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Question Number 19978 by Tinkutara last updated on 19/Aug/17

$$\mathrm{Let}f\mathrm{be}\mathrm{a}\mathrm{one}-\mathrm{to}-\mathrm{one}\mathrm{function}\mathrm{from}$$$$\mathrm{the}\mathrm{set}\mathrm{of}\mathrm{natural}\mathrm{numbers}\mathrm{to}\mathrm{itself}$$$$\mathrm{such}\mathrm{that}f\left(mn\right)=f\left(m\right)f\left(n\right)\mathrm{for}\mathrm{all}$$$$\mathrm{natural}\mathrm{numbers}m\mathrm{and}n.\mathrm{What}\mathrm{is}\mathrm{the}$$$$\mathrm{least}\mathrm{possible}\mathrm{value}\mathrm{of}f\left(999\right)?$$

Answered by mrW1 last updated on 20/Aug/17

$$\mathrm{Every}\mathrm{natural}\mathrm{number}\mathrm{N}\mathrm{can}\mathrm{be}\mathrm{written}$$$$\mathrm{as}\mathrm{the}\mathrm{product}\mathrm{of}\mathrm{a}\mathrm{serial}\mathrm{of}\mathrm{prime}$$$$\mathrm{numbers}:$$$$\mathrm{N}={\mathrm{p}}_1^{{\mathrm{q}}_1}\times {\mathrm{p}}_2^{{\mathrm{q}}_2}\times ...\times {\mathrm{p}}_{\mathrm{k}}^{{\mathrm{q}}_{\mathrm{k}}}$$$$\mathrm{p}=\mathrm{prime}\mathrm{numbers}1,2,3,5,7....$$$$$$$$\mathrm{since}\mathrm{f}\left(\mathrm{mn}\right)=\mathrm{f}\left(\mathrm{m}\right)\mathrm{f}\left(\mathrm{n}\right)$$$$\mathrm{f}\left(\mathrm{N}\right)={\left[\mathrm{f}\left({\mathrm{p}}_1\right)\right]}^{{\mathrm{q}}_1}\times {\left[\mathrm{f}\left({\mathrm{p}}_2\right)\right]}^{{\mathrm{q}}_2}\times ...\times {\left[\mathrm{f}\left({\mathrm{p}}_{\mathrm{k}}\right)\right]}^{{\mathrm{q}}_{\mathrm{k}}}$$$$$$$$\mathrm{for}\mathrm{f}\left(\mathrm{N}\right)\mathrm{to}\mathrm{be}\mathrm{a}\mathrm{one}\mathrm{to}\mathrm{one}\mathrm{funtion},\mathrm{it}\mathrm{is}$$$$\mathrm{to}\mathrm{say}\mathrm{that}\mathrm{f}\left({\mathrm{p}}_{\mathrm{i}}\right)\mathrm{with}\mathrm{i}=1,2,...\mathrm{\infty }\mathrm{must}$$$$\mathrm{be}\mathrm{different}.$$$$$$$$\mathrm{f}\left(1\right)=\mathrm{f}\left(1\times 1\right)=\mathrm{f}\left(1\right)\times \mathrm{f}\left(1\right)$$$$\Rightarrow \mathrm{f}\left(1\right)=1$$$$\mathrm{i}.\mathrm{e}.\mathrm{only}\mathrm{f}\left(1\right)\mathrm{should}\mathrm{be}1,\mathrm{f}\left({\mathrm{p}}_{\mathrm{i}}\right)\mathrm{with}\mathrm{i}>2$$$$\mathrm{can}\mathrm{be}\mathrm{freely}\mathrm{choosen}.$$$$$$$$\mathrm{With}\mathrm{N}=999=1\times 3^3\times 37$$$$\Rightarrow \mathrm{f}\left(999\right)=\mathrm{f}\left(1\right)\times {\left[\mathrm{f}\left(3\right)\right]}^3\times \mathrm{f}\left(37\right)$$$$\mathrm{to}\mathrm{keep}\mathrm{f}\left(999\right)\mathrm{as}\mathrm{small}\mathrm{as}\mathrm{possible}\mathrm{we}$$$$\mathrm{can}\mathrm{choose}$$$$\mathrm{f}\left(3\right)=2\mathrm{and}\mathrm{f}\left(37\right)=3$$$$\Rightarrow \mathrm{f}\left(999\right)=1\times 2^3\times 3=24$$

Commented by Tinkutara last updated on 20/Aug/17

$$\mathrm{Thank}\mathrm{you}\mathrm{very}\mathrm{much}\mathrm{Sir}!$$

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