
theorem
Sierp36 195,15
proof
  thus Sum digits(195,10) = 15 by Th126;
  195=13*15;
  hence 15 divides 195 by INT_1:def 3;
  let m be Nat;
  assume A1: Sum digits(m,10) = 15 & 15 divides m;
  then consider j being Nat such that
  A2: m=15*j by NAT_D:def 3;
  assume m < 195;
  then 15*j < 15*13 by A2;
  then j < 12+1 by XREAL_1:64;
  then j <= 12 by NAT_1:9;
  then j=0 or ... or j=12;
  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) = 6 by A2,Th168;
    hence contradiction by A1;
  end;
  suppose j=2;
    then Sum digits(m,10) = 3 by A2,Th21;
    hence contradiction by A1;
  end;
  suppose j=3;
    then Sum digits(m,10) = 9 by A2,Th170;
    hence contradiction by A1;
  end;
  suppose j=4;
    then Sum digits(m,10) = 6 by A2,Th27;
    hence contradiction by A1;
  end;
  suppose j=5;
    then Sum digits(m,10) = 12 by A2,Th172;
    hence contradiction by A1;
  end;
  suppose j=6;
    then Sum digits(m,10) = 9 by A2,Th33;
    hence contradiction by A1;
  end;
  suppose j=7;
    then Sum digits(m,10) = 6 by A2,Th174;
    hence contradiction by A1;
  end;
  suppose j=8;
    then Sum digits(m,10) = 3 by A2,Th39;
    hence contradiction by A1;
  end;
  suppose j=9;
    then Sum digits(m,10) = 9 by A2,Th176;
    hence contradiction by A1;
  end;
  suppose j=10;
    then Sum digits(m,10) = 6 by A2,Th45;
    hence contradiction by A1;
  end;
  suppose j=11;
    then Sum digits(m,10) = 12 by A2,Th82;
    hence contradiction by A1;
  end;
  suppose j=12;
    then Sum digits(m,10) = 9 by A2,Th51;
    hence contradiction by A1;
  end;
end;
