Problem 8.6. Vasya thought of three natural numbers with a sum of 1003. Calculating their product, Vasya noticed that it ends with $N$ zeros. What is the maximum value that
$N$ can take?
Answer: 7.

Solution. The product of the three thought-of numbers could end with 7 zeros, for example, if these were the numbers $625, 250, 128$. Indeed, $625+250+128=1003$ and

$$
625 \cdot 250 \cdot 128=5^{4} \cdot\left(2 \cdot 5^{3}\right) \cdot 2^{7}=2^{8} \cdot 5^{7}=2 \cdot 10^{7}=20000000
$$

Suppose there exist three natural numbers with a sum of 1003, the product of which ends with at least 8 zeros. This means that it is divisible by $10^{8}=2^{8} \cdot 5^{8}$. Since the sum of all three numbers is not divisible by 5, then one of them is not divisible by 5. Then the product of the other two is divisible by $5^{8}$. Then either one of them is divisible by $5^{5}$ (which is impossible, since otherwise the total sum of the three numbers would be greater than $5^{5}=3125>1003$), or they both are divisible by $5^{4}$ (which is also impossible, since otherwise the total sum of the three numbers would be greater than $\left.2^{4}+5^{4}=625+625>1003\right)$. Contradiction.