Math Goes to the Movies

Chapter 17 Problem Corner

Chapter 17 Problem Corner Thérèse is a mathematical prodigy in Antonia’s Line (1985). We see her as a young girl, her precociousness on display: ANTONIA: Since when can you do additions? THÉRÈSE: Since I was three. ANTONIA: What is 147 times 48? THÉRÈSE: 7056, of course.—Square root is 84. The square is 49,787,136. ANTONIA [to Thérèse’s parents]: I regret to have to tell you that your daughter is not normal. THÉRÈSE: I am a wunderkind. In this chapter we collect some of the problems, from math competitions and the like, that have confronted movie characters. See also chapter 7 (“Escape from the Shrinking Square”) for a movie full of puzzles. For the benefit of anyone who may not quite be a Thérèse-like wunderkind, we also supply some answers and some hints. 17.1 Problems for Wizkids, and a Wizdog We begin with some problems from Little Man Tate (1991), a movie about wizkids. The mathematical stars are little Fred and mathemagician Damon. The first few problems are from a math competition in which both participate. Problem 1: How many minutes are there in 48 years? Answer (courtesy of Damon): 25,228,800 minutes—151,368,000 seconds. It seems Damon is assuming that there are 365 days in a year, which is a trifle odd. Even then, the number of seconds should be 1,513,728,000, but nobody seems to notice. Problem 2: How many factors are there in 3067? Answer (Damon again): Come on guys. There are no factors of 3067. The number is prime. Somebody, for God’s sake, challenge me! 191 192 17 Problem Corner Problem 3: How about giving me a number, that when divided by the product of its digits the quotient is three, and if you were to add 18 to this number the digits would be inverted? Answer (Fred): 24. Assuming the number we’re looking for has two digits, we can write it as 10X + Y , with X and Y natural numbers from 0 to 9. Then the information provided gives us two equations for X and Y , and it’s easy to show that the only integer solution is X = 2, Y = 4. One can also look for solutions with more than two digits: it’s trickier, but it can be shown there are no further solutions to be found. Problem 4: What is the cube root of 3,796,466? Answer (Fred): 156. Again, there is a mistake. For Fred’s answer to be correct, the quizmaster should have asked for the cube root of 3,796,416. The following is our favorite problem in this movie. Problem 5 (Teacher) How many of the numbers 1, 2, 3, 4, 5, 6, 7, 8, 9, and 10 are divisible by 2? [Figure 17.1] Fig. 17.1 How many of these numbers are divisible by 2? Answer (Fred): Um, all of them. And one more problem from the math competition. Problem 6: What number has the following peculiarity. If its cube were added to five times its square and from the result 42 times the number and 40 is subtracted, the remainder is nothing. 17.1 Problems for Wizkids, and a Wizdog 193 Answer (Fred): 5. If we let X stand for the number we’re after, then the information provided amounts to the equation X3 + 5X2 − (42X + 40) = 0. One solution to this equation is Fred’s answer, X = 5. However, since it’s a cubic equation, we expect two more solutions. Given that we know one answer, it is not hard to factor (X−5) into the cubic, and then use the quadratic formula to obtain the two answers that Fred missed: X = −5 + √ 17 and X = −5 − √ 17. Possibly Fred only concerned himself with finding an integer solution. However, after making a fool of his teacher with the previous problem, it is only fair that we nitpick Fred here. Next, we have the questions from the School Mathletes State Championship in Mean Girls (2004). Problem 7: Twice the larger of two numbers is three more than five times the smaller, and the sum of four times the larger and three times the smaller is 71. What are . . . ? Answer: 14 and 5. Writing X and Y for the two numbers, it is easy to set up...