2. How many natural numbers exist such that the product of all digits of such a number, multiplied by their quantity, equals 2014?
2. Answer: 1008.

Solution. The number 2014 is divisible only by the digits 1 and 2, so the number can only contain the digits 1 and 2. Moreover, the digit 2 can only appear once. If the number consists only of 1s, there are 2014 of them, and there is only one such number. If there is a 2, then there are 1007 digits of 1, and the 2 can be in any position, resulting in 1007 such numbers. Hence the answer.

Grading criteria:

Correct answer with proper justification: 7 points. Incorrect answer with correct idea of justification: 2 points. Only the correct answer provided: 1 point.

3) Answer: AC can be 6 cm, 8 cm, 12 cm, 16 cm, 18 cm.
4) C is the midpoint of AB. There are two cases: D is the midpoint of AC or D is the midpoint of BC, in each case AC = 12 cm.

![](https://cdn.mathpix.com/cropped/2024_05_06_f47394c692dc6c4034adg-1.jpg?height=78&width=484&top_left_y=1185&top_left_x=178)

$\xrightarrow[A C=12]{A} B$

2) D is the midpoint of AB. There are two cases: C is the midpoint of AD or C is the midpoint of BD; in the first case AC = 6 cm, and in the second case AC = 18 cm.

![](https://cdn.mathpix.com/cropped/2024_05_06_f47394c692dc6c4034adg-2.jpg?height=122&width=1083&top_left_y=246&top_left_x=143)

3) D is the midpoint of AC and C is the midpoint of BD. In this case AC = 16 cm.
4) D is the midpoint of BC and C is the midpoint of AD. In this case AC = 8 cm.

![](https://cdn.mathpix.com/cropped/2024_05_06_f47394c692dc6c4034adg-2.jpg?height=79&width=404&top_left_y=704&top_left_x=418)
BC and C is the midpoint of AD. In this case AC = 8 cm.

![](https://cdn.mathpix.com/cropped/2024_05_06_f47394c692dc6c4034adg-2.jpg?height=83&width=438&top_left_y=843&top_left_x=274)

Grading criteria:

Correct answer and all necessary diagrams provided: 7 points. Four answers with justification provided: 4 points. Three answers with justification provided: 2 points. Only the correct answer provided with no justification: 1 point.