Find-the-number-of-ordered-triples-a-b-c-of-positive-integers-such-that-abc-108-

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

a b c = 108 .

Answered by dioph last updated on 10/Sep/17

108 = 2^{2} \times 3^{3}

pick a \in D (108) = {1, 2, 3, 4, 6, 9, 12, 18, 27, 36, 54, 108}

then b c = (108 / a) \in 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 \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}}

and the number of ordered pairs

(b, c) such that b c 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

Commented by Tinkutara last updated on 10/Sep/17

Thank you very much Sir!