Thread: December 2020
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Old 2020-12-01, 15:44   #8
Dr Sardonicus
 
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Feb 2017
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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
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