2014 Chapter Competition Solutions Are you wondering how we could have possibly thought that a Mathlete® would be able to answer a particular Sprint Round problem without a calculator? Are you wondering how we could have possibly thought that a Mathlete would be able to answer a particular Target Round problem in less 3 minutes? Are you wondering how we could have possibly thought that a Mathlete would be able to answer a particular Team Round problem with less that 10 sheets of scratch paper? The following pages provide solutions to the Sprint, Target and Team Rounds of the 2014 MATHCOUNTS® Chapter Competition. Though these solutions provide creative and concise ways of solving the problems from the competition, there are certainly numerous other solutions that also lead to the correct answer, and may even be more creative or more concise! We encourage you to find numerous solutions and representations for these MATHCOUNTS problems. Special thanks to volunteer author Mady Bauer for sharing these solutions with us and the

rest of the MATHCOUNTS community!

2014 Chapter Competition

Sprint Round

1. ( + 2)/3 = 5

+ 2 = 15

= 13 Ans.

2. A shirt sells originally for $10. 30% off is 10 × 0.7 = 7 Arturo pays $7 for the shirt. 50% more than $7 is 7 + (1/2 × 7) = 7 + 3.50 = 10.50 Ans.

3. Sally needs a cleaning solution of 1 part bleach to 8 parts water. That means she needs 1 quart of bleach for every 8 quarts of water. Therefore, she needs 24/8 = 3 quarts of bleach. 3 Ans.

4. A square is divided by its four lines of symmetry. The four lines are symmetry are the two diagonals and the horizontal and vertical bisectors of the square.

There are 8 triangles created by the bisectors.

Let’s put together two adjacent triangles that form a bigger triangle. There are 4 of those.

There’s no way to put three adjacent triangles together but we can do four adjacent

triangles. There are 4 of those. The total number of triangles is 8 + 4 + 4 =

16 Ans.

5. There are 4.5 inches of snow on the ground. Snow is falling at an average rate of 1.5 inches per hour for 9 more hours. At this rate, an additional 9 × 1.5 = 13.5 inches of snow will fall. This makes a total of 4.5 + 13.5 = 18 inches that will have fallen. 18 Ans.

6. Given xy = 100, find the positive difference between the maximum and minimum possible values of x + y, if x and y are positive. Integer pairs with a product of 100 are: 1 × 100 2 × 50 4 × 25 5 × 20 10 × 10 So, x + y has minimum and maximum values 10 + 10 = 20 and 1 + 100 = 101, respectively, and 101 – 20 = 81 Ans.

7. A bat catches 560 bugs in 7 nights. The bat catches the same number of bugs each night. Each night, the bat caught 560/7 = 80 bugs. Therefore, the bat caught 80 bugs on the fifth night. 80 Ans.

8. This problem basically asks us to find the length of the diagonal of a square with side length 20. You may recognize that the diagonal of a square also is the hypotenuse of a 45­45-90 right triangle. Recall that a 45­45-90 right

triangle with side length s, has hypotenuse of length s√2. Thus, the length of the diagonal is 20√2 Ans. If you aren’t familiar with this property of 45­45­90 right triangles, you can use the Pythagorean Theorem. Let x represent the length of the diagonal. 202 + 202 = x2 400 + 400 = x2 x2 = 800 x = √800 = √2 × √4 × √100 x = √2 × 2 × 10 = 20√2 Ans.

9. a ۞ b = ab/(a + b) 4a/(a + 4) = 3 4a = 3(a + 4) 4a = 3a + 12 a = 12 Ans.

10. A jar has 3 green, 2 red and 5 blue marbles. The jar has a total of 3 + 2 + 5 = 10 marbles. P(red) = 2/10 P(blue) = 5/10 P(red or blue) = 2/10 + 5/10

= 7/10 Ans.

11. Six points are in a plane and no three are collinear. Find the number of distinct lines that are determined by all possible pairs of points. There are 6 points and each point connects to 5 other points making a total of 6 × 5 = 30 lines. But we end up counting the same line twice, as this example illustrates:

The line from point 1 to point 4 is counted once among the lines between point 1 and the 5 other points, then again with the lines from point 4 to the other 5 points. This will be true for each line. So, 30/2 = 15 Ans.

If you are familiar with the concept of combinations, you can solve this problem by finding the number of combinations of 2 objects chosen from a collection of 6 objects. That’s 6C2 = 6!/(4! × 2!) = 30/2 = 15 Ans.

12. According to the bar graph 2 workers earn $41,000 to $50,000. 6 workers earn $51,000 to $60,000. 5 workers earn $61,000 to $70,000. 4 workers earn $71,000 to $80,000. 2 workers earn $81,000 to $90,000. To find the greatest mean salary we need to take the highest value in each range, multiply it by the number of workers in that range, add them all up and divide by the number of workers. Then we need to follow the same steps to find the lowest mean salary. Let’s see if we can avoid actually adding and multiplying those large values since we do have the aid of a calculator. Let’s write the sum of the lowest salaries as 2a + 6b + 5c + 4d + 2e = S. Since the difference between the lowest and highest salary in each range is $9000, the sum of the highest salaries is 2(a + 9000) + 6(b + 9000) + 5(c + 9000) + 4(d + 9000) + 2(e + 9000)) = 2a + 6b + 5c + 4d + 2e + 9000(2 + 6 + 5 + 4 + 2) = S + 9000(19) = S + 171,000. The difference in the means would be ((S + 171,000) − S)/19 = 171,000/19 = 9000 Ans.

13. The sum of the first n integers is (n + 1) × (n/2). So, the sum of the first 2000 positive integers is (2000 + 1) × (2000/2) = 2001 + 1000 = 2,001,000 Ans.

14. A stalactite grows downward at 0.004 inches per year. A stalagmite, directly below grows upward at 0.006 inches per year. The height of the ceiling is 10 feet (or 120 inches). Each passing year, the stalactite and stalagmite grow 0.006 + 0.004 = 0.01 inches closer. At

1

4

1

4

this rate, the number of years until they meet is 120/0.01 = 12,000 Ans.

15. The product of 3 positive integers is 144. The sum of the three integers is 26. What is the sum of their squares? Let’s look at the factor pairs of 144. 1 × 144 2 × 72 = 1 × 2 × 72 3 × 48 = 1 × 3 × 48 4 × 36 = 1 × 4 × 36 6 × 24 = 1 × 6 × 24 8 × 18 = 1 × 8 × 18 9 × 16 = 1 × 9 × 16 12 × 12 = 1 × 12 × 12 Of these triples, the one that works for us is 1 × 9 × 16 since 1 + 9 + 16 = 26. The sum of the squares is 12 + 92 + 162 = 1 + 81 + 256 = 338 Ans.

16. How many ordered triples (x, y, z) of positive integers have a sum of 6? Three triples have a sum of 6: 2, 2 and 2 yield 1 ordered triple 1, 1 and 4 yield 3!/2 = 3 ordered triples 1, 2 and 3 yield 3! = 6 ordered triples The total number of ordered triples is 1 + 3 + 6 = 10 Ans.

17. The slope of a line is k. The line passes through (0, k + 3) and (−3, 0). Find the value of k. Recall the formula for the slope of a line ∆y/∆x = (y2 − y1)/ (x2 − x1). Using the two given ordered pairs, we have x1 = −3, y1 = 0, x2 = 0 and y2 = k + 3. Using the formula, the line has slope k = [(k + 3) − 0]/[0 − (−3)] = (k + 3)/3 Ans.

18. How many permutations of the digits 1, 2, 3, 4 have 1 before 3 and 2 before 4? Let’s start with 1 as the thousands digit.

1324 1234 3 permutations 1243 When 2 is the thousands digit. 2134 2143 3 permutations 2413 When 3 is the thousands digit. 3 – uh, oh. Can’t happen because 1 must always come before 3. The same is true for 4. The number of permutations is 3 + 3 = 6 Ans.

19. What is the value of ( −3 8 )4? The cube root of −8 is −2. And (−2)4 = 16 Ans.

20. The line y = −(1/3)x + 4 intersects a perpendicular line at (6, 2). Find the y-intercept of the perpendicular line. The slope of the line, m, is −(1/3) The slope of the perpendicular line is −1/m = −1/−(1/3) = 3. Therefore, the equation for the perpendicular line is y = 3x + b. Since (6, 2) is on this line, we can substitute for x, y and m to get 2 = 3 × 6 + b 2 = 18 + b b = −16 The equation of the perpendicular line is y = 3x – 16. And when x = 0, we have y = 3(0) − 16 y = −16 The y-intercept has coordinates (0, −16) Ans.

21. Nine regions of a spinner are numbered 1 through 9. The probability of the spinner landing on a given region is proportional to the label on the region. 1 + 2 + 3 + 4 + 5 + 6 + 7+ 8 + 9 = 45. Therefore, the probability of the spinner landing on the region labeled 1 is 1/45. The probability of landing on the region

labeled 2 is 2/45, and so on. Thus, the probabilities of landing on each of the spaces labeled 3, 5, 7 and 9 are 3/45, 5/45, 7/45 and 9/45, respectively. It follows that, in a given spin, the probability that the spinner will land on a space labeled 1, 3, 5, 7 or 9 is 1/45 + 3/45 + 5/45 + 7/45 + 9/45 = 25/45 = 5/9 Ans.

22. The length of the diagonal of rectangle ABCD is half the rectangle’s perimeter less four-fifths the length of the shorter side. Find the ratio of the length of the shortest side to the length of the longest side of the rectangle. Let L and W represent the length and width of the rectangle, respectively. Four-fifths of the length of the shortest side of the rectangle is (4/5)W. The rectangle has diagonal of length √(L2 + W

2) = L + W − (4/5)W √(L2 + W

2) = L + (1/5)W L2 + W

2 = L 2 + (2/5)LW + (1/25)W

2 0 = (2/5)LW − (24/25)W

2 (2/5)L = (24/25)W 10L = 24W 5L = 12W W/L = 5/12 Ans.

23. An equilateral triangle has sides of length 2. A circular arc is centered at B. Find the length of a segment drawn parallel to side AC with endpoints D and E. Triangle ABC is an equilateral triangle, so angle B measures 60°. Let’s find the radius of the circle containing arc DE. To do so, we drop a perpendicular from B to point F on side AC.

We are told that AB = 2. Because ∆ABF is a 30-60-90 right triangle, we know that AF = ½(2) = 1 and the circle containing are

DE has radius, BF = √3. Segments BD and BE also are radii of the circle containing are DE. Note that ∆DBE ~ ∆ABC, which means that ∆DBE is an equilateral triangle. It follows, then, that BD = DE = √3 Ans.

24. On a number line, how many units apart are the two points that are twice as far from 0 as they are from 9? Let a be one of the points, and 0 < a < 9. The distances from 0 to a and from a to 9 are a and 9 − a, respectively. Since a is to be twice as far from 0 as it is from 9, we can write the equation 2 × (9 − a) = a. Solving for a, we see the first point is 18 − 2a = a → 3a = 18 → a = 6. Now, assume the second point, b, is located on the number line such that b > 9. The distances from 0 to b and from 9 to b are b and b − 9, respectively. We have the equation 2 × (b − 9) = b. Solving for b, we see the other point is 2b − 18 = b → b = 18. It follows that the distance between a and b is 18 – 6 = 12 Ans.

25. Given 3n3 + 3n2 + 4n = nn, find n. We can simplify this equation a bit by dividing both sides by n: 3n3 + 3n2 + 4n = nn → 3n2 + 3n + 4 = nn−1 n n That simplifies the equation, but it doesn’t tell us the value of n. Let’s evaluate the simplified equation for several values of n. The results are displayed in the table below.

n 3n2 + 3n + 4 nn−1 Is n a solution? 1 3 + 3 + 4 = 10 10 = 1 10 ≠ 1 → NO 2 12 + 6 + 4 = 22 21 = 2 22 ≠ 2 → NO 3 27 + 9 + 4 = 40 32 = 9 40 ≠ 9 → NO 4 48 + 12 + 4 = 64 43 = 64 64 = 64 → YES

n = 4 Ans.

26. How many different combinations of 3 numbers can be selected from the set {1, 2, 3, 4, 5, 6, 7, 8} so that the numbers could

represent the side lengths of a triangle? The sum of the lengths of two sides of a triangle must be greater than the other side. Start with a smallest side of 2: (2, 3, 4) (2, 6, 7) (2, 4, 5) (2, 7, 8) (2, 5, 6) -- 5 values Start with a smallest side of 3: (3, 4, 5) (3, 5, 6) (3, 6, 7) (3, 7, 8) (3, 4, 6) (3, 5, 7) (3, 6, 8) -- 7 values Start with a smallest side of 4: (4, 5, 6) (4, 6, 7) (4, 7, 8) (4, 5, 7) (4, 6, 8) (4, 5, 8) -- 6 values Start with a smallest side of 5: (5, 6, 7) (5, 7, 8) (5, 6, 8) -- 3 values Start with a smallest side of 6: (6, 7, 8) -- 1 value 5 + 7 + 6 + 3 + 1 = 22 Ans.

27. If 1/x + 3/y = 3/4 and 3/x − 2/y = 5/12, what is x + y? To solve this system of equations, we start by multiplying the first equation by 2 and the second equation by 3 to get 2/x + 6/y = 6/4 and 9/x − 6/y = 15/12. Next, we add the two equations and get 2/x + 9/x + 6/y − 6/y = 6/4 + 15/12 → 11/x = 33/12 → 33x = 132 → x = 4. Substituting for x in the first equation and solving for y yields 1/4 + 3/y = 3/4 → 3/y = 1/2 → y = 6. So, x + y = 4 + 6 = 10 Ans.

28. What is the smallest prime number that divides some number of the form 424242…42 + 1 or 424242…42 – 1? Let’s start checking from the smallest prime number. 2: Any number of this form will have a ones digit that is odd, namely 3 or 1, and therefore will not be divisible by 2. 3: For a number to be divisible by 3, the sum of its digits must be divisible by 3. The sum of

the numbers in 424242…42 is divisible by 3, so adding or subtracting 1 results in a number that is not divisible by 3. 5: For a number to be divisible by 5, it must end in 0 or 5. As previously stated, a number of this form will have a ones digit of 3 or 1, and is therefore not divisible by 5. 7: As with 3, the number 424242…42 is divisible by 7. Therefore, adding or subtracting 1 will result in a number that is not divisible by 7. 11: We can’t immediately rule out 11 as we did the other primes, so let’s divide and see if it is a factor of some number of the given form. Dividing, we find that 424242424243/11 = 38567493113 and 4242424241/11 = 385674931. The smallest prime that divides a number of this form is11 Ans.

29. Point E lies within rectangle ABCD. IF AE = 7, BE = 5 and CE = 8, what is DE?

Draw horizontal segment JK through point E and parallel to sides AB and DC. Then draw vertical segment GH through point E and parallel to sides AD and BC. As shown, we’ll let EG = x, EH = y, EK = w and EJ = z. It follows that JK = z + w and GH = x + y. Segments GH and JK are perpendicular to the sides of the rectangle and several right triangles are formed. Applying the Pythagorean Theorem, we have : x2 + z2 = 72 → x2 + z2 = 49 w2 + x2 = 52 → w2 + x2 = 25 w2 + y2 = 82 → w2 + y2 = 64 y2 + z2 = (DE)2 → DE = √(y2 + z2) We can manipulate the first three equations to get a new equation relating y and z.

Subtracting the second equation from the first yields the following: (x2 + z2) − (w2 + x2) = 49 − 25 z2 − w2 = 24 To this we can add the third equation to get: (z2 − w2) + (w2 + y2) = 24 + 64 y2 + z2 = 88 So, DE = √88 = 2√22 Ans.

30. What is the value of

�1 + �2 + �2 + √2 +… where all

subsequent numbers in the expression are 2s? Start by finding the value of the square of the expression. The square is

1 +�2 + �2 + �2 + √2 +…

Let x = �2 + �2 + �2 + √2 +…

√2 + x = x 2 + x = x2 x2 − x − 2 = 0 (x + 1)(x − 2) = 0 x = −1 or x = 2 Since x must be positive, x = 2.

1 +�2 + �2 + �2 + √2 +… = 1 + 2 = 3

Recall that the original expression is the square root of this expression. So, the value of the original expression is √3 Ans.

Target Round

1. Box A has 142 marbles. Box B has 152 marbles. Box C has 136 marbles. Moving marbles only from B to C, what’s the least number of marbles that need to be transferred so that C contains more marbles

than each of the other two boxes? If we transfer 6 marbles from B to C, then B will have 146 marbles and A and C will each have 142 marbles. If we transfer 2 more marbles from B to C then C will have 144 marbles as will B. If we transfer one more marble from B to C we will have: Box A with 142 marbles Box B with 143 marbles. Box C with 145 marbles. The total number of marbles transferred 6 + 2 + 1 = 9 Ans.

2. What is the largest prime that divides both 20! + 14! and 20! – 14!? The expression 20! + 14! can be rewritten as 20 × 19 × 18 × 17 × 16 × 15 × 14! + 14! = 14!((20 × 19 × 18 × 17 × 16 × 15) + 1). Similarly, 20! – 14! can be rewritten as 20 × 19 × 18 × 17 × 16 × 15 × 14! − 14! = 14!((20 × 19 × 18 × 17 × 16 × 15) − 1). The greatest common factor for 20! + 14! and 20! – 14! is 14!, and 13 is the largest prime factor of 14! So the greatest common, prime factor of these two numbers is 13 Ans.

3. In how many ways can the letters in the word NINE be arranged? If all four letters were different the answer would be 4! or 24. But since two of the letters are the same, we have to account for duplicates. So the total number of arrangements is 24/2 = 12 Ans.

4. A vendor offers discounts for orders of 11 or more shirts as follows: 11-25 – 10% off 26-50 – 15% off 51-100 – 20% off 101-250 – 30% off 251 or more – 35% off For how many different quantities of shirts would the cost exceed the cost of buying the

least number of shirts at the next discount level? Let x = the cost of a shirt. We will consider each discount level separately to count the number of quantities for which the cost exceeds the cost of buying the least number of shirts at the next discount level.

11 for 11(0.9x) = 9.9x

10 for 10x 9 for 9x 26 for 26(0.85x) = 22.1x

25 for 25(0.9x) = 22.5x 24 for 24(0.9x) = 21.6x 51 for 51(0.8x) = 40.8x

50 for 50(0.85x) = 42.5x 49 for 49(0.85x) = 41.65x 48 for 48(0.85x) = 40.8x 101 for 101(0.7x) = 70.7x

100 for 100(0.8x) = 80x 90 for 90(0.8x) = 72x 89 for 89(0.8x) = 71.2x 88 for 88(0.8x) = 70.4x 251 for 251(0.65x) = 163.15x

250 for 250(0.7x) = 175x 240 for 240(0.7x) = 168x 235 for 235(0.7x) = 164.5x 234 for 234(0.7x) = 163.8x 233 for 233(0.7x) = 163.1x

In all the number of quantities for which this is true is 1 + 1 + 2 + 12 + 17 = 33 Ans.

5. Each term in the sequence 13, 9, 18, … is the sum of three times the tens digit and two times the units digit of the previous term. Find the greatest value of a term in the sequence. Let’s look at the next several terms to see if this is a repeating sequence. After 18 we get, 19, 21, 8, 16, 15, 13 and it does repeat. The highest value in this sequence is 21 Ans.

6. In square ABCD, sector BCD is drawn with center C and BC = 24. A semicircle with diameter AE is drawn tangent to the sector BCD. Points A, E and D are collinear. Find AE. Since BC = 24, the radius of the sector is 24. When two circles are tangent to each other a line can be drawn from the center of one circle to the center of the other one. (In the figure, the entire circles are displayed for clarity.)

Right triangle CDF has leg CD (a side of the square). It has hypotenuse CF, which is the line drawn from the circle with center C. And is has leg DF, which is the portion of side AD with one endpoint at the center of the circle, F, with diameter AE. Let r represent the radius of the circle with center at point F. Again, we know that AD = 24. It follows that FD = AD – AF = 24 – r, and CF = 24 + r. Using the Pythagorean Theorem, we have

(24 − r)2 + 242 = (24 + r)2 242 − 48r + r2 + 242 = 242 + 48r + r2

−48r + 576 = 48r 96r = 576

r = 6 Therefore, AE = 2r = 2 × 6 = 12 Ans.

7. How many distinct unit cubes are there with two faces painted red (R), two painted green (G) and two painted blue (B)? We’ll consider a cube with pairs of opposite faces numbered as follows: 1 ↔ 2 (front/back), 3 ↔ 4 right/left) and 5 ↔ 6 (top/bottom).

No discount

LEVEL 1-10

20% OFF

LEVEL 51-100

10% OFF

LEVEL 11-25

20% OFF

LEVEL 26-50

30% OFF

LEVEL 101-250

17 quantities (234-250 shirts)

12 quantities (89-100 shirts)

2 quantities (49-50 shirts)

1 quantity (25 shirts)

1 quantity (10 shirts)

CUBE 1: All opposite faces of the same color. 1G ↔ 2G 3B ↔ 4B 5R ↔ 6R Note: Each color has is arbitrarily assigned to a pair of opposite faces and when rotated the result is still CUBE 1. CUBE 2: One pair of opposite faces both red. Other opposite faces one blue, one green. 1G ↔ 2B 3B ↔ 4G 5R ↔ 6R

CUBE 3: One pair of opposite faces both blue. Other opposite faces one red, one green. 1G ↔ 2R 3R ↔ 4G 5B ↔ 6B

CUBE 4: One pair of opposite faces both green. Other opposite faces one red, one blue. 1B ↔ 2R 3R ↔ 4B 5G ↔ 6G

CUBE 5: No opposite faces of the same color. Left face green and right face blue. 1R ↔ 2B 3G ↔ 4B 5R ↔ 6G

CUBE 6: No opposite faces of the same color. Left face blue and right face green. 1R ↔ 2B 3B ↔ 4G 5R ↔ 6G

CUBE 5 and CUBE 6 are different because no rotation of CUBE 5 results in any rotation of CUBE 6. In fact, if you turn them upside down (6 on top and 5 on the bottom), the lateral faces (1-4) of CUBE 5 are now in the original order of the lateral faces of CUBE 6, and vice versa. As we have shown, the number of different cubes is 6 Ans.

8. What is the greatest possible area of a triangle with vertices on or above the x-axis and on or below the parabola y = −(x − ½)2 + 3 y = −(x2 − x + ¼) + 3 y = −x2 + x − ¼ + 3 y = −x2 + x + 11/4 For a parabola given by the equation y = ax2 + bx + c, the axis of symmetry is

x = −b/2a. In this case, a = −1 and b = 1. So, the parabola is symmetric across the line x = 1/2, and its vertex occurs at the point (½, 3). Since this parabola opens down the vertex is also the highest point. Now let’s determine where this parabola crosses the x­axis. We will use the quadratic formula to solve 0 = −x2 + x + 11/4. Since a = −1, b = 1 and c = 11/4, we have

2 11241 1 4( 1)( )4

2 2( 1)b b acx

a− ± − −− ± −= =

−

1 1 ( 11) 1 12 1 2 3 1 32 2 2 2

− ± − − − ± − ±= = = = ±− − −The parabola crosses the x-axis at (½ − 3,

0) and (½ + 3, 0). The triangle shown has vertices on the parabola and the x­axis at A(½, 3), B(½ − 3, 0) and C(½ + 3, 0).

Point D(½, 0) is the midpoint of segment BC.

The length of base BC is ½ + 3 − (½ − 3 )

= 2 3 , and the height, which is the length of segment AD, is 3. The area of the triangle is

½ × 3 × 2 3 = 3 3 Ans.

Team Round

1. The area of square F is 16. The area of square B is 25. The area of square H is 25. Find the area of square D.

Let a, b, c, d, e, f, g and h represent the side lengths of squares A through H, respectively. Square F has area 16, and square B and H have area 25. So, it follows that f = 4, b = 5 and h = 5. From the figure we see that b = f + e. So 5 = 4 + e and e = 1. We also can see that b + c = f + d. So 5 + c = 4 + d and c = d − 1. Since a + b + c = a + f + d, a = b + f. So a = 5 + 4 and a = 9. Next, since h + g = a + f, it follows that 5 + g = 9 + 4 and g = 8. According to the figure, e + b = c. That means 1 + 5 = c and c = 6. Finally, we have d = c + 1, so d = 7. The area of square D is 7 × 7 = 49 Ans.

2. Nine consecutive positive even integers are entered into a 3×3 grid. The sums of the three numbers in each row, column and diagonal are the same. Find the average value of the five missing numbers.

We’ll let u, v, w, x and y represent the missing numbers, then calculate their average value, which is (u + v + w + x + y)/5. The sums of the numbers in each row, each column and each diagonal must be equal, so we have the following equivalent sums: 38 + u + v = w + x + 40 = 46 + y + 50 = 38 + w + 46 = u + x + y = v + 40 + 50 = 38 + x + 50 = 46 + x + v. That means 38 + x + 50 = 46 + x + v 88 + x = 46 + x + v 88 = 46 + v v = 42. And 38 + u + v = v + 40 + 50 38 + u = 90 u = 52. In addition, 38 + w + 46 = v + 40 + 50 84 + w = 42 + 40 + 50 84 + w = 132 w = 48. Finally, 38 + w + 46 = u + x + y 84 + 48 = 52 + x + y 132 = 52 + x + y 80 = x + y. It follows that, (u + v + w + x + y)/5 = (52 + 42 + 48 + 80)/5 = 222/5 = 44.4 Ans.

3. To provide direct flights between four cities takes 6 routes. We must find the number of routes needed to connect 15 cities. Each city can be connected to the other 14 cities by 14 unique routes. However, we will end up counting each route twice. That means the number of routes needed is actually (15 × 14)/2 = 105 Ans.

4. Plan A costs $250 to set up and $4.25 per shirt. Plan B costs $150 to set up and $5.25 per shirt. Find the number of shirts to print

that cost the same with either plan. Let x represent the number of shirts for which the cost is the same using either plan. Now, we solve 250 + 4.25x = 150 + 5.25x and get x = 100 Ans.

5. $1000 is in an account that earns 3% interest, compounded annually. The account now is worth $1125.51. How many years has the money been in the account? The table shows the year-end balance of an account with an initial balance of $1000 that earns 3% interest, compounded annually.

Year Ending Balance 1 1.03 × 1000 = 1030 2 1.03 × 1030 = 1060.90 3 1.03 × 1060.90 = 1092.727 4 1.03 × 1092.73 = 1125.50881

That means, 4 years after the account balance was $1000, the account earned enough interest to have a balance of $1125.51. 4 Ans.

6. Abigail, Bartholomew and Cromwell play a game in which they take turns adding 1, 2, 3, or 4 to a sum in order to create an increasing sequence of primes. We'll let A, B and C stand for Abigail, Cromwell and Bartholomew, respectively. Let's suppose the game proceeds as follows: A chooses 2 B adds 1 to make 3 C adds 2 to make 5 A chooses 2 to make 7 B chooses 4 to make 11 C chooses 2 to make 13 A chooses 4 to make 17 B chooses 2 to make 19 C chooses 4 to make 23 A chooses... Well, A can’t choose, since there are no primes between 24 and 27. That means

Cromwell chose the last prime. But we want Bartholomew to end up with the last prime. So let's suppose, instead, that the game proceeds this way: A chooses 2 B adds 3 to make 5 C adds 2 to make 7 A chooses 4 to make 11 B chooses 2 to make 13 C chooses 4 to make 17 A chooses 2 to make 19 B chooses 4 to make 23 This time, Bartholomew has chosen the last prime, and the total number of primes made was 8 Ans.

7. How many positive integers less than 1000 do not have 7 as any digit? Let’s start with positive 3-digit integers. The possible values for the digit the hundreds place are 1 through 6, 8 or 9. That’s 8 choices. The possible values for the digit the tens and ones places are of 0 through 6, 8 or 9. That’s 9 choices for the tens place value and 9 choices for the ones place value. Thus, there are 8 × 9 × 9 = 648 positive 3-digit integers. For positive 2-digit integers, the possible values for the digit in the tens place are 1 through 6, 8 or 9. That’s 8 values. And again, there are 9 choices for the ones place value. So there are 8 × 9 = 72 positive 2­digit integers. Finally, the positive 1-digit integers are 1, 2, 3, 4, 5, 6, 8 and 9. That’s 8 positive 1-digit integers. That brings the total number of integers to 648 + 72 + 8 = 728 Ans.

8. In the game, vowels are worth 1 point and consonants are worth 2 points. When more than one letter of the same type appears consecutively, each letter is worth twice as much as the one before. Find the absolute difference between the values of QUEUEING and SYZYGY. Following are the point values for each letter in QUEUEING.

Q is a consonant worth 2. U is a vowel worth 1. E is an adjacent vowel worth 2. U is an adjacent vowel worth 4. E is an adjacent vowel worth 8. I is an adjacent vowel worth 16. N is a consonant worth 2. G is a consonant worth 4. QUEUEING is worth a total of 2 + 1 + 2 + 4 + 8 + 16 + 2 + 4 = 39 points. Following are the point values for each letter in SYZYGY. S is a consonant worth 2. Y is an adjacent consonant worth 4. Z is an adjacent consonant worth 8. Y is an adjacent consonant worth 16. G is an adjacent consonant worth 32. Y is an adjacent consonant worth 64. SYZYGY is worth a total of 2 + 4 + 8 + 16 + 32 + 64 = 126 points. That gives us an absolute difference of 126 – 39 = 87 Ans.

9. Using each of the digits 1 to 6 exactly once, how many six-digit integers can be formed that are divisible by 6? An integer is divisible by 6 if it is divisible by both 2 and 3. So, for an integer to be divisible by 6, it must be even (divisible by 2), and the sum of its digits must be divisible by 3. Well, 1 + 2 + 3 + 4 + 5 + 6 = 21, which is divisible by 3. But only half of the integers composed of these six digits are even. There are a total of 6! = 720 different ways to arrange the digits 1, 2, 3, 4, 5 and 6 to form a six-digit integer. Therefore, the number of these integers that are divisible by 6 is 720/2 = 360 Ans.

10. Points D, E and F lie along the perimeter of ∆ABC such that AD, BE and CF intersect at point G. AF = 3, BF = BD = CD = 2 and AE = 5. Find BG/EG.

This problem can be solve using mass-point geometry to solve this problem. Let D serve as the fulcrum balancing segments BC. Likewise, E and F each serve as the fulcrum balancing segments AC, and AB, respectively. So the product of the mass at B and the length of segment BF must be equal to the product of the mass at A and the length of segment AF. Since AF = 3 and BF = 2, we assign a mass of 3 to B, which leads to assigning a mass of 2 to A. Now since the mass at C must be such that the product of the mass and the length of segment CD must equal the product of the mass at B and the length of segment BD. Since BD = CD = 2, we assign a mass of 3 to C. Therefore, the mass at E (the sum of the masses at A and C) is 5.

Finally, if G serves as a fulcrum balancing segment BE, the product of the mass at B and the length of segment BG must equal the product of the mass at E and the length of segment EG. While we don’t know the exact lengths of segments BG and EG, we do know that in order for segment BE to remain balanced on G, the segments must be in the ratio BG/EG = 5/3 Ans.