## Task 2 - 190612

Ulrike wants to specify four natural numbers in a certain order so that the following conditions are met:

The second number is 1 less than double the first number, the third number is 1 less than double the second number, the fourth number is 1 less than double the third number, and the sum of the four specified numbers is 79.

Show how to find all the numbers that meet these conditions! Verify that the numbers found satisfy all the conditions!
Assume four numbers satisfy the given conditions.

The first number is denoted by $a$. The second number is then $2 a-1$, the third number is $2(2 a-1)-1=$ $4 a-2-1=4 a-3$, and the fourth number is $2(4 a-3)-1=8 a-6-1=8 a-7$. Therefore, the sum of the four numbers is

$$
a+2 a-1+4 a-3+8 a-7=15 a-11
$$

From $15 a-11=79$ it follows that $15 a=90$. This implies that the first number $a=6$, and the second, third, and fourth numbers are 11, 21, and 41, respectively.

Therefore, only the numbers 6, 11, 21, 41 (in this order) can satisfy the conditions. The following verification shows that they meet all the required conditions:

It holds that: $2 \cdot 6-1=11, 2 \cdot 11-1=21, 2 \cdot 21-1=41$ and $6+11+21+41=79$.