Examination PuzzleDate: 02/14/2003 at 20:41:44 From: Brian Subject: Math puzzle Below is a puzzle I found. I figured out the answer, but only by trial and error. 3 students are taking an examination of N subjects. Points are awarded according to the positions in each subject they took. The person who gets 1st will get the highest points, followed by 2nd and lastly 3rd. (So let 1st = X points, 2nd = Y points, 3rd = Z points.) Student B obtained X points in English. Student A obtained 22 points total. Students B and C obtained 9 points total each. How many subjects did they take in the examination, and what are the values of X, Y, Z? The answer is N = 5, X = 5, Y = 2, Z = 1. I was told that this is the only solution with nonzero integers. I was wondering how you would go about solving this in a more mathematical way. The only thing I could think of doing was to try to solve the system of equations: aX+bY+cZ = 22 dX+eY+fZ = 9 gX+hY+iZ = 9 for X, Y, and Z in terms of a through i, and hoping it would be apparent from there what to do next. This just got insanely complicated and I gave up. Date: 02/21/2003 at 23:51:39 From: Doctor Peterson Subject: Re: Math puzzle Hi, Brian. My solution involves intelligent trial and error, so there aren't too many things to try. I have to assume that in each subject there were no ties or other problems, so that one got X, one got Y, and one got Z. Then in each subject the total score is X+Y+Z, and the total of all scores is N(X+Y+Z). Since this total is 22+9+9 = 40, N and X+Y+Z must be a pair of factors of 40. The smallest value for X+Y+Z would be 1+2+3, since they must be different. So the greatest N can be is 40/6, and N<=6. The only possible values for N (factors of 40 up to 6) are 1, 2, 4, and 5. If there were only one subject, B and C could not get the same total, so that is out. We are left with N X+Y+Z --- ----- 2 20 4 10 5 8 Now we can look at the details: [1] B got X, the largest, in one subject [2] A's total is 22 (which can't be as great as NX) [3] B and C both got 9 Suppose that N=2. Then the average of X, Y, and Z is 10; but then X>10, which contradicts [1] and [3]. Suppose that N=4. For B to get 9, X can't be larger than 9-(1+1+1) = 6. For A to get 22, 4X must be greater than 22, so X is 6. Then Y+Z = 4, and we have Y = 3 and Z = 1. But if A got 3 6's, the fourth score must have been 4. So this is out too. So N = 5. For B to get 9, X can't be larger than 9-(1+1+1+1) = 5. For A to get 22, 5X must be greater than 22, so X is 5. Then Y+Z = 3; and since they are different, Y = 2 and Z = 1. The scores therefore are A: 5+5+5+5+2 = 22 B: 1+1+1+1+5 = 9 C: 2+2+2+2+1 = 9 It works! I could probably refine this, but it's too late to do any more. Is this logical enough for you? - Doctor Peterson, The Math Forum http://mathforum.org/dr.math/ |
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