Calculating Trailing Zeroes in Factorials

Trailing zeroes in a factorial $N!$ are determined by the number of times 10 is a factor in its prime factorization. Since $10 = 2 \times 5$, and factors of 2 are always more frequent than factors of 5 in $N!$, the count of trailing zeroes equals the count of factors of 5.

We can find the number of factors of 5 using Legendre's formula:

The number of trailing zeroes in a product of numbers (like $2000! \times 1200!$) is the sum of the number of trailing zeroes in each number.

Zeroes in 2000!

Applying Legendre's formula for $N = 2000$:

• $ \left\lfloor \frac{2000}{5} \right\rfloor = 400 $
• $ \left\lfloor \frac{2000}{25} \right\rfloor = 80 $
• $ \left\lfloor \frac{2000}{125} \right\rfloor = 16 $
• $ \left\lfloor \frac{2000}{625} \right\rfloor = 3 $

The total number of factors of 5 in $2000!$ is $ 400 + 80 + 16 + 3 = 499 $. So, $2000!$ has 499 trailing zeroes.

Zeroes in 1200!

Applying Legendre's formula for $N = 1200$:

• $ \left\lfloor \frac{1200}{5} \right\rfloor = 240 $
• $ \left\lfloor \frac{1200}{25} \right\rfloor = 48 $
• $ \left\lfloor \frac{1200}{125} \right\rfloor = 9 $
• $ \left\lfloor \frac{1200}{625} \right\rfloor = 1 $

The total number of factors of 5 in $1200!$ is $ 240 + 48 + 9 + 1 = 298 $. So, $1200!$ has 298 trailing zeroes.

Total Zeroes in the Product $2000! \times 1200!$

To find the total number of trailing zeroes in the product $2000! \times 1200!$, we add the number of zeroes from each factorial:

$ \text{Total Zeroes} = (\text{Zeroes in } 2000!) + (\text{Zeroes in } 1200!) $

$ \text{Total Zeroes} = 499 + 298 $

$ \text{Total Zeroes} = 797 $

Conclusion

Therefore, the product $2000! \times 1200!$ has a total of 797 zeroes at the end.