6. permutation and combination — ગણિત ધોરણ 11 સાયન્સ — Question

Hence, the sum of the divisors

$ = \sum\limits_{a = 0}^5 {\sum\limits_{b = 0}^4 {\sum\limits_{c = 0}^2 {{2^a}{3^b}{5^c}}  = \left( {1 + 2 + ... + {2^5}} \right)\left( {1 + 3 + .... + {3^4}} \right)\left( {1 + 5 + {5^2}} \right)} } $

$ = \left( {\frac{{{2^6} - 1}}{{2 - 1}}} \right)\left( {\frac{{{3^5} - 1}}{{3 - 1}}} \right)\left( {\frac{{{5^3} - 1}}{{5 - 1}}} \right)$

$ = 63 \times 121 \times 31 = {3^2}{.7^1}{.11^2}.31$