Mongolian Mathematical Olympiad

Denote $S(a)$ sum of digits of number $a$. Consider the polynomial $f(x)={\sum }_{a\text{–}55\text{digit numbers}}{x}^{S(a)}$. It is obvious that $f(x)=(x+x^2+\dots +x^9)$.

$(1+x+x^2+\dots +x^9{)}^{54}=\frac{x(x^9-1)(x^{10}-1{)}^{54}}{(x-1{)}^{55}}$.

If we suppose that one can divide 4 parts that satisfy the given condition then it follows root of unit of the degree is also root of the polynomial $f(x)$. ${ϵ}^3+{ϵ}^2+ϵ+1=0$. On the other hand $f(x)=\frac{x(x^9-1)(x^{10}-1{)}^{54}}{(x-1{)}^{55}}$ and it follows that $f(x)$ has no such roots. It is a contradiction. Therefore it is impossible to divide set of 55 digit numbers so that satisfy the given conditions.