
theorem
Sierp36 198,18
proof
  thus Sum digits(198,10) = 18 by Th88;
  198=11*18;
  hence 18 divides 198 by INT_1:def 3;
  let m be Nat;
  assume A1: Sum digits(m,10) = 18 & 18 divides m;
  then consider j being Nat such that
  A2: m=18*j by NAT_D:def 3;
  assume m < 198;
  then 18*j < 18*11 by A2;
  then j < 10+1 by XREAL_1:64;
  then j <= 10 by NAT_1:9;
  then j=0 or ... or j=10;
  then per cases;
  suppose j=0;
    then Sum digits(m,10) = 0 by A2,Th6;
    hence contradiction by A1;
  end;
  suppose j=1;
    then Sum digits(m,10) = 9 by A2,Th269;
    hence contradiction by A1;
  end;
  suppose j=2;
    then Sum digits(m,10) = 9 by A2,Th97;
    hence contradiction by A1;
  end;
  suppose j=3;
    then Sum digits(m,10) = 9 by A2,Th271;
    hence contradiction by A1;
  end;
  suppose j=4;
    then Sum digits(m,10) = 9 by A2,Th273;
    hence contradiction by A1;
  end;
  suppose j=5;
    then Sum digits(m,10) = 9 by A2,Th33;
    hence contradiction by A1;
  end;
  suppose j=6;
    then Sum digits(m,10) = 9 by A2,Th275;
    hence contradiction by A1;
  end;
  suppose j=7;
    then Sum digits(m,10) = 9 by A2,Th151;
    hence contradiction by A1;
  end;
  suppose j=8;
    then Sum digits(m,10) = 9 by A2,Th189;
    hence contradiction by A1;
  end;
  suppose j=9;
    then Sum digits(m,10) = 9 by A2,Th277;
    hence contradiction by A1;
  end;
  suppose j=10;
    then Sum digits(m,10) = 9 by A2,Th51;
    hence contradiction by A1;
  end;
end;
