Thread: December 2020 View Single Post 2020-12-01, 15:44 #8 Dr Sardonicus   Feb 2017 Nowhere 2×11×227 Posts I don't see anything about "the" solution being unique, so why not multiple solutions? I'm not sure exactly what "67 'unique' solutions" means; it seems like an oxymoron. I'm guessing "unique" refers to a specific ordering of the 5 population numbers, so you don't have two solutions with the same five population numbers. Nonuniqueness of solutions is no major defect, but geez -- they can't even give a proper example. Pathetic. Struggling to derive an interesting puzzle from the conditions... It occurred to me to wonder how many ways there are of expressing 1001 as the sum of 5 odd positive integers. Subtracting 1 from each summand gives 5 non-negative even integers. Dividing through by 2, we have the the problem of expressing 498 as the sum of 5 non-negative integers. This is well-known to be equal to the number of ways of expressing 498 as the sum 498 = x1 + 2*x2 + 3*x3 + 4*x4 + 5*x5 where the x's are non-negative integers; the corresponding 5 summands of 498 are x5, x5 + x4, x5 + x4 + x3, x5 + x4 + x3 + x2, and x5 + x4 + x3 + x2 + x1. The number of such 5-tuples is approximately the volume of the 5-dimensional "simplex" in Euclidean 5-space bounded by the coordinate axes and the hyperplane given by the above equation. This volume is 498^5/(5!*5!), which is approximately 2,000,000,000. Last fiddled with by Dr Sardonicus on 2020-12-01 at 16:14 Reason: Forgot extra factor of 5! in denominator  