1. Find the largest 65-digit number, the product of whose digits equals the sum of its digits.
1. Answer: $991111 \ldots 1$.

Notice that this number fits. Suppose there is a larger number. Then the sum of its digits does not exceed 9.65, and the first two digits are nines. Then the product of all other digits does not exceed $9 \cdot 65 /(9 \cdot 9)<8$. This means that, apart from the nines, there are no more than two digits greater than 1 in the number. Then, in fact, the sum of the digits does not exceed $9 \cdot 4+61=97$, and the product of all digits, except for the first nines, does not exceed $97 /(9 \cdot 9)<2$, that is, all other digits are ones.