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Question Number 35071 by Rasheed.Sindhi last updated on 15/May/18

$$\mathrm{A}20-\mathrm{digit}\mathrm{decimal}\mathrm{number}\mathrm{has}\mathrm{been}$$$$\mathrm{converted}\mathrm{into}\mathrm{octal}\mathrm{system}.\mathrm{Say}\mathrm{it}\mathrm{has}$$$$\mathrm{n}\mathrm{digits}.\mathrm{What}\mathrm{can}\mathrm{be}\mathrm{minimum}\mathrm{and}$$$$\mathrm{maximum}\mathrm{possible}\mathrm{values}\mathrm{of}\mathrm{n}?$$$$$$

Answered by candre last updated on 15/May/18

$$m=l{\alpha }^k+d;0<l⩽\alpha -1\wedge 0⩽d<{\alpha }^k\wedge (l,d,k,\alpha )\in {\mathbb{N}}^4\wedge \alpha >1$$$$m=(l+d{\alpha }^{-k}){\alpha }^k$$$${\log }_{\alpha }m={\log }_{\alpha }\left[(l+d{\alpha }^{-k}){\alpha }^k\right]$$$$={\log }_{\alpha }{\alpha }^k+{\log }_{\alpha }(l+d{\alpha }^{-k})$$$$=k{\log }_{\alpha }\alpha +{\log }_{\alpha }(l+d{\alpha }^{-k})$$$$=k+{\log }_{\alpha }(l+d{\alpha }^{-k})$$$$0⩽d<{\alpha }^k\Rightarrow 0⩽d{\alpha }^{-k}<1$$$$0<l+d{\alpha }^{-k}<\alpha$$$${\log }_{\alpha }(l+d{\alpha }^{-k})<1$$$${\alpha }^k⩽m<{\alpha }^{k+1}\Rightarrow k⩽{\log }_{\alpha }m<k+1$$$$m=a_k...a_0\Rightarrow k-0+1=k+1$$$$min\Rightarrow num={10}^{19}$$$$min=\lfloor {\log }_8{10}^{19}\rfloor +1=\lfloor {19\log }_{8}10\rfloor +1=22$$$$max\Rightarrow num={10}^{20}-1$$$$max=\lfloor {\log }_8({10}^{20}-1)\rfloor +1=23$$

Commented by Rasheed.Sindhi last updated on 15/May/18

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