Problem 3. Intersperse the number $\overline{x y}$ between the digits of the number 111, so that a perfect square of five digits is obtained.
Problem 3. $104^{2}=10816 ; 105^{2}=11025 ; 141^{2}=19881 ; 142^{2}=20164$ ..... $2 p$
11011,19911 are not perfect squares ..... $.1 p$
The base of the perfect square power can be one of the numbers: $109,111,119,121,129,131,139,141$ .....  $.2 p$
By trial, the only possibility is $109^{2}=11881$ ..... $.2 p$