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S-N-S-2-0-2-1-2-13-7-0-7-1-7-15-11-0-11-1-11-100-determinate-the-sum-of-positive-divisors-of-S-




Question Number 124696 by mathocean1 last updated on 05/Dec/20
S ∈ N.  S=(2^0 ×2^1 ×...×2^(13) )(7^0 ×7^1 ×...×7^(15) )(11^0 ×11^1 ×...×11^(100) )  determinate the sum of  positive  divisors of S.
$${S}\:\in\:\mathbb{N}. \\ $$$${S}=\left(\mathrm{2}^{\mathrm{0}} ×\mathrm{2}^{\mathrm{1}} ×…×\mathrm{2}^{\mathrm{13}} \right)\left(\mathrm{7}^{\mathrm{0}} ×\mathrm{7}^{\mathrm{1}} ×…×\mathrm{7}^{\mathrm{15}} \right)\left(\mathrm{11}^{\mathrm{0}} ×\mathrm{11}^{\mathrm{1}} ×…×\mathrm{11}^{\mathrm{100}} \right) \\ $$$${determinate}\:{the}\:{sum}\:{of}\:\:{positive} \\ $$$${divisors}\:{of}\:{S}. \\ $$
Commented by mr W last updated on 05/Dec/20
see Q124332  S=2^(0+1+2+...+13) ×7^(0+1+2+...+15) ×11^(0+1+2+...+100)   =2^(91) ×7^(120) ×11^(5050)   sum of all divisors of S is  (((2^(92) −1)(7^(121) −1)(11^(5051) −1))/((2−1)(7−1)(11−1)))  =(((2^(92) −1)(7^(121) −1)(11^(5051) −1))/(60))
$${see}\:{Q}\mathrm{124332} \\ $$$${S}=\mathrm{2}^{\mathrm{0}+\mathrm{1}+\mathrm{2}+…+\mathrm{13}} ×\mathrm{7}^{\mathrm{0}+\mathrm{1}+\mathrm{2}+…+\mathrm{15}} ×\mathrm{11}^{\mathrm{0}+\mathrm{1}+\mathrm{2}+…+\mathrm{100}} \\ $$$$=\mathrm{2}^{\mathrm{91}} ×\mathrm{7}^{\mathrm{120}} ×\mathrm{11}^{\mathrm{5050}} \\ $$$${sum}\:{of}\:{all}\:{divisors}\:{of}\:{S}\:{is} \\ $$$$\frac{\left(\mathrm{2}^{\mathrm{92}} −\mathrm{1}\right)\left(\mathrm{7}^{\mathrm{121}} −\mathrm{1}\right)\left(\mathrm{11}^{\mathrm{5051}} −\mathrm{1}\right)}{\left(\mathrm{2}−\mathrm{1}\right)\left(\mathrm{7}−\mathrm{1}\right)\left(\mathrm{11}−\mathrm{1}\right)} \\ $$$$=\frac{\left(\mathrm{2}^{\mathrm{92}} −\mathrm{1}\right)\left(\mathrm{7}^{\mathrm{121}} −\mathrm{1}\right)\left(\mathrm{11}^{\mathrm{5051}} −\mathrm{1}\right)}{\mathrm{60}} \\ $$

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