The product of the ages of Mr. Multiplier's children is 1408. The age of the youngest child is equal to half the age of the oldest child.

How many children does Mr. Multiplier have and how old are they?

(L. Hozová)

Hint. How can you systematically navigate the divisors of a given number?
The prime factorization of the product of the children's ages is $1408=2^{7} \cdot 11$. This means that the age of exactly one of the $z$ children is a multiple of 11, and the ages of the others are powers of 2. Since the age of the oldest is twice that of the youngest, the age of neither is a multiple of 11.

Therefore, there are at least three siblings, with the oldest certainly being over 11 years old. From the initial factorization, we find that the oldest is 16 years old, the youngest is 8 years old, and that there are no other siblings:

$$
1408=(2 \cdot 2 \cdot 2) \cdot 11 \cdot(2 \cdot 2 \cdot 2 \cdot 2)
$$

Mr. Multiple has three children aged 8, 11, and 16 years.