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Question Number 20983 by Tinkutara last updated on 09/Sep/17

$$\mathrm{Find}\mathrm{the}\mathrm{number}\mathrm{of}\mathrm{ordered}\mathrm{triples}$$$$(a,b,c)\mathrm{of}\mathrm{positive}\mathrm{integers}\mathrm{such}\mathrm{that}$$$$abc=108.$$

Answered by dioph last updated on 10/Sep/17

$$108=2^2\times 3^3$$$$\mathrm{pick}a\in D\left(108\right)=\{1,2,3,4,6,9,12,18,27,36,54,108\}$$$$\mathrm{then}bc=\left(108/a\right)\in D\left(108\right)$$$$\mathrm{if}\mathrm{we}\mathrm{write}\mathrm{the}\mathrm{members}\mathrm{of}D\left(108\right)$$$$\mathrm{in}\mathrm{their}\mathrm{factored}\mathrm{form},\mathrm{we}\mathrm{get}:$$$$D\left(108\right)=\{1,2^1,3^1,2^2,2\times 3,3^2,2^2\times 3,2\times 3^2,3^3,2^2\times 3^2,2\times 3^3,2^2\times 3^3\}$$$$\mathrm{and}\mathrm{the}\mathrm{number}\mathrm{of}\mathrm{ordered}\mathrm{pairs}$$$$(b,c)\mathrm{such}\mathrm{that}bc\mathrm{is}\mathrm{each}\mathrm{of}\mathrm{those}\mathrm{is}:$$$$N\left(D\left(108\right)\right)=\{1,2,2,3,4,3,6,6,4,9,8,12\}$$$$\mathrm{so}\mathrm{the}\mathrm{answer}\mathrm{is}\mathrm{the}\mathrm{sum}\mathrm{of}\mathrm{the}\mathrm{elements}$$$$\mathrm{from}\mathrm{the}\mathrm{set}\mathrm{above}$$$$=60\mathrm{triplets}$$

Commented by Tinkutara last updated on 10/Sep/17

$$\mathrm{Thank}\mathrm{you}\mathrm{very}\mathrm{much}\mathrm{Sir}!$$

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