4.1. In the product

$$
24^{a} \cdot 25^{b} \cdot 26^{c} \cdot 27^{d} \cdot 28^{e} \cdot 29^{f} \cdot 30^{g}
$$

the seven exponents $a, b, c, d, e, f, g$ were replaced by the numbers $1, 2, 3, 5, 8, 10, 11$ in some order. Find the maximum number of zeros that the decimal representation of this product can end with.
Answer: 32

Solution. The prime factor 5 enters the factorization of this product with multiplicity $2 b+g$, so there are no more than $2 b+g$ zeros in this product. Since $b \leqslant 11$ (the largest of the given exponents) and $b+g \leqslant 11+10$ (the sum of the two largest of the given exponents), then $2 b+g \leqslant 11+11+10=32$.

On the other hand, for example, the product $24^{8} \cdot 25^{11} \cdot 26^{1} \cdot 27^{2} \cdot 28^{3} \cdot 29^{5} \cdot 30^{10}$ ends with exactly 32 zeros, since the prime number 5 enters the factorization of this product with multiplicity $2 \cdot 11+10=32$, and the prime number 2 enters the factorization of this product with multiplicity $3 \cdot 8+1+3+10>32$.