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

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

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

Answered by dioph last updated on 10/Sep/17

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

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

Commented by Tinkutara last updated on 10/Sep/17

Thank you very much Sir!

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

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