## Task 3 - 030513

Klaus has come up with an interesting problem for the "Puzzle Corner":

In the multiplication problem

$$
13 * \cdot \quad 7 *=1 * * * *
$$

all * should be replaced by digits so that all three numbers end with the same digit, and that in the result, the digit in the tens place is the same as the digit in the hundreds place.

What did you consider when solving this problem?
Since the last digits of the two factors and the product are the same digit $x$, with $0 \leq x \leq 9$, the product of $x$ with itself must also end in $x$.

This is only true for the digits $x=0, x=1, x=5$ and $x=6$. Therefore, the two factors are either 130, 70 or 131, 71 or 135, 75 or 136, 76. The multiplication yields:

$130 \cdot 70=9100$ is four digits and thus not a solution;

$131 \cdot 71=9301$ is four digits and also not a solution;

$135 \cdot 75=10125$ is five digits, but the tens and hundreds places are not the same;

$136 \cdot 76=10336$ meets all the requirements and is therefore the only solution to the problem.