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

How many divisors has the positive integer n which verify n^n = 2027^(2027^(2028) ) ?

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

Answered by MJS_new last updated on 04/Nov/21

it′s easy to see that the solution of n^n =a^a^(a+2) is n=a^a . in this case n=2027^(2027) . 2027 is prime ⇒ 2027^j ∣2027^(2027) ; 0≤j≤2027 ⇒ we have 2028 divisors

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

Commented by HongKing last updated on 04/Nov/21

thank you so much my dear Ser perfect

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

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