The Advantage Testing Foundation 2019 Solutions Problem 1 In the USA, standard letter-size paper is 8:5 inches wide and 11 inches long. What is the largest integer that cannot be written as a sum of a whole number (possibly zero) of 8:5's and a whole number (possibly zero) of 11's? Answer: 159 Solution: If 8:5a+11B is an integer, where a and B are whole numbers, then 8:5a must be an integer, and therefore a must be even. Let a = 2A, where A is a whole number, so that 8:5a+11B = 17A+11B. It is a well-known result of Sylvester1 that if n and m are relatively prime positive integers, then the largest integer that cannot be expressed in the form An + Bm, where A and B are whole numbers, is given by nm − n − m. Applying this result in the case n = 17 and m = 11 yields the answer 17(11) − 17 − 11 = 159 . Problem 2 Let a1, a2,..., a2019 be a sequence of real numbers. For every five indices i, j, k, `, and m from 1 through 2019, at least two of the numbers ai, aj, ak, a`, and am have the same absolute value. What is the greatest possible number of distinct real numbers in the given sequence? Answer: 8 Solution: If 5 or more distinct absolute values of numbers exist in the sequence, then we could choose the 5 indices so that each has a different absolute value, contradicting the hypothesis. So there are at most 4 distinct absolute values of numbers in the sequence. By the pigeonhole principle, for any 5 terms in the sequence, some two have the same absolute value. Since at most 2 distinct real numbers can have the same absolute value, the answer is 2 × 4 = 8 . 1Look up the \Frobenius coin problem" for more information. 1 Math Prize for Girls 2019 Solutions Problem 3 The degree measures of the six interior angles of a convex hexagon form an arithmetic sequence (not necessarily in cyclic order). The common difference of this arithmetic sequence can be any real number in the open interval (−D; D). Compute the greatest possible value of D. Answer: 24 Solution: Without loss of generality, we may assume that the degree mea- sures of the interior angles are a + kd, where 0 ≤ k ≤ 5 is an integer and d ≥ 0. Since the hexagon is convex, we also know that a > 0 and a+5d < 180. The degree measures of the interior angles add up to 180(6−2), thus 6a + 15d = 4(180). Solving for a and substituting the result into the inequality, we find (4(180) − 15d) + 30d < 6(180), or d < 24. For such d, 6a = 4(180) − 15d > 360, hence a is positive. Therefore, the answer is 24 . Problem 4 A paper equilateral triangle with area 2019 is folded over a line parallel to one of its sides. What is the greatest possible area of the overlap of folded and unfolded parts of the triangle? Answer: 673 Solution: We make frequent use of the fact that the area of an equilateral triangle of side length s is Es2, for some positive constant E. Let the side length of the triangle be s. Let l be the length of the crease. If 0 < l ≤ s=2, then the area of overlap is an equilateral triangle with side length l. The area of the overlap is El2. For 0 < l ≤ s=2, this area is maximized with l = s=2 which corresponds to an area of Es2=4. If s=2 < l < s, then the area of the overlap is an isosceles trapezoid that can be thought of as an equilateral triangle of side length l with an equilateral triangular tip removed. So construed, let t be the side length of the equilateral triangular tip. Observe that the nonoverlapping parts of the folded model consist of three equilateral triangles: the tip with side length t and two congruent equilateral triangles whose side lengths we will label x. Then 2x+t = s and x+t = l. Therefore, t = s−2x = s−2(l−t), from which we find t = 2l−s. Hence, the are of the overlap is E(l2 −t2) = E(3l−s)(s−l). This is a quadratic in l with maximum value attained at l = 2s=3, which is a value that satisfies the defining conditions of this case. The area when l = 2s=3 is Es2=3, which is larger than Es2=4. We conclude that the maximum area of overlap is 1=3 the area of the original piece of paper, or 2019=3 = 673 . 2 Math Prize for Girls 2019 Solutions Problem 5 Two ants sit at the vertex of the parabola y = x2. One starts walking northeast (i.e., upward along the line y = x) and the other starts walking northwest (i.e., upward along the line y = −x). Each time they reach the parabola again, they swap directions and continue walking. Both ants walk at the same speed. When the ants meet for the eleventh time (including the time at the origin), their paths will enclose 10 squares. What is the total area of these squares? Answer: 770 Solution: Suppose an ant walks northeast from the point (−a; a2) on the parabola, where a ≥ 0. Where will the ant meet the parabola again? It will meet it where the line y = x + a + a2 meets the parabola y = x2 in the first quadrant. We solve the equation x2 = x + a + a2 to find the x-coordinate, which turns out to be a + 1. Thus, the ant meets the parabola again at the point (a + 1; (a + 1)2). By symmetry, an ant that walks from the point (a; a2) in the northwesterly direction meets the parabola again at the point (−(a + 1); (a + 1)2). Also, by symmetry, an ant walking northwest from the point (a; a2) will meet an ant walking northeast from the point (−a; a2) at the y-intercept of the line y = x + a + a2, which is (0; a + a2). Putting all this together, we see that the ants will meet at the points (0; n + n2) for each integer n ≥ 0. Therefore, the diagonals of the squares enclosed by the ant trails have length (n+1)+(n+1)2 −(n+n2) = 2(n+1). The area of a square with diagonal length 2(n + 1) is 2(n + 1)2, so the total area of the first 10 squares is 9 X (10)(10 + 1)(2(10) + 1) 2(n + 1)2 = 2 = 770 ; 6 n=0 where we have applied the well-known formula for the sum of the first n Pn 2 k(k+1)(2k+1) perfect squares, k=1 k = 6 . Problem 6 For each integer from 1 through 2019, Tala calculated the prod- uct of its digits. Compute the sum of all 2019 of Tala's products. Answer: 184;320 Solution: We group together numbers with the same number of digits to compute this sum. The single-digit numbers 1 through 9 contribute 45 to the sum. 3 Math Prize for Girls 2019 Solutions P9 P9 The double-digit numbers 10 through 99 contribute t=1 u=0 tu = P9 P9 2 ( t=1 t)( u=0 u) = 45 = 2;025 to the sum. P9 P9 P9 The triple-digit numbers 100 through 999 contribute h=1 t=0 u=0 htu = P9 P9 P9 3 ( h=1 h)( t=0 t)( u=0 u) = 45 = 91;125 to the sum. For the four-digit numbers, note that the numbers from 2000 to 2019 all have a zero digit and do not contribute to the sum. However, since the numbers from 1000 to 1999 all have a 1 in the thousands place, these contribute an amount equivalent to the contribution from the three-digit numbers. Thus, the answer is 45 + 2025 + 2(91;125) = 184;320 . Problem 7 Mr. Jones teaches algebra. He has a whiteboard with a pre- drawn coordinate grid that runs from −10 to 10 in both the x and y coor- dinates. Consequently, when he illustrates the graph of a quadratic, he likes to use a quadratic with the following properties: I The quadratic sends integers to integers. II The quadratic has distinct integer roots both between −10 and 10, in- clusive. III The vertex of the quadratic has integer x and y coordinates both between −10 and 10, inclusive. How many quadratics are there with these properties? Answer: 478 Solution: A quadratic with distinct roots a < b can be uniquely written as c(x−a)(x−b), where c 6= 0. It has vertex ((a+b)=2; −c(b−a)2=4). Condition II forces a and b to be integers between −10 and 10, inclusive. Condition III requires (a + b)=2 and −c(b − a)2=4) to be integers between −10 and 10, inclusive. Thus a ≡ b (mod 2) and jc(b − a)2=4j = k where k is a positive integer less than 11. In particular, 0 < jcj ≤ 40=(b − a)2. Condition I means2 that there are integers A, B, and C such that c(x − a)(x − b) = Ax(x − 1)=2 + Bx + C: 2George P´olya showed that polynomials that send integers to integers are integral linear combinations of binomial coefficient polynomials. See P´olya, G. (1915), \Uber¨ ganzwertige ganze Funktionen," 40: 1-16. 4 Math Prize for Girls 2019 Solutions Comparing coefficients, we have 2c = A, −c(a+b) = −A=2+B, and cab = C.