Task 4. The arithmetic mean of ten different natural numbers is 12.5. What is the greatest possible value of the smallest of these numbers? Explain your answer.
Solution. Let the given numbers be $a_{1}<a_{2}<\ldots<a_{10}$. Their arithmetic mean is $\frac{a_{1}+a_{2}+\ldots+a_{10}}{10}=12.5$ from which it follows that

$$
a_{1}+a_{2}+\ldots+a_{10}=125
$$

Since the numbers $a_{1}, a_{2}, \ldots, a_{10}$ are natural, we have

$$
a_{2} \geq a_{1}+1, a_{3} \geq a_{2}+1 \geq a_{1}+2 \ldots, a_{10} \geq a_{1}+9
$$

Now we have

$$
125=a_{1}+a_{2}+\ldots+a_{10} \geq a_{1}+\left(a_{1}+1\right)+\ldots+\left(a_{1}+9\right)=10 a_{1}+45
$$

i.e., $80 \geq 10 a_{1}$, from which it follows that $a_{1} \leq 8$. According to this, the smallest of the mentioned numbers can be at most 8.