First of all, we have the conclusion from Euclid that define an integer n, and if 2^n – 1 is a prime, the number 2^(n-1)*(2^n – 1) must be a Perfect Number.<\/p>\n
Here follows my source code on finding 5 the smallest Perfect Number.It\u2019s written by C++.<\/p>\n
using namespace std;\nbool IsPrime (long long s)\n{\n\tfor(long long i=2;i<sqrt((long double)s);i++)\n\t{\n\t\tif(s%i==0) return false;\n\t}\n\treturn true;\n}\nlong long Power(int n)\n{\n\tif(n!=0) return 2*Power(n-1);\n\telse return 1;\n}\nint main()\n{\n\tint cnt=0;\n\tfor(int i=2;cnt<9;i++)\n\t{\n\t\tif(IsPrime(Power(i)-1)) cout<<++cnt<<") "<<power(i-1)*(Power(i)-1)<<endl;\n\t}\n\treturn 0;\n}<\/pre>\n","protected":false},"excerpt":{"rendered":"First of all, we have the conclusion from Euclid that define an integer n, and if 2^n – 1 is a prime, the number 2^(n-1)*(2^n – 1) must be a Perfect Number. Here follows my source code on finding 5 the smallest Perfect Number.It\u2019s written by C++. using namespace std; bool IsPrime (long long s) … <\/p>\n