National Olympiad Final Round

Call digits $0,1,2,3,4$ small and digits $5,6,7,8,9$ large. Denote the digits of $k$-digit number $n$ from right to left by $(n{)}_0,(n{)}_1,\dots ,(n{)}_{k-1}$. Denote by $\Sigma (n)$ the sum of all digits of $n$ and by $l(n)$ the number of large digits of $n$.

a) If $(a{)}_i$ is small then $(2a{)}_i=2(a{)}_i$ or $(2a{)}_i=2(a{)}_i+1$ depending on whether $(a{)}_{i-1}$ is small or large. Similarly if $(a{)}_i$ is large then $(2a{)}_i=2(a{)}_i-10$ or $(2a{)}_i=2(a{)}_i-9$ depending on whether $(a{)}_{i-1}$ is small or large. In other words, when multiplying by $2$, each large digit necessitates decreasing of the digit at its place by $10$ and increasing of the preceding digit by $1$ in comparison with a small digit. Hence, for every natural number $n$, $\Sigma (2n)=2\Sigma (n)-9l(n)$. As $\Sigma (a)=\Sigma (b)$ and $l(a)=l(b)$, we have $\Sigma (2a)=\Sigma (2b)$.

b) If $a=34$ and $b=43$ then $\Sigma (3a)=1+0+2=3$ but $\Sigma (3b)=1+2+9=12$.

c) Obviously $\Sigma (10a)=\Sigma (10b)$. By part a), on the other hand, $\Sigma (10n)=\Sigma (2\cdot 5n)=2\Sigma (5n)-9l(5n)$, holding for every natural number $n$. Hence $2\Sigma (5a)-9l(5a)=2\Sigma (5b)-9l(5b)$. The digit $(5n{)}_i$ is large if and only if the digit $(n{)}_i$ is odd, because a carry during multiplying by $5$ can be at most $4$. Thus $l(5n)$ is the number of odd digits of $n$, implying $l(5a)=l(5b)$. Consequently, $\Sigma (5a)=\Sigma (5b)$.