Menu Close

Question-163161

Tinku Tara 0 Comments
Pin




Question Number 163161 by kdaramaths last updated on 04/Jan/22
Answered by JDamian last updated on 05/Mar/23
there is no β satisfying those constraints n=5^α 119^β =5^α 7^β 17^β S stands for the sum of all divisors of n S=((5^(α+1) −1)/(5−1))×((7^(β+1) −1)/(7−1))×((17^(β+1) −1)/(17−1))=3720 S=((5^(α+1) −1)/2^2 )×((7^(β+1) −1)/(2∙3^ ))×((17^(β+1) −1)/2^4 )=3720= (((5^a −1)(7^b −1)(17^b −1))/(3∙2^7 )) = 2^3 ∙3∙5∙31 (5^a −1)_(F_5 ) (7^b −1)_(F_7 ) (17^b −1)_(F_(17) ) = 2^(10) ∙3^2 ∙5∙31 The 5 in the right must be from one of three factors in the left. But we know F_5 =(5^a −1)mod 5 ≠ 0 Then; F_7 =(7^b −1)mod 5 = 0 ⇒ b=4k but then (17^(4k) −1)mod 5 = 0 ⇒ S=5^2 m′≠3720
$$\mathrm{there}\:\mathrm{is}\:\mathrm{no}\:\beta\:\:\mathrm{satisfying}\:\mathrm{those}\:\mathrm{constraints} \\ $$$$ \\ $$$${n}=\mathrm{5}^{\alpha} \mathrm{119}^{\beta} =\mathrm{5}^{\alpha} \mathrm{7}^{\beta} \mathrm{17}^{\beta} \\ $$$$ \\ $$$${S}\:\mathrm{stands}\:\mathrm{for}\:\mathrm{the}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{all}\:\mathrm{divisors}\:\mathrm{of}\:{n} \\ $$$${S}=\frac{\mathrm{5}^{\alpha+\mathrm{1}} −\mathrm{1}}{\mathrm{5}−\mathrm{1}}×\frac{\mathrm{7}^{\beta+\mathrm{1}} −\mathrm{1}}{\mathrm{7}−\mathrm{1}}×\frac{\mathrm{17}^{\beta+\mathrm{1}} −\mathrm{1}}{\mathrm{17}−\mathrm{1}}=\mathrm{3720} \\ $$$${S}=\frac{\mathrm{5}^{\alpha+\mathrm{1}} −\mathrm{1}}{\mathrm{2}^{\mathrm{2}} }×\frac{\mathrm{7}^{\beta+\mathrm{1}} −\mathrm{1}}{\mathrm{2}\centerdot\mathrm{3}^{\:} }×\frac{\mathrm{17}^{\beta+\mathrm{1}} −\mathrm{1}}{\mathrm{2}^{\mathrm{4}} }=\mathrm{3720}= \\ $$$$\:\:\:\:\:\:\:\frac{\left(\mathrm{5}^{{a}} −\mathrm{1}\right)\left(\mathrm{7}^{{b}} −\mathrm{1}\right)\left(\mathrm{17}^{{b}} −\mathrm{1}\right)}{\mathrm{3}\centerdot\mathrm{2}^{\mathrm{7}} }\:=\:\mathrm{2}^{\mathrm{3}} \centerdot\mathrm{3}\centerdot\mathrm{5}\centerdot\mathrm{31} \\ $$$$\underset{{F}_{\mathrm{5}} } {\underbrace{\left(\mathrm{5}^{{a}} −\mathrm{1}\right)}\underset{{F}_{\mathrm{7}} } {\left(\mathrm{7}^{{b}} −\mathrm{1}\right)}\underset{{F}_{\mathrm{17}} } {\left(\mathrm{17}^{{b}} −\mathrm{1}\right)}}\:=\:\mathrm{2}^{\mathrm{10}} \centerdot\mathrm{3}^{\mathrm{2}} \centerdot\mathrm{5}\centerdot\mathrm{31} \\ $$$$ \\ $$$$\mathrm{The}\:\mathrm{5}\:\mathrm{in}\:\mathrm{the}\:\mathrm{right}\:\mathrm{must}\:\mathrm{be}\:\mathrm{from}\:\mathrm{one} \\ $$$$\mathrm{of}\:\mathrm{three}\:\mathrm{factors}\:\mathrm{in}\:\mathrm{the}\:\mathrm{left}.\:\mathrm{But}\:\mathrm{we}\:\mathrm{know} \\ $$$${F}_{\mathrm{5}} =\left(\mathrm{5}^{{a}} −\mathrm{1}\right)\mathrm{mod}\:\mathrm{5}\:\neq\:\mathrm{0}\:\:\:\:\:\:\mathrm{Then}; \\ $$$${F}_{\mathrm{7}} =\left(\mathrm{7}^{{b}} −\mathrm{1}\right)\mathrm{mod}\:\mathrm{5}\:=\:\mathrm{0}\:\:\Rightarrow\:\:{b}=\mathrm{4}{k} \\ $$$$\mathrm{but}\:\mathrm{then} \\ $$$$\left(\mathrm{17}^{\mathrm{4}{k}} −\mathrm{1}\right)\mathrm{mod}\:\mathrm{5}\:=\:\mathrm{0}\:\:\Rightarrow\:\:{S}=\mathrm{5}^{\mathrm{2}} {m}'\neq\mathrm{3720} \\ $$$$ \\ $$

Pin

Leave a Reply

Your email address will not be published. Required fields are marked *