The problem asks to determine the largest value of $k$ such that $2^k$ is a factor of the product $1 \times 2 \times 3 \times \cdots \times 30$, which is the factorial $30!$.

To find the largest value of $k$ for which $2^k$ divides $30!$, we need to count how many factors of 2 are in $30!$. This can be done by using Legendre's formula, which states that the exponent of a prime $p$ in $n!$ is given by:

$k = \left\lfloor \frac{n}{p} \right\rfloor + \left\lfloor \frac{n}{p^2} \right\rfloor + \left\lfloor \frac{n}{p^3} \right\rfloor + \cdots$

Here, $n = 30$ and $p = 2$. Let's calculate $k$:

$k = \left\lfloor \frac{30}{2} \right\rfloor + \left\lfloor \frac{30}{4} \right\rfloor + \left\lfloor \frac{30}{8} \right\rfloor + \left\lfloor \frac{30}{16} \right\rfloor + \left\lfloor \frac{30}{32} \right\rfloor$

Now, evaluating each term:

$\left\lfloor \frac{30}{2} \right\rfloor = 15$ $\left\lfloor \frac{30}{4} \right\rfloor = 7$ $\left\lfloor \frac{30}{8} \right\rfloor = 3$ $\left\lfloor \frac{30}{16} \right\rfloor = 1$ $\left\lfloor \frac{30}{32} \right\rfloor = 0$

Thus, the total $k$ is:

$k = 15 + 7 + 3 + 1 = 26$

So, the largest value of $k$ such that $2^k$ is a factor of $30!$ is 26.

Would you like a detailed breakdown or further clarification?

Here are five related questions for practice:

1. What is the largest $k$ such that $2^k$ divides $50!$?
2. How would you find the largest $k$ such that $3^k$ divides $30!$?
3. Can you calculate the number of factors of 5 in $30!$?
4. What is the sum of the factors of 2 in the factorial of any number $n!$?
5. How would this approach change if the prime $p$ was 7 instead of 2?

Tip: Legendre's formula is a powerful tool for finding the highest powers of any prime that divide a factorial.