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Question Number 145359 by imjagoll last updated on 04/Jul/21
How many digits will there be  in 875^(16)  ?
$$\mathrm{How}\:\mathrm{many}\:\mathrm{digits}\:\mathrm{will}\:\mathrm{there}\:\mathrm{be} \\ $$$$\mathrm{in}\:\mathrm{875}^{\mathrm{16}} \:? \\ $$
Answered by Olaf_Thorendsen last updated on 04/Jul/21
If N is a n−digit number :  N = a_(n−1) a_(n−2) a_(n−3) ....a_2 a_1 a_0   N = a_(n−1) ,a_(n−2) a_(n−3) ...a_3 a_2 a_1 ×10^(n−1)   log_(10) N = log_(10) (a_(n−1) ,a_(n−2) ...a_3 a_2 a_1 )+n−1  ⇒ n = ⌊log_(10) N⌋+1    If N = 875^(16)  :  n = ⌊log_(10) 875^(16) ⌋+1  n = ⌊16log_(10) 875⌋+1  n = ⌊47,72...⌋+1  n = 48
$$\mathrm{If}\:\mathrm{N}\:\mathrm{is}\:\mathrm{a}\:\mathrm{n}−\mathrm{digit}\:\mathrm{number}\:: \\ $$$$\mathrm{N}\:=\:{a}_{{n}−\mathrm{1}} {a}_{{n}−\mathrm{2}} {a}_{{n}−\mathrm{3}} ….{a}_{\mathrm{2}} {a}_{\mathrm{1}} {a}_{\mathrm{0}} \\ $$$$\mathrm{N}\:=\:{a}_{{n}−\mathrm{1}} ,{a}_{{n}−\mathrm{2}} {a}_{{n}−\mathrm{3}} …{a}_{\mathrm{3}} {a}_{\mathrm{2}} {a}_{\mathrm{1}} ×\mathrm{10}^{{n}−\mathrm{1}} \\ $$$$\mathrm{log}_{\mathrm{10}} \mathrm{N}\:=\:\mathrm{log}_{\mathrm{10}} \left({a}_{{n}−\mathrm{1}} ,{a}_{{n}−\mathrm{2}} …{a}_{\mathrm{3}} {a}_{\mathrm{2}} {a}_{\mathrm{1}} \right)+{n}−\mathrm{1} \\ $$$$\Rightarrow\:{n}\:=\:\lfloor\mathrm{log}_{\mathrm{10}} \mathrm{N}\rfloor+\mathrm{1} \\ $$$$ \\ $$$$\mathrm{If}\:\mathrm{N}\:=\:\mathrm{875}^{\mathrm{16}} \:: \\ $$$${n}\:=\:\lfloor\mathrm{log}_{\mathrm{10}} \mathrm{875}^{\mathrm{16}} \rfloor+\mathrm{1} \\ $$$${n}\:=\:\lfloor\mathrm{16log}_{\mathrm{10}} \mathrm{875}\rfloor+\mathrm{1} \\ $$$${n}\:=\:\lfloor\mathrm{47},\mathrm{72}…\rfloor+\mathrm{1} \\ $$$${n}\:=\:\mathrm{48} \\ $$

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