call wont work in order to update another contract's storage

Question

Storage(variable n) of E won't get updated when I use call from contract D.

addressOfE = "0x13fac93069c10e977204b9b64502439740cbe46a"; //Contract E's address that exist on my private chain.

call does not update E's storage:

D.callSetN(addressOfE, 100); //I have waited transaction to be deployed.
D.getN(); //returns 0;
E.getN(); //returns 0; 

---

delegatecall seems working:

D.delcallCfunc(addressOfE, 101);  //I have waited transaction to be deployed.
D.getN(); //returns 101; 
E.getN(); //returns 0; 

---

callcode seems working:

D.callcodeSetN(addressOfE, 102);  //I have waited transaction to be deployed.
D.getN(); //returns 102; 
E.getN(); //returns 0; 

[Q] What should I do to update E's storage by calling contract D's function? Is it possible?

I have tried the exact same tutorial on the following guide: https://ethereum.stackexchange.com/a/3672/4575

contract D {
  uint public n;
  address public sender;
  function callSetN(address _e, uint _n) {
    _e.call(bytes4(sha3("setN(uint256)")), _n); // E's storage is set, D is not modified 
  }

  function callcodeSetN(address _e, uint _n) {
    _e.callcode(bytes4(sha3("setN(uint256)")), _n); // D's storage is set, E is not modified 
  }

  function delegatecallSetN(address _e, uint _n) {
    _e.delegatecall(bytes4(sha3("setN(uint256)")), _n); // D's storage is set, E is not modified 
  }
  function getN() constant returns(uint){
      return n;
  }
  function getSender() constant returns(address){
       return sender;
   }
}

contract E {
  uint public n;
  address public sender;
  function setN(uint _n) {
    n = _n;
    sender = msg.sender;
    // msg.sender is D if invoked by D's callcodeSetN. None of E's storage is updated
    // msg.sender is C if invoked by C.foo(). None of E's storage is updated
  }
  function getN() constant returns(uint){
      return n;
  }
    function getSender() constant returns(address){
       return sender;
   }
}

Thank you for your valuable time and help.