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Question Number 203566 by BaliramKumar last updated on 22/Jan/24

$${}^{\prime }{\mathrm{P}}^{\prime }\mathrm{is}\mathrm{a}\mathrm{prime}\mathrm{number}(\mathrm{P}>1000).$$$$\mathrm{If}\mathrm{P}\equiv \mathrm{r}\left(\mathrm{mod}1000\right).\mathrm{how}\mathrm{many}\mathrm{value}\mathrm{of}{}^{\prime }{\mathrm{r}}^{\prime }.$$

Answered by MM42 last updated on 22/Jan/24

$$p=1000k+r\mathrm{\&}0<r<1000$$$$\Rightarrow ‘‘p^{″}isprimenumber$$$$\Rightarrow ‘‘r^{″}mustbeanoddnumber\mathrm{\&}r\ne 5k^{\prime }$$$$A=\{1,3,5,...,999\}\Rightarrow n\left(A\right)=500$$$$B=\{5,15,...,995\}\Rightarrow n\left(B\right)=100$$$$ans=500-100=400✓$$$$forexample$$$$3001,2003,4007,...,1993,6997,1999,..$$$$$$

Commented by BaliramKumar last updated on 23/Jan/24

$$\mathrm{Can}\mathrm{we}\mathrm{use}{\mathrm{Euler}}^{\prime }\mathrm{s}\mathrm{totient}\mathrm{function}?$$$$\mathrm{I}{\mathrm{don}}^{\prime }\mathrm{t}\mathrm{know}\mathrm{if}{\mathrm{it}}^{\prime }\mathrm{s}\mathrm{true}\mathrm{forever}.$$$$\varphi \left(1000\right)=\varphi \left[{\left(2\times 5\right)}^3\right]=(2-1)(5-1)2^{3-1}\times 5^{3-1}=400$$

Commented by MM42 last updated on 23/Jan/24

$$I{don}^{\prime }tkhow$$

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