Subject 3. Consider 5 natural numbers with an arithmetic mean equal to 24. Dividing the first number by the sum of the second and third, then the second number by the sum of the third and fourth, and finally the third number by the sum of the fourth and fifth, in each case the quotient is 2 and the remainder is 1. Knowing that the last two numbers are consecutive, find the numbers.
## Solution:

Let the first number be $x$, the second number be $y$, the third number be $z$, the fourth number be $t$, and the fifth number be $u$.

We have $(x+y+z+t+u): 5=24$ from which we get the sum $x+y+z+t+u=5 \cdot 24=120$, (1)

Using the division theorem with remainder, we obtain:

$$
\begin{array}{ll}
x:(y+z)=2 \text { remainder } 1 \Rightarrow x=2(y+z)+1, & 1 associated |
| :---: | :---: |
| ![](https://cdn.mathpix.com/cropped/2024_06_07_62ba5c7a5ea058b319eag-2.jpg?height=110&width=1585&top_left_y=583&top_left_x=185) | $1 \text{p}$ |
| From $(x+y+z+t+u): 5=24$ deduce $x+y+z+t+u=120$, (1) | $1 \text{p}$ |
| Use the division theorem and obtain the relations:  $x=2(y+z)+1, y=2(z+t)+1, z=2(t+u)+1 \ldots \ldots \ldots \ldots \ldots$ | $1 \text{p}$ |
| Specify that $u=t+1$ and deduce that $z=4 t+3, y=10 t+7$ and $x=28 t+21$ | $1 \text{p}$ |
| Use in (1) the obtained relations and determine $t=2$ | $1 \text{p}$ |
| Calculate $x=77, y=27, z=11$ and $u=3 \ldots \ldots$ | $1 \text{p}$ |
| The case $u=t-1$ does not work. | $1 \text{p}$ |
$$