Solution: Let x be the amount of paint in one stripe. Let r and w be the amount of red and white paint respectively in a pink stripe. Then x = r + w.

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1 1. Before starting to paint, Paul had 130 ounces of blue paint, 164 ounces of red paint and 188 ounces of white paint. Paul painted four equally sized stripes on the wall, making a blue stripe, a red stripe, a white stripe and a pink stripe. Pink is a mixture of red and white, not necessarily in equal amounts. When Paul had finished, he had equal amounts of blue, red and white paint left. Find the total number of ounces of blue paint Paul had left. Solution: Let x be the amount of paint in one stripe. Let r and w be the amount of red and white paint respectively in a pink stripe. Then x = r + w. Since there is an equal amount of paint left in all three colors, we must have 130 x = 164 x r = 188 x w. Then 130 x = 164 x r implies r = 34, and 130 x = 188 x w implies w = 58. Hence x = r + w = 92. Since x = 92, Paul must have had = 38 ounces of blue paint left.

2 2. Suppose three fair six-sided dice, one red, one green, and one blue, are rolled. How many ways can the numbers on the dice sum to 10? Solution: The ways that 10 can be formed from the sum of the numbers on three six-sixed dice are: 1, 3, 6 1, 4, 5 2, 2, 6 2, 3, 5 2, 4, 4 3, 3, 4 Note that 1, 3, 6 appearing on the red die, green die and blue die respectively is different from them appearing on the green die, blue die and red die respectively. There are six number combinations. There are 3! = 6 ways of ordering each combination of distinct numbers on the red, green and blue dice, and 3 ways of ordering each combination where two of the numbers are repeated so there are = 27 ways that the number 10 can be formed.

3 3. Assume that x and y are real numbers such that Find cos(2x) cos(2y) Solution: in lowest terms. sin(x) sin(y) = 3 and cos(x) cos(y) = 1 2 First, note our given assumptions imply sin(x) = 3 sin(y) and cos(x) = 1 cos(y). This implies 1 = sin 2 (x) + cos 2 (x) = 9 sin 2 (y) cos2 (y). Since cos 2 (y) = 1 sin 2 (y), we have 2 1 = 9 sin 2 (y) (1 sin2 (y)) = 35 4 sin2 (y) Solving for sin2 (y) gives sin 2 (y) = Now, sin 2 (x) = 9 sin 2 3 (y) = 9 35 = 27. Then a double-angle identity for cos(x) gives: 35 cos(2x) cos(2y) = 1 2 sin2 (x) 1 2 sin 2 (y) = =

4 4. The lengths of the sides of a triangle with positive area are log 10 (12), log 10 (75) and log 10 (n) where n is a positive integer. Find the number of possible values of n. Solution: Since these three numbers are the sides of a triangle we must have the following true for all positive integer n: By properties of logarithms this gives: log 10 (12) < log 10 (75) + log 10 (n); log 10 (75) < log 10 (12) + log 10 (n); and log 10 (n) < log 10 (12) + log 10 (75). log 10 (12) < log 10 (75n); log 10 (75) < log 10 (12n); and log 10 (n) < log 10 (900). and so 12 < 75n 75 < 12n and n < 900. The last two inequalities give: < n < 900. Now, = 6.25 and since n is an integer n 899. Hence there are = 893 possible values of n.

5 5. Find the number of positive integers that are divisors of at least one of 10 7, 15 5 and 9 3. Solution: First, note the prime factorizations of each number are: 10 7 = , 15 5 = and 9 3 = 3 6. Let A be the set of divisors of , B the set of divisors of and C the set of divisors of 3 6. We wish to find A B C where X denotes the number of elements of set X. By the Principle of Inclusion-Exclusion, A B C = A + B + C A B A C B C + A B C. Computing each individually, A = 8 8 = 64 (8 choices for the power of 2 and 8 choices for the power of 5.) B = 6 6 = 36 C = 7 A B = 6 (The only possible divisors of and are of the form 5 k for 0 k 5) A C = 1 B C = 6 A B C = 1 So A B C = = 95.

6 6. A group of children held a grape-eating contest. When the contest was over, the winner had eaten n grapes and the child in k th place had eaten n + 2 2k grapes for 1 k l. There were a total of 69 grapes eaten in the contest. Find the smallest possible value of n. Solution: If the winner eats n grapes, then the second place winner eats n 2 grapes, the third place winner eats n 4 grapes, etc. Hence if there are l children in the contest, the number of grapes eaten is. Now n + (n 2) + (n 4) +... (n + 2 2l) = l (n + 2 2k) = 69. k=1 ( l l ) ( l ) ( l ) (n + 2 2k) = n + 2 2k k=1 k=1 k=1 l(l + 1) = nl + 2l 2 2 = nl + 2l l(l + 1) = nl + 2l l 2 l = l(n l + 1). k=1 Hence l divides 69. The possible divisors of 69 are 1, 3, 23, 69. If l = 1, then n l + 1 = 69 and so n = 69. If l = 3, then n l + 1 = 23 and so n = 25. The other two cases are analogous, so the smallest possible value of n is 25.

7 7. The numbers 1447, 1005 and 1231 have something in common: each is a four digit number beginning with 1 that has exactly two identical digits. How many such numbers are there? Solution: To start, let s count the number of four digit numbers beginning with 1 that have exactly two 1 s with the other two distinct. The first 1 is fixed, so we have three choices for the other 1. Then there are 9 digits remaining for the third digit and 8 digits remaining for the fourth digit, so there are = 216 such numbers. Now, if there is only one 1 with a different digit repeated, then we have 3 choices for the places in which that digit can occupy. There are 9 choices for that digit. Then there are 8 choices for the last digit. Again, there are = 216 such numbers. In total, there are = 432 four digit numbers beginning with 1 that have exactly two identical digits.

8 8. Three of the vertices of a cube of side length x are P = (7, 12, 10), Q = (8, 8, 1) and R = (11, 3, 9). Find x. P Q R Solution: The surface area of a cube of side length x is A = 6x 2. Let s find the distance between the given points to determine their positions within the cube. P Q = (7 8) 2 + (12 8) 2 + (10 1) 2 = 98 P R = (7 11) 2 + (12 3) 2 + (10 9) 2 = 98 QR = (8 11) 2 + (8 3) 2 + (1 9) 2 = 98 Hence P, Q and R are equidistant. Thus, they lie on opposite corners of two faces that share a common edge. By the Pythagorean Theorem, this gives x 2 + x 2 = 98. Hence x 2 = 49 and x = 7.

9 9. When a right triangle is rotated about one leg, the volume of the cone produced is 800π cm 3. When the triangle is rotated about the other leg, the volume of the cone produced is 1920π cm 3. What is the length in cm of the hypotenuse of the triangle? Solution: Suppose the legs of the right triangle have legs a and b. After rotating about the leg of length a, we obtain a cone with height a and radius b. If we rotate about the leg of length b, we obtain a cone with height b and radius a. Using the formula for the volume of a cone, V = 1 3 πr2 h, we obtain 1 3 πb2 a = 800π and 1 3 πa2 b = 1920π. Hence b 2 a = 2400 and a 2 b = This implies 2400 b = ab = 5760 a 2400 = b 2 a = b 2 ( 12b 5 and so a = 5760b 2400 = 12b. This gives 5 ) = 12b3 5. Therefore b = 1000 giving b = 10 and so a = = Therefore, the length of the hypotenuse is c = a 2 + b 2 = = 676 = 26 cm.

10 10. A string is wound symmetrically around a circular rod. The string goes exactly 4 times around the rod. The circumference of the rod is 4cm and its length is 12cm. Find the length of the string. Show all of your work. Hint: You may need to unwrap your mind a bit for this one. 4 a a a a Solution: Cut the circular rod lengthwise and unfold it into a rectangle as shown above. This breaks the rod into a series of right triangles whose legs have lengths 3 cm and 4 cm respectively. By the Pythagorean Theorem = a 2 and so a = = 25 = 5 cm. Therefore, the string has length 4a = 20 cm.