## Task 1 - 200611

Petra, an avid reader of the mathematical student magazine alpha, presents the following problem to her classmates in a study group:

In the illustrated figure, natural numbers are to be entered for $a, h, l, p$ such that the sum in each of the two diagonals $D_{1}, D_{2}$ is 80.

![](https://cdn.mathpix.com/cropped/2024_06_06_b85a894aa8dbf722a7b3g-0228.jpg?height=272&width=414&top_left_y=475&top_left_x=1369)

The number $a$ should be twice as large as the number $p$; the number $l$ should be a prime number, and the number $h$ should be a prime number that is greater than ten times $l$.

Determine all entries that satisfy these conditions! Explain how you found them!
If an entry of natural numbers for $a, h, l, p$ satisfies the conditions, then it follows: $a = 2p$.

In the diagonal $D_{1}$, the sum is therefore $2p + p + 2p = 5p$.

According to the problem, $5p = 80$, so $p = 16$ and $a = 32$. Furthermore, it follows that $l + 16 + h = 80$, so $l + h = 64$.

If the prime number $l$ were greater than or equal to 7, then $h$ would be greater than ten times this, i.e., $h > 70$ and thus $l + h > 77$, which contradicts $l + h = 64$. Therefore, $l$ can only be one of the prime numbers 2, 3, 5. For $l = 2$, this would result in $h = 62$, which is not a prime number. Thus, the only possibilities remaining are that either $l = 3, h = 61$ or $l = 5, h = 59$.

Therefore, only the entries in the figure can satisfy all the required conditions.

![](https://cdn.mathpix.com/cropped/2024_06_06_b85a894aa8dbf722a7b3g-0228.jpg?height=212&width=423&top_left_y=1353&top_left_x=268)
These entries satisfy the conditions; because: $32 + 16 + 32 = 80, 3 + 16 + 61 = 80, 5 + 16 + 59 = 80, 32$ is twice as large as 16, the numbers 3, 61, 5, 59 are prime numbers, 61 is greater than ten times 3, and 59 is greater than ten times 5.

Thus, exactly the two given entries satisfy the required conditions.