Question

Determine the largest value of \( n \) for which \[ 2^{n} \mid (75)! \]

• A. 71
• B. 72
• C. 73
• D. 74

Hint:

To find the highest power of a prime \(p\) dividing \(n!\), always use Legendre’s formula: \[ \sum \left\lfloor \frac{n}{p^k} \right\rfloor. \]

Answer 1

The Correct Option is B

To determine the largest value of \( n \) such that \( 2^{n} \mid (75)! \), we need to find the highest power of 2 that divides \( 75! \). This can be calculated using the formula for the highest power of a prime \( p \) (in this case \( p = 2 \)) dividing \( n! \):

\(e_p(n!) = \left\lfloor \frac{n}{p} \right\rfloor + \left\lfloor \frac{n}{p^2} \right\rfloor + \left\lfloor \frac{n}{p^3} \right\rfloor + \cdots\) 

Applying this formula to \( n = 75 \) and \( p = 2 \), we calculate as follows:

1. \(\left\lfloor \frac{75}{2} \right\rfloor = 37\)
2. \(\left\lfloor \frac{75}{4} \right\rfloor = 18\) (since 75 divided by 4 equals 18 with a remainder)
3. \(\left\lfloor \frac{75}{8} \right\rfloor = 9\)
4. \(\left\lfloor \frac{75}{16} \right\rfloor = 4\)
5. \(\left\lfloor \frac{75}{32} \right\rfloor = 2\)
6. \(\left\lfloor \frac{75}{64} \right\rfloor = 1\)
7. For higher powers of 2 such as 128, the floor division will be zero because 128 > 75.

Adding these values together gives us the highest power of 2:

\(37 + 18 + 9 + 4 + 2 + 1 = 71\)

Therefore, the largest value of \( n \) for which \( 2^{n} \mid (75)! \) is actually 71, not 72 as previously noted in the correct answer section. The correct computation gives us 71.