theturning points of the graph. We can now determine that the graph lookssomething like Figure 2. Though we haven�t determined where the tip of theother curve occurs, we can see that once we look to the right of x = �

�, the

graph is always above the x-axis. In other words, if x > ��

, then y is alwayspositive, and so the value of a is �

�. Similarly, if x < − �

�, then the y -value

is always negative.

2x � 1 = 0 2x + 1 = 0 2x + 1 = 0 2x = 1 2x = � 1 2x = � 1 x = �

� x = − �

� x = − �

�

Figure 1

Figure 2

MATHCOUNTS 2004-200566

Warm-Up 16Answers

1. 16 (C, F, M, P)

2. 6 (C, E, P, T)

3. 5 (C, F, G, M)

4. 2 (C, F)

5. 11 (C, E, F, G, M)

6.�� (F, P, S)

7. 15 (C, P, T)

8. � � (C, F, M)

9.�� (C, M, P, T)

10. 36 (C, M, P, S, T)

Solution - Problem #3From the figure, we can see the relationship of the two semi-

circles even though we don�t know the radius of them. We see theyare tangent to each other in order to use as much space as possible.(We also are assuming that no wood is lost when cutting the wood.)We can see on the top edge that the little space left over after thethree radius-lengths are identified is (8 - 3r ) feet. We also canconnect the centers of the two semi-circles, and this segment will gothrough the point of tangency. We now have a right triangle(shaded) with legs measuring 4 feet and [(8 - 3r ) + r ] =8 - 2r feet. The length of the hypotenuse is 2r feet. Using thePythagorean Theorem we have (2r )2 = (8 - 2r )2 + (4)2;4r 2 = 64 - 16r - 16r + 4r 2 + 16; 0 = 80 - 32r ; and finally r = ��

��

�

�= = 2.5 feet. The diameter of

the table would then be 5 feet.

Solution/Multiple Representations - Problem #6

If we let x = � �

�

�

�

� �

�

�+ + + +� � � � � � ��� , then ��

x = ��

(� �

�

�

�

� �

�

�+ + + +� � � � � � ��� ). This can be

rewritten as [�

�

�

�

�

� �

�

� �

�

�= + + + +� � � � � � ��� . By substituting into the very first equation we arrive at

[ [= +��

. Multiplying both sides of this equation by three leads to 3x = 3 + x ; 2x = 3; and x = ��

.

A visual representation may help us to make sense of the answer wefound in the solution above. Let�s take out the first term of 1 for right now.It will not be represented in our figure, but we�ll know to add it back inlater. So we are essentially rewriting the series in two parts:[1] + [ � � � � � � ����

�

�

�

� �

�

�+ + + ]. Consider the isosceles trapezoid in the firstfigure with one-third of the area shaded. This represents the initial termof �

� in the second part of our rewritten series. The next step is to shade a

third of one of the unshaded thirds. This is equivalent to adding (shading)�

� of �

� or the second term ( �

�)2. This step is shown in the second figure.

Again, taking a third of one of the newly created ninths, we shade a portionthat is equivalent to ( �

�)3, as seen in the third figure. We can see that we

are getting closer and closer to half of the trapezoid being shaded. Infact, we can see that every shaded region on the left of the trapezoid willhave a congruent unshaded region on the right of the trapezoid.Remembering that we started with 1 and added on this entire shaded regionthat is becoming �

�, we see that the sum of the series is 1 + �

� = �

�.

r r

r

r

r8-3r

MATHCOUNTS 2004-200570

Warm-Up 17Answers

1. 12 (C, S)

2. 10 (E, G, T)

3. 7 (C, E, G, T)

4. 72 (E, G, P, T)

5.�� (C, F, M)

6. 60 (C,F, M, S)

7. 50 (C, E, G, T)

8. 3 (C, F)

9. 144 (C, G, P, T)

10. 164 (C, E, F, G, S, T)

Solution - Problem #5We need the circle to completely lie within the three unit squares. If we make M the center of

our circle, then the largest possible radius of the circle would be ��

unit since that is the distancefrom M to A or B. Any larger radius would result in a circle that extends beyondthe squares. Notice, though, that as this circle with radius �

� unit is lowered

(Figure 1), there is more and more area around it. This shows that the circle�sradius could be enlarged if we move away from M as the center.

We must find the point Q such thatQA = QB = QD (Figure 2). Every point on MD isequidistant from A and B, so we really only need tobe sure QA = QD. We know that AB = 1 unit, soAM = �

� unit. Since MD = 1 unit, too, we can assign

MQ = x and then QD = 1 - x. Since QD = QA, wecan use the Pythagorean Theorem with right

triangle AMQ and see that (1 - x )2 = ( �

�)2 + x 2; 1 - 2x + x 2 = �

� + x 2;

1 - 2x = �

�; -2x = - �

�; and finally x = �

�. Plugging this into our

expression for radius QD, we see that the radius of this circle is 1 - x= 1 - �

� = �

� units.

Solution - Problem #10We know that one doll takes 20 minutes and results in $7 profit. One train takes 15 minutes

and results in $5 profit. The elf has eight hours and must make at least as many trains as dolls(# Trains ³ # Dolls). Let�s start by having the elf spend four hours on each type of toy. We then haveRow 1 in our table below. We still have as many trains as dolls, too, as seen in the last column of thetable. We make more money for charity with dolls, so let�s see what happens if we spend five hours ondolls and three hours on trains (Row 2). We made more money, but our number of trains is not greaterthan or equal to the number of dolls. Altering this scenario by giving up one doll (20 minutes) isn�tgoing to be enough time to make two more trains (30 minutes), which would make the number of eachtoy equal. So let�s take away two dolls (40 minutes) and add in two trains (30 minutes). We�ll have10 minutes left over, but our profit might still be greater than the $164 in Row 1 of the table. Thisnew data is in Row 3, and we see that the profit is only $161. The greatest possible profit is thenstill $164.

Figure 2

x

1 - x

Figure 1

Doll Train

T # $ T # $ Total $ T ³ D?4 hr 12 84 4 hr 16 80 $164 yes5 hr 15 105 3 hr 12 60 $165 no

4 �

� hr 13 91 3 �

� hr 14 70 $161 yes

MATHCOUNTS 2004-200572

Warm-Up 18Answers

1. 7 (C, E, M, P)

2. 30 (C, F, M)

3. 5 (E, G, P, T)

4.��

��

��

+ + (E, G, P)

5. 330 (C, F, M)

6.���� (C, E, P, S, T)

7. 12 (C, F, M, P)

8. 3 (E, G, P, T)

9. 7 (C, G, M)

10.�� (C, F, M, T)

Solution - Problem #1Let�s imagine that there is one English class, one Spanish class and one French class. They are

each offered at a different time during the school day. There is one big room that has 90 desks forthe English class . There is also a room with 75 desks for the Spanish classand a room with 42 desks for the French class. This is a total of90 + 75 + 42 = 207 desks. However, there are only 100 students in theschool. We are asked to find the smallest number of students who could betaking all three classes. Is it possible for this number to be zero? Even ifeach student is taking two of the three language classes, only 200 of the207 desks are being occupied each day. Since there are seven desks leftover, seven of the students must be taking a third language class. Thisscenario is demonstrated in the Venn diagram shown here.

Solution - Problem #5We need to draw in rectangle DFGH, but we need to

determine which of the three options shown in this first figure itwould resemble. We do know that FG = 12 and if P is the midpoint ofDF, determining PE might help us estimate how to draw rectangleDFGH. From the Pythagorean Theorem, DF = 40. (Triangle CDF is amultiple of a 3-4-5 right triangle.) This tells us PD = 20 and PE = 20.Since FG is considerably less than PE, we can guess that rectangleDFGH does not extend past point E and is positioned as shown in thefigure below. (In fact, it is possible to determine that if FG = 19.2,the rectangle would extend out just enough for segment GH to passthrough E.) We need to find the area of the shaded region in the figurebelow. The angles with dots in them are congruent, and the angles with double-loops are congruent. (They are pairs of alternate interior angles formed by a transversal intersectingparallel lines.) Additionally, angle H and angle G are right angles. Since ∆ DEF consists of a rightangle, a dot-angle and a double-loop-angle, the sum of these three anglesis 180 degrees. This means ∆ IHD is missing a double-loop in angle D and∆ FGJ is missing a dot in angle F. Now triangles DEF, IHD and FGJ areeach similar to each other. Using side proportions for ∆ DEF and ∆ IHD,we have �� ��

����

[ \= = , so x = 9 and y = 15. This means IE = 24 - 15 = 9.

Additionally, using ∆ DEF and ∆ FGJ, we have ����

�� ��= =Q P

, so n = 16 andm = 20. This means JE = 12. Since the shaded area is equal to half of thearea of rectangle CDEF minus the area of ∆ IEJ, we see that our desiredarea is > � �@ > � �� �@�

�

�

��� �� � ��× − = 384 - 54 = 330 square units.

(Solution is NOT unique!)