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Question Number 158443 by HongKing last updated on 04/Nov/21

$$\mathrm{How}\mathrm{many}\mathrm{divisors}\mathrm{has}\mathrm{the}\mathrm{positive}$$$$\mathrm{integer}\mathbf{n}\mathrm{which}\mathrm{verify}$$$${\mathrm{n}}^{\mathbf{n}}={2027}^{{2027}^{2028}}?$$

Answered by MJS_new last updated on 04/Nov/21

$${\mathrm{it}}^{\prime }\mathrm{s}\mathrm{easy}\mathrm{to}\mathrm{see}\mathrm{that}\mathrm{the}\mathrm{solution}\mathrm{of}{n}^n=a^{a^{a+2}}$$$$\mathrm{is}n=a^a.\mathrm{in}\mathrm{this}\mathrm{case}n={2027}^{2027}.2027\mathrm{is}$$$$\mathrm{prime}\Rightarrow {2027}^j\mid {2027}^{2027};0⩽j⩽2027\Rightarrow$$$$\mathrm{we}\mathrm{have}2028\mathrm{divisors}$$

Commented by HongKing last updated on 04/Nov/21

$$\mathrm{thank}\mathrm{you}\mathrm{so}\mathrm{much}\mathrm{my}\mathrm{dear}\mathbf{S}\mathrm{er}\mathrm{perfect}$$

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