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Question Number 213261 by lmcp1203 last updated on 02/Nov/24

$$ifbetween{10}^{4}and{10}^{n}thereare9999000coprimenumberswith20.findn.please.thanks$$

Answered by MrGaster last updated on 02/Nov/24

$${10}^4<x<{10}^n,(x,20)=1$$$$20=2^2\times 5$$$$\varphi \left(20\right)=20\times (1-\frac{1}{2})\times (1-\frac{1}{5})=8$$$$\frac{({10}^n-{10}^4)}{20}\times \varphi \left(20\right)=9999000$$$$\frac{({10}^n-{10}^4)}{20}\times 8=9999000$$$$({10}^n-{10}^4)=\frac{9999000\times 20}{8}$$$${10}^n-{10}^4=24997500$$$${10}^n=25000000+{10}^4$$$${10}^n=25010000$$$$n={\log }_{10}(2.501\times {10}^7)$$$$n={\log }_{10}(2.501)+{\log }_{10}\left({10}^7\right)$$$$n={\log }_{10}(2.501)+7$$$$n\approx 0.398+7$$$$n\approx 7.398$$$$\mathrm{since}n\mathrm{must}\mathrm{be}\mathrm{an}\mathrm{integer},$$$$\overline{)\begin{array}{c}n=8\end{array}}$$

Commented by mr W last updated on 02/Nov/24

$$butbetween{10}^{4}and{10}^{8}thereare$$$$39996000coprimeswith20,not$$$$9999000.$$

Answered by lmcp1203 last updated on 02/Nov/24

$$thankuou.$$

Answered by mr W last updated on 02/Nov/24

$$20=2^2\times 5$$$$acoprimewith20shouldnotcontain$$$$theprimefactor2and5.thatmeans$$$$allnumbersbetween{10}^{4}and{10}^n,$$$$whichareamultipleof2orof5or$$$$ofbothmustbeeliminated.$$$$({10}^n-{10}^4)(1-\frac{1}{2}-\frac{1}{5}+\frac{1}{10})=9999000$$$${10}^n=25007500$$$$thisisnotpossible,sincenisnatural$$$$number.$$$$thatmeansitisnotpossiblethat$$$$thereareexactly9999000coprimes$$$$with20between{10}^{4}and{10}^n!$$

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