Some positive integers sum up to 65. What is the maximum possible product of these numbers, given that the product is divisible by 100?
Decompose 65 into the sum of positive integers in all possible ways. In each case, multiply the addends, then select the largest multiple of 100 from among the products (or one of the largest if it occurs more than once). Suppose we obtained this maximum from the decomposition

$$
x_{1}+x_{2}+\ldots+x_{n}=65
$$

Since $2+2=2 \cdot 2=4$, we can replace every 4 in the decomposition with two 2s without changing the sum or the product - thus we can assume that 4 does not appear. The numbers $\left\{x_{i}\right\}$ cannot include any number greater than 14. Indeed, if, for example, $x_{n}>14$, we can replace $x_{n}$ with the numbers $2,2,5,5,\left(x_{n}-14\right)$. The sum remains unchanged, while the product

$$
x_{n}4$ would be, replacing $x_{n}$ with 2 and $\left(x_{n}-2\right)$ would result in the product

$$
5 \cdot 5 \cdot x_{3} \ldots x_{n-1} \cdot 2\left(x_{n}-2\right)
$$

still being divisible by 100, but larger than the product of the $\left\{x_{i}\right\}$ numbers. Given our condition, the decomposition does not include 4, so only 1, 2, and 3 can appear.

There must be at least two 2s to ensure divisibility by 100, but there cannot be five or more, because replacing three 2s with two 3s increases the product:

$$
2+2+2=3+3 \quad \text { and } \quad 2 \cdot 2 \cdot 2<3 \cdot 3
$$

There can be at most one 1, and

$$
5+5+2+2+2+2+1=19<65
$$

implies that 3 must also appear in the decomposition. However, replacing a 3 and a 1 with two 2s would again increase the product, so there is no 1 among the $\left\{x_{i}\right\}$.

In summary, among the numbers $x_{3}, x_{4}, \ldots, x_{n}$, there are at least two but no more than four 2s, the rest are 3s, and

$$
5+5+x_{3}+\ldots+x_{n}=65
$$

This is only possible if the decomposition includes two 2s and 17 3s, and in this case the product, which also gives the maximum, is:

$$
5^{2} \cdot 2^{2} \cdot 3^{17}=12914016300
$$

From our considerations, it also follows that the maximum is given by the following two decompositions:

$$
\begin{gathered}
5+5+2+2+\underbrace{3+3+\ldots+3}_{17-\text { times }} \\
5+5+4+\underbrace{3+3+\ldots+3}_{17-\text { times }}
\end{gathered}
$$