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The problem asks to determine the largest value of \( k \) such that \( 2^k \) is a factor of the product:

\[
1 \times 2 \times 3 \times 4 \times \cdots \times 30
\]

This product is \( 30! \) (30 factorial). The largest value of \( k \) is the highest power of 2 that divides \( 30! \). To find this, we use Legendre's formula for finding the power of a prime \( p \) dividing \( n! \):

\[
k = \sum_{i=1}^{\infty} \left\lfloor \frac{n}{2^i} \right\rfloor
\]

where \( n = 30 \) and \( p = 2 \).

Let's compute it step by step:

1. \( \left\lfloor \frac{30}{2} \right\rfloor = 15 \)
2. \( \left\lfloor \frac{30}{4} \right\rfloor = 7 \)
3. \( \left\lfloor \frac{30}{8} \right\rfloor = 3 \)
4. \( \left\lfloor \frac{30}{16} \right\rfloor = 1 \)
5. \( \left\lfloor \frac{30}{32} \right\rfloor = 0 \) (no need to go further)

Now, sum these values:

\[
k = 15 + 7 + 3 + 1 = 26
\]

Therefore, the largest value of \( k \) such that \( 2^k \) divides \( 30! \) is **26**.

Do you want more details or have any questions?

Here are 5 related questions:
1. How would the method change if the question asked for powers of 3 dividing \( 30! \)?
2. Can you explain Legendre’s formula in more detail?
3. How do we extend this method to find the power of other primes dividing \( n! \)?
4. Why do higher powers of 2 contribute smaller values to the sum in Legendre’s formula?
5. What is the general strategy to calculate factors of other products similar to \( 30! \)?

**Tip:** Always look for patterns in factorials and prime factorization to simplify large product computations!

Determine the largest value of k such that 2^k is a factor of the product 1 × 2 × 3 × 4 × ... × 30.

Learn how to find the largest value of k such that 2^k is a factor of 30! using Legendre's formula. This step-by-step solution explores factorials, prime factorization, and the application of Legendre's theorem. Suitable for high school students preparing for advanced mathematics.

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