Today I came through this question when one of my friends asked. I don't know if this math.stackexchange community is the right place. I am new but active on other SO communities. Take me easy if this is not the right place to ask the question.
$$\begin{array} &A&B&C&D&E\\ &B&C&D&E\\ & &C&D&E\\ & & &D&E\\ + & & & &E\\\hline A&A&A&A&A \end{array}$$
What is the solution?
Scratch paper 2
Scratch paper 3
The given equation can be written as $10000A + 2000B + 300C + 40D + 5E = 11111A$.
Since the left side is a multiple of $5$, so is the right side. Thus, $A = 5$ (since $A = 0$ isn't possible).
Plugging in $A = 5$ yields $2000B + 300C + 40D + 5E = 5555$, i.e. $400B + 60C + 8D + E = 1111$.
If $B \le 1$, we have $400B + 60C + 8D + E \le 400 \cdot 1 + 60 \cdot 9 + 8 \cdot 9 + 9 = 1021 < 1111$, a contradiction. If $B \ge 3$, then $400B + 60C + 8D + E > 400 \cdot 3 = 1200 > 1111$, a contradiction. Thus $B = 2$.
Plugging in $B = 2$ yields $60C + 8D + E = 311$. Using similar logic as above, we can easily find the solutions $(C,D,E) = (5,1,3)$ and $(4,8,7)$.
Thus, $(A,B,C,D,E) = (5,2,5,1,3)$ or $(5,2,4,8,7)$.