6. This sum can be rewritten as 4( ). We then recall the formula n =

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1 . c = 9b = 3 b = 3 a 3 = a = = 6.. (3,, ) = = = We see that this is equal to. 3 = ( +.) 3. Using the fact that (x + ) 3 = x 3 + 3x + 3x + and replacing x with., we find that. 3 = x + = 7 means that either 8x + = 7 or 8x + = 7. Finding the solutions to both of these, x = or 9. These are both perfectl valid, so there are solutions.. Let the square have side length s. Since the area of a square is equal to s and its perimeter is equal to s, the question tells us that s = s. Since s 0, we get that s =. 6. This sum can be rewritten as ( ). We then recall the formula n = So, because = (6)() n(n + ). = 3, our desired answer is 3 = Since the letters that the crowd chant repeat ever three letters, ever third position will be A. 03 is divisible b 3, so the 03 th position will be A. 8. Let F be the number of fish that AJ gives to Soonho. After the transfer, AJ has F fish and Soonho has 3 + F fish. Since AJ has exactl double the amount of fish that Soonho has, F = (3 + F ). Solving this equation for F gives F =. 9. We observe that for an x, (x ) 3 = (x x + )(x ) = x 3 3x + 3x. (This is an example of the Binomial Theorem.) So, plugging in x = ields = ( ) 3 = 3 = We see that there are 36 das in a non-leap ear, and that 36 leaves a remainder of when divided b 7, which is the number of das in a week. Thus, after each non-leap ear, we see that October 3th will move ahead b one da. (For instance, in 0 it will be a Monda.) However, on a leap ear, there are 366 das, which leaves a remainder of when divided b 7. So this tells us that in a leap ear, October 3th will move ahead b two das. Since the ear 06 is a leap ear, we see that in 0 it will be a Tuesda; in 06, Thursda; in 07, Frida; in 08, Saturda; in 09, Sunda. Thus the answer is 09.. If x denotes the amount of mone Kelvin had in the beginning, then we get that x+0 = 3(x 0). Solving this equation ields x = $0.. Let the radii of the solids be r and the height of the solids be h. Recall that the volume of a cone is 3 πr h, while the volume of a clinder is πr h. Thus, the clinder can hold πr h 3 πr h = 3 times as much water as the cone. 3. Let the common ratio of the geometric sequence be x. Thus, since the first term is 8 and the fourth term is, we have 8x 3 = which can be simplified to x 3 = 8 7. Thus, x =, and so ( ) 3 the sixth term is 8 =

2 . The dimensions of the sheet of paper can be converted to 36 inches b 60 inches. So, if both the sheet of paper and the cards are positioned verticall, we can completel cover the paper b placing 6 cards across and cards down. Therefore, our desired answer is 6 = 90.. Since onl one person is ling, Sabeenee must be telling the truth (Steven is not ling) as if he were ling, Steven is ling and then two people are ling. So that means Steven is not ling either, and therefore AJ did not eat the cake. Kelvin claims Alex ate the cake. If he is telling the truth, then Alex and AJ must be ling. So Kelvin must be telling a lie, and therefore, Kelvin ate the cake. 6. Let the probabilit that the product of the numbers is even be p. Then p is the probabilit that the product is odd. The product of 3 integers can be odd onl if each of the 3 integers are also odd. The probabilit of each integer being odd is =, so p = ( )3 = 8. Therefore p = Let the perimeter of the square and equilateral triangle be x. Thus, each side of the square has length 3x, while each side of the equilateral triangle has length x. Since the area of a square with side length s is s, the area of the square is thus 9x. Also, since the area of an equilateral triangle with side length s is equal to s 3, the area of the equilateral triangle is (x) 3 = x 3. The ratio of the area of the triangle to the square is thus x 3 9x = Suppose a and b are diametricall opposite such that a < b. Then there are equal amounts of numbers from a to b on the circle and from b to a. Since there are n total numbers excluding a and b, there must be n numbers from a to b. And since the numbers are placed in order, we have that b a = n + = n. Specificall, since a = 7 and b = 38, we see that b a = 38 7 = = n, and so n =. 9. Note that x = x = x = This can go on forever. However, if it goes on forever, then we can sa that x = 6 + x. Squaring both sides and rearranging ields that x x 6 = 0 so x is either 3 or. Clearl, x is positive, so x = Note that the maximum sum of an two numbers from to is just + = 9. Thus, the sum of an two adjacent numbers must be,, 9, 6,, 36, or 9. So, the pair of adjacent numbers including 8 must sum to either, 36, or 9. However, the can t sum to 36 as 36 = and a number can t be repeated. Also, the can t sum to 9 as 9 = and 3 doesn t fall between and. However, = 8 + 7, and so the onl number that can be next to 8 is 7.. Let s sa Soonho holds up all ten fingers x times. Then the other 0 x times, he can hold at most 9 fingers. So in total, he can hold up at most (0)(x) + (9)(0 x) fingers, which is equal to 90 + x. So since we want to maximize x, clearl x = does the trick.. The left-hand side can be written as ( n ) = n+. Therefore, n + = 03 so n = If he picked the fair coin, there is a chance of it being heads times. If he chose the unfair 6 one, then there is a chance that it would happen. Thus, using conditional probabilit, the

3 chance that he chose the unfair coin are 6 + = On each edge of the cube, there are 0 cubes. The first and last cubes are corner cubes, and have 3 faces painted. The middle 8 have two faces painted. There are 8 such cubes per edge and edges, thus 96 cubes in total. The onl other cubes are the cubes part of the 8 8 center square on each face, which onl have one face painted. Thus, the answer is 96.. If we want to find the probabilit that event A happens given that event B happens, we have to divide the probabilit of both events happening b the probabilit of event B happening. (This is how we can calculate such conditional probabilities.) So, in this case, event A is choosing a student named Jim, while event B is choosing a student whose name begins with J. Hence, the chance of A and B both occurring is equal to, since there are students whose 8 name is Jim and Jim starts with J. Also, the chance of B occurring is equal to, since there 8 are students whose name begins with J. Therefore, our answer is equal to Chance of A and B Chance of B = 8 8 =. Alternativel, note that there are students whose names start with J, of whom are named Jim. Therefore our probabilit is. 6. If an integer ends in a zero, then it must be divisible b 0, and therefore also divisible b. Therefore, the smallest such number is simpl = 0. 0 = 3 3. Therefore, the number of factors of 0 is (3 + )( + )( + ) = Because ABX is isosceles, we have BAX = ABX = 0. So, we have BXA = = 0. In addition, AXC is isosceles, and thus AXB = ACX = 80 0 = 0. Thus finall, XAC = = The volume of a clinder with radius r and height h is πr h. Therefore, if both the radius and height are doubled, the volume will be = 8 times greater. The new cake now feeds 6 people, or 6 more than the original 8 people. 9. The sum of the columns and rows takes into account ever entr twice, so no matter what arrangement the numbers are in, the sum of all the columns and all the rows is ( ) == 90. Thus, the sum of the diagonals is 90 = If gosling is born out of ever 0 duck eggs, then 00 0 = 0 goslings are born out of 00 duck eggs. One-third of these is an ugl duckling, so there are 0 3 = 3 ugl ducklings. 3. Note that + n = n +. Thus, we get that n ( + ) ( + ) (... + ) = = 0 = 0, since all of the numerators and denominators except for the first and last cancel. 3. If both numbers were odd, then their sum would be even. Therefore, one of the numbers must be even. The onl even prime is, so the larger number is 03 = 0. 3

4 33. Let the side length be x. Then its surface area is 6x and it s side length is x, so using the fact that the surface area is 6 times its side length, then 6x = 6x = x =. The volume is x 3 which in this case is just 3 =. 3. The diagonal of this rectangle will be the diameter of the circle. Since the side lengths are 6 and 8, b the Pthagorean Theorem, the diameter will be = 0. The area of the circle is ( ) 0 then πr = π = π. 3. Suppose the three numbers chosen are a, b, c with a < b < c. Then notice that even if a is the largest term less than b and that if c were the least term greater than b, then a + b = c. This fails the Triangle Inequalit, which sas that if a, b, c are the sides of a triangle, then a + b > c. So no such numbers work and the probabilit is We can break this problem into cases depending on the number of elements in each subset. We will have different cases since a subset can contain either element, 3 elements, elements, 7 elements, or 9 elements. Case : Each subset contains onl element. There are 0 was to create a set with onl elements. Case : Each subset contains 3 elements. There are 0 was to pick the first element, 9 was to pick the second element and 8 was to pick the third element. However since order does not matter, we have over counted. Thus we must divide b the number of was to arrange 3 elements, which is 3. So our answer is or 0. 3 Case 3: Each subset contains elements. There are 0 was to pick the first element, 9 was to pick the second element, 8 was to pick the third element, 7 was to pick the fourth, and 6 was to pick the fifth element. However since order does not matter, we have over counted once again. Thus we must divide b the number of was to arrange elements, which is 3. So our answer is or. Case : Each subset contains 7 elements. Note that we can compute this case similar to how we did the previous cases, but we can exploit smmetr to get the answer more quickl. Note that ever time ou pick a subset containing 3 elements, ou leave behind a subset containing 7 elements. Thus the number of subsets of 7 elements is equal to the number of subsets of 3 elements, which is 0. Case : Each subset contains 9 elements. B the logic of Case, we see that this case has the same number of subsets as Case so this case has 0 subsets. Summing up the sums of all of the cases, we have , which is, our final total. Remark : Did ou notice that = 0 = 0, and that 0 was the number of elements in our original set? Do ou think that this is a coincidence? 37. If we unfold the cube to form a flat net, we remember that the shortest distance from one corner to the opposite corner is a straight line. In this case, this line is the hpotenuse of a right triangle with legs and, thus the minimum distance is. 38. Consider 7 = 9. 7 = 9 ends in 0. Note that the hundreds digit never matters. We have that 7 ends in 0, so since 7 6 = (7 ), our desired answer is also This is a classic stars and bars problem. We want to distribute 0 indistinguishable balls into 3 distinguishable boxes, each of which has at least one ball. This is because a, b, and c are positive, not just nonnegative. Therefore, put ball into each box first. Now, we need to distribute 7

5 balls into 3 distinguishable ( ) ( boxes. ) In order to do this, we use 7 stars and bars. Therefore, we have a total of = = 36 distributions. This means that there are 36 triplets (a, b, c) in positive integers to the equation a + b + c = Clearl x = 0 does not satisf the equation. Thus, we can divide b x to get x+ x =. Squaring this equation gives x + x + =, and multipling both sides b x ields x + x + = 0. Therefore, a =.. Note that the coordinates of the reflection of his house with respect to the river is (, ). If we call this point H, then we see that from an point on the river, Dennis will be equidistant from both his house and the point H. Therefore, the minimum distance he must travel from the origin, to the river, then to his house is equal to the distance from the origin, to the river, to H. Lastl, we observe that the shortest path between two points is a straight line. So, this minimum distance is achieved if Dennis travels from the origin to H in a straight line, meeting the river along the wa. And b the Distance Formula, this distance is equal to ( 0) + ( 0) = 3.. Let the sum be S. Then S = Multipling b S on both sides and rearranging, we S obtain S 3S 3 = 0. B the quadratic formula, the answer, which is clearl positive, must be Let AB = x, and extend the lines BC, DE, F A. We see that we form an equilateral triangle with side length x. Since the area of an equilateral triangle with side length s is equal to s 3, the area of the hexagon is then (x) 3 3( x 3 ) = 3x 3 = 3. Solving reveals that x =, and so the perimeter is x + x + x + x + x + x = 9x = 36.. We recall that the number of divisors of n with prime factorization p e pe k k is equal to (e + )(e + ) (e k + ). Thus, for an integer to have positive divisors, it must be of the form a 0 b or a, with a, b prime. Clearl, the former choice will ield a smaller answer, and so we can choose a = and b = 3 in order to minimize the product. Thus our answer is 0 3 = Instead of the process the problem describes, consider the following process: each square ou draw, count up the amount of unit squares it contains. Do that for ever square and add up all the numbers. In both processes, ou ll end up with the same amount. First let s look at the 6 6 squares. There s one of them so in total that takes up 36 unit squares. Now onto the squares. Let s look at the top left unit square of each square. It can onl be within the upper left unit squares, otherwise the big square wouldn t fit. So that gives is = 00 more unit squares. Now we move onto. The top left unit square of the will onl fit within the upper left 9 unit squares, so this gives us 9 = unit squares. Continuing this process, we get that the 3 3 gives us =, the gives = 00 and the gives 36 = 36. Adding it all up, we get = Let there be a n good subsets of the set {,,..., n}. If is in the subset, then cannot be in the subset, and there are a n was to choose the remaining numbers in the subset. If is not in the subset, then there are a n was to choose the remaining numbers in the subset. Therefore, a n = a n + a n. Furthermore, a = and a = 3. Therefore a 3 =, a = 8, a = 3, a 6 =, a 7 = 3, a 8 =, a 9 = 89, a 0 =. Note that these numbers follow the Fibonacci sequence.

6 7. The can either split into three ( )( groups )( ) of, two groups of 3 or one group of and one ( ) group of. 6 6 In the first case, there are = 90 was. In the second case, there are = 0 and ( ) 3 6 in the final case there are = was. Adding this up, we get a total of = was. 8. This is a classic stars and bars problem. We want to order 3 bars and 0 stars. The stars to the left of the first bar will represent the balls that go into the first box. The stars to the right of the first bar but left of the second bar will represent the balls that go into the second box. The stars to the right of the second bar but left of the third bar will represent the balls that go into the third box. The stars ( to the ) right ( of ) the third bar will represent the balls that go into the fourth box. There are = = 86 such orderings, which is the desired result Consider a coordinate plane where the x-axis goes from 0 to 60 and represents the amount of time in minutes after that AJ arrives. Define the -axis similarl for Mark. Each point in this coordinate plane represents two points in time: when AJ arrives and when Mark arrives. The region that represents the points in which AJ and Mark will be able to meet is bounded b the lines x = 0, = 0, x = + 0, = x + 0, x = 60, and = 60. For example, x = + 0 is a line that bounds the region because x, which is the time that AJ arrives, can be at most 0 after the time that Mark arrives. The area of this region divided b the area of the entire square will be the desired answer. We can subtract the two unwanted regions, which are both isosceles right triangles with leg lengths of 0, from the entire square, in order to find the area of the desired region. This is = 00. Therefore, the desired answer is = Because 9 = x + + z we can multipl the left side of the second equation b x + + z and the xz right hand side b z. The equation now becomes 6x +9 x + + z + x + + z = 8. xz z x Rewriting this gives us that (6 x z + z x )+(9x + x )+(6 z +9z ) =. Now here comes the ver trick part. Note that no matter what real numbers a and b, we alwas have ( a b) 0. Rearranging, this means that a + b ab. So now let s go back to our equation. In each pair of parentheses, let the expression left of the plus sign be a and the expression to the right be b. Now appl our cool result that a + b ab. For example, we could see that 6 x z + z x 6 z x = (8) = 6. Repeating this for the second and third expression, x z we can see that 9 x + x and 6 z + 9z. Thus, their sum is greater than or equal to =. But wait. We alread knew that their sum was equal to, so how does this help? Well, in each of those inequalities we had, ou ll remember that we got them all from using the fact that ( a b) 0. But when is it exactl 0? When a = b. So since our sum is exactl, we need each of our a s and b s to be equal. So we can solve the equations 6 x z = z x, 9x = x and 6 z = 9z. To solve the equations completel though, we must also use x + + z = 9. This is a lot of tedious algebra, but in the end we should get that x =, = 3, z = which means that xz =. 6