## Task 4 - 020514

In this multiplication problem, some digits are illegible. You are to fill them in.

Describe how you found the missing digits!





翻译完成，但根据您的要求，最后一句是中文，我已将其转换为英文以保持一致性。
From the last two columns, it follows that the result ends in 68 and the third intermediate product ends in 48.

The last digit of the left factor can only be 1 or 6 to get a 2 as the last digit of the second intermediate product. However, the 1 in the tens place of this product must result from a carry, which means the last digit of the left factor must definitely be 6. For the tens place of this factor, either 0 or 5 is possible.

![](https://cdn.mathpix.com/cropped/2024_06_06_b85a894aa8dbf722a7b3g-0018.jpg?height=299&width=254&top_left_y=2055&top_left_x=1392)

$*$ * 2 *

$* 3 * *$

* 12
* $* 4$ *

$* * * * * 8$

The digit 0 can be excluded because then the 3 in the first intermediate product would have to follow from a multiplication of 4 with another digit without a carry, which is not possible:

$456 \cdot * 2 *$

$* 3 * *$

912

$* * 48$

$* * * * 68$

The unit digit of the right factor can only be 3 or 8. The 3 can be ruled out because the last product would then not end in 48.

$456 \cdot * 28$

* $3 * *$

912

3648

*     * **6 8

The remaining first multiplication is solved by trial and error, where only 3 can be the first digit of the second factor. When all intermediate results are added, the final result is:

456 . 328

1368

912

3648

149568