S62 - 1/x 1/y 1/z> 1/2 As x y z are primes and 1/5 1/7 1/11< 1/2 you conclude that x = 3 Now yz< 1000/3 so 999< 2(3y 3z yz< 2(3y 3z

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Solution to Problem 62 Congratulations to this week’s winner Dilip Chaudhary Correct solutions were also received from Tracy Thatcher, Ray Kremer, and Bradley University professor David Quigg. Further correct solutions were also received from William Webb, Robert McQuaid, Thomas Teo, Burkart Venzke, Daniel J. Statman, Brian Laughlin, Nacho Echevarri. Three incorrect solutions were submitted. The block is of dimension 3 by 5 by 61. This is the only solution, as Dilip Chaudhary also showed. Let the dimensions be x, y, z with x < y < z and write V for the volume and S for the surface area. Then V = xyz , and S = 2(xy + xz + yz) = 2xyz(1/x + 1/y + 1/z) . Since S > V , you have
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Unformatted text preview: 1/x + 1/y + 1/z > 1/2 . As x, y, z are primes and 1/5 + 1/7 + 1/11 < 1/2 you conclude that x = 3 . Now yz < 1000/3 so 999 < 2(3y + 3z + yz) < 2(3y + 3z + 1000/3) which gives 55 < y + z . On the other hand 1/6 < 1/y + 1/z , so if y is greater than 5, then z < 42 which makes y + z too small. So y = 5 . The equation for volume gives 15x < 1000 so z < 66 , while the equation for surface area gives 2(15 + 3z + 5z) > 999 so z > 60 . The only option is z = 61 . Brian Laughlin points out there are 29 more solutions if you allow 2 as one of the dimensions. By the way, the number 1 is not considered a prime by (most) mathematicians. this page....
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This note was uploaded on 03/01/2010 for the course MATH 301 taught by Professor Albertodelgado during the Spring '10 term at Bradley.

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S62 - 1/x 1/y 1/z> 1/2 As x y z are primes and 1/5 1/7 1/11< 1/2 you conclude that x = 3 Now yz< 1000/3 so 999< 2(3y 3z yz< 2(3y 3z

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