Day 3: The Skating Rink
Rick is building a rectangular skating rink in his backyard. He wants the dimensions of the rink to be optimal, for the best skating experience. This means that Rick wants the area of the rink to be equal to minus the sum of the two sides. If we can assume both the length and the width of the rink are positive integers, what are all the different ways that Rick can build his rink?
If is the sum of all possible lengths and widths the the rink, and is the number of possible solutions, give your answer as .
This problem is part of The 12 Days of Math-Mas 2018
The answer is 64.
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1 solution
We can arrange this relationship like this:
l w = 2 3 − ( l + w ) = 2 3 − l − w
Moving l to the other side of the equation, we get:
l ( w + 1 ) = 2 3 − w
We can now set w + 1 = c :
l c = 2 4 − c ⇒ l = c 2 4 − c = c 2 4 − 1
Since we know that both the length and the width of the rink are integers, we know that c must be an integer factor of 2 4 . This means that the set of solutions to c is { 1 , 2 , 3 , 4 , 6 , 8 , 1 2 , 2 4 } . We can then find that the set of solutions for l is { 2 3 , 1 1 , 7 , 5 , 3 , 2 , 1 , 0 } . Since we know that w + 1 = c , we know the set of solutions for w is { 0 , 1 , 2 , 3 , 5 , 7 , 1 1 , 2 3 } . We can eliminate the solutions ( 2 3 , 0 ) and ( 0 , 2 3 ) , since 0 is not a positive integer, so we get the sum of the lengths of the widths to be 5 8 , and the number of solutions is 6 , so we get the answer to be 5 8 + 6 = 6 4 .