
#1
Nov1406, 09:44 PM

P: 110

A couple of years back, I came across a nice problem in a puzzle magazine.
PROBLEM: Weedy Willie was getting too old to work the land alone so he decided to divide his cornfield between himself and his four sons in proportion to their five work rates. He knew that Rastus, Wig, Twig and Swig together could plant a field of corn in five hours whereas Wig, Twig and Swig and himself together could manage the same task in six hours. So Weedy divided his field into a twodigit square number of parts and kept just one part for himself. Now, Wig, Twig and Swig were identical triplets so each received the same whole number of parts. How many parts did each son get? ( Methodology: Let the total number of parts dividing the field be p^2; 1<=p<=9. Let the fraction of the field received by each of the triplets, Rastus and Weedy be a/p^2, b/p^2 and 1/p^2 respectively. Then, 3a+b+1 = p^2...(i) Let the total number of hours taken by each of the triplets, Rastus and Weedy be x, y and z. Then the proportion of land received by each of the triplets, Rastus and Weedy are in the proportion 1/x : 1/y: 1/z ....(ii) Comparing (i) and (ii): x = z/a; y = z/b Now, the fraction of the work finished by each of the Triplets, Rastus and Weedy separately in 1 hour are 1/x, 1/y and 1/z. So, by conditions of the problem 3/x + 1/z = 1/6; 3/x + 1/y = 1/5 Or, (3a+1)/z=1/6; (3a+b)/z = 1/5 Or, (b1)/z = 1/5  1/6 = 1/30; giving, b = z/30 + 1 But (3a+1)/z = 1/6; (3a+b)/z = 1/5 and 3a+b+1 = p^2 yields: a = (5*p^2  11)/18; b = (p^2 + 5)/6 Since a and b must be positive integers; it follows that both (5*p^2 11)/18 and (p^2+5)/6 must be integers. Employing either number theoretic techniques or by trial and error, p is 7 in the interval 1<=p<=9. This gives (a,b) = (13,9) Hence, the field was divided into 49 equal parts. The # parts received by each of the triplets (Wig, Twig, Swig) = (13,13,13) and the # parts received by Rastus is 9. This result is also supported by the solution given in the location http://homepage.ntlworld.com/barry.r...e/zadvpuzz.htm (NOTE: However, in terms of the solution provided in the said puzzle periodical: The field was divided into 16 equal parts. The # parts received by the triplets (Wig, Twig, Swig) = (3,3,3) and the # parts received by Rastus is 6. In terms of the methodology given above, this would appear to contravene the tenets corresponding to the problem; since: for p=4; (p^2 + 5)/6 = 3.5, which is not an integer.) Consequently, I solicit comments from the members on the foregoing matter. 


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