Question Number 213261 by lmcp1203 last updated on 02/Nov/24
![if between 10^4 and 10^n there are 9999000 coprime numbers with 20. find n. please. thanks](https://www.tinkutara.com/question/Q213261.png)
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)))](https://www.tinkutara.com/question/Q213265.png)
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.](https://www.tinkutara.com/question/Q213310.png)
Answered by lmcp1203 last updated on 02/Nov/24
![thank uou.](https://www.tinkutara.com/question/Q213266.png)
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 !](https://www.tinkutara.com/question/Q213309.png)