To determine the number of consecutive zeros at the end of $77! \times 42!$, we calculate the number of factors of 5 in the factorials since factors of 2 are more abundant than factors of 5. Consecutive zeros in a factorial arise from the product of $2$ and $5$ factors.

Step 1: Formula for counting factors of 5

The number of factors of 5 in $n!$ is given by: $\text{Factors of 5 in } n! = \left\lfloor \frac{n}{5} \right\rfloor + \left\lfloor \frac{n}{25} \right\rfloor + \left\lfloor \frac{n}{125} \right\rfloor + \cdots$

Calculate factors of 5 in $77!$:

$\left\lfloor \frac{77}{5} \right\rfloor = 15, \quad \left\lfloor \frac{77}{25} \right\rfloor = 3, \quad \left\lfloor \frac{77}{125} \right\rfloor = 0$ $\text{Total factors of 5 in } 77! = 15 + 3 = 18$

Calculate factors of 5 in $42!$:

$\left\lfloor \frac{42}{5} \right\rfloor = 8, \quad \left\lfloor \frac{42}{25} \right\rfloor = 1, \quad \left\lfloor \frac{42}{125} \right\rfloor = 0$ $\text{Total factors of 5 in } 42! = 8 + 1 = 9$

Step 2: Add the factors of 5

$\text{Total factors of 5 in } 77! \times 42! = 18 + 9 = 27$

Thus, the number of consecutive zeros at the end of $77! \times 42!$ is 27.

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Would you like details on any step or further clarification?

Follow-up Questions:

1. How do factors of 2 and 5 determine trailing zeros?
2. Why do we only count factors of 5 for trailing zeros in factorials?
3. What happens if $77!$ is multiplied by a different factorial, such as $30!$?
4. How does the presence of prime factors other than 2 and 5 affect the result?
5. Can this method be applied to large numbers outside factorials, like $10^{77} \times 42^{77}$?

Tip: To find trailing zeros in a product of factorials, always add the total number of 5-factors from each factorial, since 2-factors are guaranteed to exceed them.