(From a translated Russian literature)
The Mathematics Circle in our school had this custom: Each applicant was given a simple problem to solve — a little mathematical nut to crack, so to speak. You became a full member only if you solved the problem.
An applicant named Vitia was given this array:
He was asked to replace 12 digits with zeros so that the sum would be 20. Vitia thought a little, then wrote rapidly:
and
He smiled and said, “If you substitute just ten zeros for digits, the sum will be 1111. Try it!”
The Circle’s president was taken aback briefly. But, he not only solved Vitia’s problem, but improved upon it:
“Why not replace only nine digits with 0’s —- and still get 1111?”
As the debate continued, ways of getting 1111 by replacing 8, 7, 6, and 5 digits with zeros were found.
Solve the six forms of this problem.
More duels for you to prove your mettle are coming !!
Nalin Pithwa
Mathematics Hothouse 
One Comment
Currently, I don’t know which is tougher: finding one with less number of zeros or more:-
10 zeros:
001+070+50+990+000=1111
9 zeros: Got three of them:
101+303+707+(000+000)=1111
100+005+007+999+(000)=1111
111+030+070+900+(000)=1111
8 zeros: Got two:
099+707+005+300+100=1111
777+333+001+(000+000)=1111
7 zeros: Got two again:
990+077+033+011+(000)=1111
009+707+055+330+010=1111
Hard pressed to find answers with 6 or 5 zeros. I guess less no. of zeros is harder…..