1 (Before beginning, I d like to thank USAMTS for the template, which I modified to get this template) It would be beneficial to assign each square a value, and then make a few equalities. a b 3 c d e X 4 5 As we know that each row and diagonal has the same value, we know: a+b+3 = a+c+x = x+9 = e+8 = c+d+e = a+d+5 = d+3+x. Using the simplest equations(x+9 = e+8), we can say e = x + 1, and substituting, we can say c + d + x + 1 = a + c + x, or d + 1 = a. Now, plugging these back in to a + d + 5 = d + 3 + x, we get 2d + 6 = d + 3 + x, or d + 3 = x. Knowing the 3 boxed equations, (and knowing that the second and third can be used to show x = a + 2), we shall put this back in the magic square: x-2 b 3 c x-3 x-1 X 4 5 Now, using the diagonal and the bottom row, we get 2x = x + 9 = x = 9 Solutions- 9 pages long. 1
2 To get as small of a number as possible, first we must try to get as few digits as possible, which we get by combining as many prime factors as possible per digit. We can prime factorize 10!: 2 3 2 2 5 2 3 7 2 2 2 3 3 2 5, or 2 8 3 4 5 2 7. It s obvious that we can t combine the 5s or 7 with anything else, as multiplying 5 or 7 by another integer gives it more than one place value. So we have to combine 8 2s and 4 3s into as few digits possible. We can combine 3 2s at a time to get an 8, so that gives us 2 8s, with 2 2s and 4 3s remaining. At this point, we can t combine numbers 3 at a time, so with 6 factors remaining, we re going to use 3 digits, so our task should be be to make the 8-digit number as small as possible with the digits 55788 already fixed. We can make the 3 last digits 669 or 499. The latter will make the starting digit, which is the first digit looked at when comparing numbers, smaller, so we will use that option. This makes our desired number 45578899. Solutions- 9 pages long. 2
3 First, we should see what combinations of 6 digits other than 6 of the decimal system can be added up to equal 36. We try the largest example: 9 + 8 + 7 + 5 + 4 + 3 = 36. It should be obvious that making any of these numbers smaller will not work; that is: this example is the only example! Thus, we must show that (9-6)(8-6)(7-6)(5-6)(4-6)(3-6) = -36 and we will be done. 3 2 1 1 2 3 = 6 6 = 36. As this is the only solution, our proof is complete. Solutions- 9 pages long. 3
4 The first thing we should notice is: all squares that have an even-numbered side-length will have an even number of 1*1 squares inside of them. As a black 1*1 square is equally likely as a white 1*1 square, squares with even-numbered side-lengths will have an equal number of white and black squares they won t have more than half their area colored black. Thus, we only have to count squares that have odd-numbered side lengths. We can use the fact that a black 1*1 square is equally likely as a white 1*1 square to say that there are as many 1*1, 3*3, 5*5, 7*7 squares with more than half their area black as their are 3*3 squares with more than half their area white (since there are no 1*1, 3*3, 5*5, 7*7 squares who have an equal black and white area). So, we can just count the number of each type of square, and halving this will give the number of squares with more than half of their area colored black. There are 64 1*1 squares, 36 3*3 squares, 16 5*5 squares, and 4 7*7 squares. Adding this up, we get 120, and dividing by 2, we get 60. Solutions- 9 pages long. 4
5 The number of games that occured between the 5 participants who withdrew after 5 games is not known, as they could have all played within the group of 5, leading to just 5 games played (AB, AC, BC, DE, DE), or they could have all played different games, leading to 10 games, or a little of both. So all we know is that these 5 players contributed to 5 to 10 games. Thus, the other x players played between 90 and 95 games within themselves. A round robin tournament means that everyone plays each other once. So other than the 5 players who left, the x players all played each other for one game, so each participant played each of the other x 1 participants once. This led to (x)(x 1) 2 games (dividing by 2 to eliminate the repetitions). We know that 180 (x)(x 1) 190. Although this can be solved using inequalties and/or graphing, some good trial-and-error will solve this faster. Trying x = 10 leads to 90, which is too small. Trying x = 15 leads to 210, which is slightly too large. Trying x=14 gives 192, which is between 180 and 190. Thus, there were 14 other participants, and 19 participants initially. Solutions- 9 pages long. 5
6 The first thing we should try when looking at this problem is bring all the terms to the LHS and factor. (x)(x 3) + (y)(y 3) + (z)(z 3) + (w)(w 3) = 0 is one possible factorization, but this would be very tedious with many numbers to check and we wouldn t know if we had all the solutions. Instead, we shall try completing the square: (x 2 3x+ 9 4 )+(y2 3y+ 9 4 )+(z2 3z+ 9 4 )+(w2 3w+ 9 4 ) = 9 = (x 3 2 )2 +(y 3 2 )2 +(z 3 2 )2 +(w 3 2 )2 = 9 Now, we know that x,y,z, and w are all integers and all squares are nonnegative, so the largest value of x,y,z, and w is 4 (as any integer greater than 4 will lead to the LHS of the equation to be greater than 9). If we square each of (4-3 2 ), (3-3 2 ), (2-3 2 ), and (1-3 2 ), we get 6.25, 2.25,.25, and.25. Combinations using these that add up to 9 are: 6.25, 2.25,.25,.25; and 2.25, 2.25, 2.25, 2.25. For the second combination (4 2.5s), there is only 1 way to make that: (3,3,3,3) because 3 is the only number that equals 2.25 (1.5) when 1.5 is subtracted. However, there are numerous variations of the first combination. Since subtracting 1.5 from and squaring both 2 and 1 gives.25, possible combinations are (4,3,2,2); (4,3,2,1); and (4,3,1,1). The first quadruple can be arranged 4! 2! = 12 ways, as can the third quadruple, and the second quadruple can be arranged 4! = 24 ways. Thus, there are 48 quadruples using 6.25, 2.25,.25,.25 and one quadruple using 2.25, 2.25, 2.25, 2.25, giving us a total of 49 combinations. These 49 solutions are permutations of (4, 3, 2, 2), (4, 3, 2, 1), (4, 3, 1, 1), and(3, 3, 3, 3). Solutions- 9 pages long. 6
7 Some points have been labeled for ease in discussion. As we are trying to prove the 3 lines are concurrent, we can use Ceva s theorem, which in this case states that MK KW W L LS SJ JM = 1 Seeing these ratios, it would be a good idea to look for similar triangles. It is known that EMW is similar to ASW is similar to SOM because all three are equilateral. Thus, there is most likely a pair of similar triangles whose ratios can be found which will make proving that the lines are concurrent using the theorem very simple. This may be facilitated by drawing in some lines such as in this diagram: This seems to me the most straightforward approach to proving the concurrency of the lines. Solutions- 9 pages long. 7
8 If we let a equal the original number of participants, and b equal the intial average age, then clearly ab equals the combined age of all the participants at AwesomeMath. If the average age increased by one month if three additional 18 year olds had participated, then we can say (a + 3)(b + 1 12 ) = ab + 54. Similarly, if the average age would have increased by one month if three 12 year olds had not particicpated, then we can say (a 3)(b + 1 12 ) = ab 36. Expanding the first equation and simplifying, we get 3b + a 12 = 53 3 4. Doing the same for the second equation gives 3b + a 12 = 35 3 4. Adding the two equations gives a 6 = 18. Multiplying both sides by 6 gives a = 108, and since we are looking for the number of participants, the answer is 108 Solutions- 9 pages long. 8
9 First, we should notice that a set of all the integers from 1025-2007 does not have two elements that add together to make a power of two. This is because 1024 and 2048 are powers of two, and since adding the smallest number of the set twice is slightly more than a power of 2, and adding the largest number of the set twice is not greater than the next power of 2. (Thus, any sum of two elements of the set of integers 1025 to 2007 inclusive would fall between 2048 and 4096.)We now must find as many other numbers (less than 1025) as possible so that we still won t have two numbers that add together to make a power of two. We know that these numbers must all be less than 41 (since 41 + 2007 = 2 11 ). We can begin small and eliminate numbers (larger than that number) that would otherwise add up to a power of 2 (We will start at 5, because 1,2,and 4 won t work [as they are already powers of two], and 3 would eliminate more numbers than 5): Element Numbers that cannot be added 5 11 (5 + 11 = 16), 27 (5 + 27 = 32) 6 10, 26 7 9, 25 12 20 13 19 14 18 15 17 21 43 22 42 23 41...... 30 34 31 33 (Note that eliminated numbers and powers of two were skipped in the numbers column) Any number larger would either have been already eliminated, or would be greater than 41 (and would thus match up with a number from 1025-2007 to make a power of 2). Thus, we have found all solutions. There are 983 integers from 1025 to 2007 inclusive, and our table gives us 18 integers, and since we are looking for the smallest set that DEFINITELY contains two elements that add up to a power of 2, the answer is 983 + 18 + 1 = 1002. Solutions- 9 pages long. 9