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Question Number 163161 by kdaramaths last updated on 04/Jan/22

Answered by JDamian last updated on 05/Mar/23

$$\mathrm{there}\mathrm{is}\mathrm{no}\beta \mathrm{satisfying}\mathrm{those}\mathrm{constraints}$$$$$$$$n=5^{\alpha }{119}^{\beta }=5^{\alpha }{7}^{\beta }{17}^{\beta }$$$$$$$$S\mathrm{stands}\mathrm{for}\mathrm{the}\mathrm{sum}\mathrm{of}\mathrm{all}\mathrm{divisors}\mathrm{of}n$$$$S=\frac{5^{\alpha +1}-1}{5-1}\times \frac{7^{\beta +1}-1}{7-1}\times \frac{{17}^{\beta +1}-1}{17-1}=3720$$$$S=\frac{5^{\alpha +1}-1}{2^2}\times \frac{7^{\beta +1}-1}{2\cdot 3^{}}\times \frac{{17}^{\beta +1}-1}{2^4}=3720=$$$$\frac{(5^a-1)(7^b-1)({17}^b-1)}{3\cdot 2^7}=2^3\cdot 3\cdot 5\cdot 31$$$$\underset{F_5}{\underbrace{(5^a-1)}\underset{F_7}{(7^b-1)}\underset{F_{17}}{({17}^b-1)}}=2^{10}\cdot 3^2\cdot 5\cdot 31$$$$$$$$\mathrm{The}5\mathrm{in}\mathrm{the}\mathrm{right}\mathrm{must}\mathrm{be}\mathrm{from}\mathrm{one}$$$$\mathrm{of}\mathrm{three}\mathrm{factors}\mathrm{in}\mathrm{the}\mathrm{left}.\mathrm{But}\mathrm{we}\mathrm{know}$$$$F_5=(5^a-1)\mathrm{mod}5\ne 0\mathrm{Then};$$$$F_7=(7^b-1)\mathrm{mod}5=0\Rightarrow b=4k$$$$\mathrm{but}\mathrm{then}$$$$({17}^{4k}-1)\mathrm{mod}5=0\Rightarrow S=5^2{m}^{\prime }\ne 3720$$$$$$

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