Hellenic Mathematical Olympiad

$$A+B=(a+d)\cdot 10^3+(b+c)\cdot 10^2+(b+c)\cdot 10+(a+d).$$ All digits of $A+B$ are odd. However, in order to find the digits of $A+B$ we must know if the integers $a+d$ and $b+c$ are less than $10$. Hence we have the cases:

(α) Let $a+d\ge 10$ and $b+c\ge 10$. Then, because $a>b>c>d>0$, we have: $$\begin{array}{rl}a+d& =10+k,k=0,1,2,...,5,\\ b+c& =10+\ell ,\ell =0,1,2,...,5.\end{array}$$ Hence we have: $$\begin{array}{rl}A+B& =(10+k)\cdot 10^3+(10+\ell )\cdot 10^2+(10+\ell )\cdot 10+(10+k)\\ & =10^4+(k+1)\cdot 10^3+(\ell +1)\cdot 10^2+(\ell +1)\cdot 10+k,\end{array}$$ that is $A+B$ has digits $1,k+1,\ell +1,\ell +1,k$, (all must be odd), absurd.

(β) Let $a+d\ge 10$ and $b+c<10$. Then, since $a>b>c>d>0$, we have: $a+d=10+k,k=0,1,2,...,5$ and: $$\begin{array}{rl}A+B& =(10+k)\cdot 10^3+(b+c)\cdot 10^2+(b+c)\cdot 10+(10+k)\\ & =10^4+k\cdot 10^3+(b+c)\cdot 10^2+(b+c+1)\cdot 10+k.\end{array}$$ Therefore we have the cases: If $b+c=9$, then $A+B$ has the digit of decades $0$, (absurd) If $b+c<9$, then $A+B$ has as digits the integers $b+c$ and $b+c+1$ which is not possible both to be odd.

(γ) Let $a+d<10$ and $b+c\ge 10$. Then, since $a>b>c>d>0$, we have: $b+c=10+\ell ,\ell =0,1,2,...,5$ and $A+B$ is written $$\begin{gathered} A+B=(a+d)\cdot 10^3+(10+\ell )\cdot 10^2+(10+\ell )\cdot 10+(a+d) \\ =(a+d+1)\cdot 10^3+(\ell +1)\cdot 10^2+\ell \cdot 10+(a+d), \end{gathered}$$ which means that $\ell$ and $\ell +1$ both are odd digits, (absurd)

(δ) Let $a+d<10$ and $b+c<10$. Then $a+d$ and $b+c$, must be odd. Since $a>b>c>d>0$ and $a\ge 7$, it follows that $a+d=9$ and since $5\ge c\ge 2$, $6\ge b\ge 3$, it follows $10>b+c\ge 5$, i.e. $b+c\in \{5,7,9\}$ Therefore we have the cases: $a+d=9$ with $a=8,d=1$ and $b+c=9$ with $b=7,c=2$ or $b=6,c=3$ or $b=5,c=4$ Hence: $A=8721$ or $A=8631$ or $A=8541$. $a+d=9$ with $a=7,d=2$ and $b+c=9,b=6,c=3$ or $b=5,c=4$. Hence: $A=7632$ or $A=7542$ $a+d=9$ with $a=8,d=1$ and $b+c=7$ with $b=5,c=2$ or $b=4,c=3$. Hence: $A=8521$ or $A=8431$. $a+d=9$ with $a=7,d=2$ and $b+c=7$ with $b=4,c=3$. Hence: $A=7432$. $a+d=9$ with $a=8,d=1$ and $b+c=5$ with $b=3,c=2$. Hence: $A=8321$.