The number of trailing zeros in a factorial $n!$ is determined by the number of times 5 is a prime factor in its prime factorization. This is because trailing zeros are formed by factors of 10 ($2 \times 5$), and factors of 2 are always more numerous than factors of 5.

Calculating Trailing Zeros using Legendre's Formula

We use Legendre's formula to find the number of trailing zeros in $n!$: $ \text{Number of zeros} = \sum_{k=1}^{\infty} \left\lfloor \frac{n}{5^k} \right\rfloor $ Here, $n = 15620$. We sum the integer part of $n$ divided by powers of 5 ($5, 25, 125,$ etc.) until the power of 5 exceeds $n$.

Step-by-Step Calculation for 15620!

1. Divide by 5: $ \left\lfloor \frac{15620}{5} \right\rfloor = 3124 $
2. Divide by 25 ($5^2$): $ \left\lfloor \frac{15620}{25} \right\rfloor = 624 $
3. Divide by 125 ($5^3$): $ \left\lfloor \frac{15620}{125} \right\rfloor = 124 $
4. Divide by 625 ($5^4$): $ \left\lfloor \frac{15620}{625} \right\rfloor = 24 $
5. Divide by 3125 ($5^5$): $ \left\lfloor \frac{15620}{3125} \right\rfloor = 4 $
6. Divide by 15625 ($5^6$): $ \left\lfloor \frac{15620}{15625} \right\rfloor = 0 $ Since $5^6 > 15620$, we stop here.

Total Number of Trailing Zeros

Sum the results from the divisions:

Therefore, there are 3900 trailing zeros in $15620!$.