10.3. The sum $1+\frac{1}{2}+\frac{1}{3}+\ldots+\frac{1}{45}$ is represented as a fraction with the denominator $45!=1 \cdot 2 \cdots 45$. How many zeros (in decimal notation) does the numerator of this fraction end with?
Answer: 8 zeros. Solution. The numerator of the fraction is the sum of numbers of the form $1 \cdot 2 \cdot \ldots \cdot(k-1)(k+1) \cdot \ldots .45$ (the product lacks one of the natural numbers from 1 to 45). Let's denote such a term as $c_{k}$. Note that $45!=5^{10} \cdot 2^{n} \cdot \boldsymbol{p}$, where $\boldsymbol{p}$ is coprime with 10, and $n>$ 15 (in fact, $\boldsymbol{n}=\left[\frac{45}{2}\right]+\left[\frac{45}{2^{2}}\right]+\ldots+\left[\frac{45}{2^{5}}\right]=22+11+5+2+1=41$), i.e., 45! ends with 10 zeros. In the numerator, in the given sum, the terms $c_{k}$ for $k$ not divisible by five will be divisible by $5^{10} \cdot 2^{10}$, i.e., such terms end with 10 zeros. The terms $c_{k}$ for $k$ divisible by five but not equal to 25 will be divisible by $5^{9} \cdot 2^{9}$ and, therefore, will end with 9 zeros. There is only one term, namely, $c_{25}$, in which 5 enters to the 8th power, and only this term ends with 8 zeros. Therefore, the numerator ends with 8 zeros.