if-between-10-4-and-10-n-there-are-9999000-coprime-numbers-with-20-find-n-please-thanks-

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

$${if}\:{between}\:\:\mathrm{10}^{\mathrm{4}} \:{and}\:\mathrm{10}^{{n}} \:{there}\:{are}\:\mathrm{9999000}\:{coprime}\:{numbers}\:{with}\:\mathrm{20}.\:{find}\:{n}.\:\:{please}.\:{thanks} \\ $$

Answered by MrGaster last updated on 02/Nov/24

10^4 <x<10^n ,(x,20)=1 20=2^2 ×5 φ(20)=20×(1−(1/2))×(1−(1/5))=8 (((10^n −10^4 ))/(20))×φ(20)=9999000 (((10^n −10^4 ))/(20))×8=9999000 (10^n −10^4 )=((9999000×20)/8) 10^n −10^4 =24997500 10^n =25000000+10^4 10^n =25010000 n=log_(10) (2.501×10^7 ) n=log_(10) (2.501)+log_(10) (10^7 ) n=log_(10) (2.501)+7 n≈0.398+7 n≈7.398 since n must be an integer, determinant (((n=8)))

$$\mathrm{10}^{\mathrm{4}} <{x}<\mathrm{10}^{{n}} ,\left({x},\mathrm{20}\right)=\mathrm{1} \\ $$$$\mathrm{20}=\mathrm{2}^{\mathrm{2}} ×\mathrm{5} \\ $$$$\phi\left(\mathrm{20}\right)=\mathrm{20}×\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}\right)×\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{5}}\right)=\mathrm{8} \\ $$$$\frac{\left(\mathrm{10}^{{n}} −\mathrm{10}^{\mathrm{4}} \right)}{\mathrm{20}}×\phi\left(\mathrm{20}\right)=\mathrm{9999000} \\ $$$$\frac{\left(\mathrm{10}^{{n}} −\mathrm{10}^{\mathrm{4}} \right)}{\mathrm{20}}×\mathrm{8}=\mathrm{9999000} \\ $$$$\left(\mathrm{10}^{{n}} −\mathrm{10}^{\mathrm{4}} \right)=\frac{\mathrm{9999000}×\mathrm{20}}{\mathrm{8}} \\ $$$$\mathrm{10}^{{n}} −\mathrm{10}^{\mathrm{4}} =\mathrm{24997500} \\ $$$$\mathrm{10}^{{n}} =\mathrm{25000000}+\mathrm{10}^{\mathrm{4}} \\ $$$$\mathrm{10}^{{n}} =\mathrm{25010000} \\ $$$${n}=\mathrm{log}_{\mathrm{10}} \left(\mathrm{2}.\mathrm{501}×\mathrm{10}^{\mathrm{7}} \right) \\ $$$${n}=\mathrm{log}_{\mathrm{10}} \left(\mathrm{2}.\mathrm{501}\right)+\mathrm{log}_{\mathrm{10}} \left(\mathrm{10}^{\mathrm{7}} \right) \\ $$$${n}=\mathrm{log}_{\mathrm{10}} \left(\mathrm{2}.\mathrm{501}\right)+\mathrm{7} \\ $$$${n}\approx\mathrm{0}.\mathrm{398}+\mathrm{7} \\ $$$${n}\approx\mathrm{7}.\mathrm{398} \\ $$$$\mathrm{since}\:{n}\:\mathrm{must}\:\mathrm{be}\:\mathrm{an}\:\mathrm{integer}, \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\begin{array}{|c|}{{n}=\mathrm{8}}\\\hline\end{array} \\ $$

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

but between 10^4 and 10^8 there are 39 996 000 coprimes with 20, not 9 999 000.

$${but}\:{between}\:\mathrm{10}^{\mathrm{4}} \:{and}\:\mathrm{10}^{\mathrm{8}} \:{there}\:{are} \\ $$$$\mathrm{39}\:\mathrm{996}\:\mathrm{000}\:{coprimes}\:{with}\:\mathrm{20},\:{not} \\ $$$$\mathrm{9}\:\mathrm{999}\:\mathrm{000}. \\ $$

Answered by lmcp1203 last updated on 02/Nov/24

$${thank}\:{uou}. \\ $$

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

20=2^2 ×5 a coprime with 20 should not contain the prime factor 2 and 5. that means all numbers between 10^4 and 10^n , which are a multiple of 2 or of 5 or of both must be eliminated. (10^n −10^4 )(1−(1/2)−(1/5)+(1/(10)))=9999000 10^n =25 007 500 this is not possible, since n is natural number. that means it is not possible that there are exactly 9999000 coprimes with 20 between 10^4 and 10^n !

$$\mathrm{20}=\mathrm{2}^{\mathrm{2}} ×\mathrm{5} \\ $$$${a}\:{coprime}\:{with}\:\mathrm{20}\:{should}\:{not}\:{contain} \\ $$$${the}\:{prime}\:{factor}\:\mathrm{2}\:{and}\:\mathrm{5}.\:{that}\:{means} \\ $$$${all}\:{numbers}\:{between}\:\mathrm{10}^{\mathrm{4}} \:{and}\:\mathrm{10}^{{n}} , \\ $$$${which}\:{are}\:{a}\:{multiple}\:{of}\:\mathrm{2}\:{or}\:{of}\:\mathrm{5}\:{or} \\ $$$${of}\:{both}\:{must}\:{be}\:{eliminated}. \\ $$$$\left(\mathrm{10}^{{n}} −\mathrm{10}^{\mathrm{4}} \right)\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{5}}+\frac{\mathrm{1}}{\mathrm{10}}\right)=\mathrm{9999000} \\ $$$$\mathrm{10}^{{n}} =\mathrm{25}\:\mathrm{007}\:\mathrm{500} \\ $$$${this}\:{is}\:{not}\:{possible},\:{since}\:{n}\:{is}\:{natural} \\ $$$${number}. \\ $$$${that}\:{means}\:{it}\:{is}\:{not}\:{possible}\:{that} \\ $$$${there}\:{are}\:{exactly}\:\mathrm{9999000}\:{coprimes} \\ $$$${with}\:\mathrm{20}\:{between}\:\mathrm{10}^{\mathrm{4}} \:{and}\:\mathrm{10}^{{n}} ! \\ $$

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