19/04/26 20:52:53.25 RWx0fLA2.net
2^a*3^b*5^c*(1/2^a+1/3^b+1/5^c) mod 2^a*3^b = X
a,b,c,,N,Xは1以上の整数
2^a*3^b*5^c*(1/2^a+1/3^b+1/5^c)=N*2^a*3^b+X
Nが2,3,5の素数のみで構成される際Xは必ず素数になる
(2^2*3^4*5^5*(1/2^2+1/3^4+1/5^5)) mod (2^2*3^4) =269
(2^2*3^4*5^5*(1/2^2+1/3^4+1/5^5))=820*(2^2*3^4)+269