Sa se afle de cate ori apar cifrele 0,1,4,7,8,9, in rezultatul urmatorului calcul;
12+132+1332+13332+...+133...332.
                                                 |_____|
                                                   2008

Solutia ar fi urmatoarea:
12=3+9
132=33+99
1332=333+999
          etc
133....332=3333...33 + 9999...99
                     |______|     |______|
                         2009          2009

Si obtinem expresia:
3+33+333+...+33...33+9+99+999+....+99...99
                           |____|                              |____|
                            2009                                2009
Iar expresia:
9+99+999+...+99...99=10-1+100-1+1000--1+...+100...00-1
                          |_____|                                                 |_____|
                           2009                                                      2009
Si este Egal in continuare:
10+100+1000+...+100...00-2009=11111....110- 2009
                                   |____|               |________|
                                    2009                     2009
Care este egal mai de parte....dupa ce efectuam scaderea cu:
111...119101
|______|
  2005

De aici rezulta ca 3+33+333+... este egal cu 1/3 din acest numar
Facem deci impartirea si obtinem:
370370...3706367
|__| |__|   |__|
|__________|
501 grupuri


Deci trebuie sa adunam:
111...119101+370370...3706367
|______|           |_________|
  2005                3*501=1503

Ca sa fie adunarea mai usoare...preferam sa scriem asa:

111 111...111 111 111 ... 1119101+370370370...3706367=
 |_____________|  |_______|            |____________|
          1503                     502                       1503

= 481 481...481 481 7478 111...111 9101
    |_____________|            |_______|
             1503                                498


Deci pentru scrierea numarului se folosesc:
1001 de 1
503 de 7
502 de 4
502 de 8
un 9
un 0