Question

The number of odd factors of \(24!\) is

• A. \(23450\)
• B. \(23\)
• C. \(10560\)
• D. \(76954\)

Hint:

To find the number of odd factors of \(n!\), \[ \boxed{\text{Ignore the exponent of }2} \] and multiply \[ \boxed{(e_1+1)(e_2+1)\cdots} \] using only the exponents of the odd prime factors.

Answer 1

The Correct Option is C

Step 1: Find the prime factorization of \(24!\).& nbsp;

The odd prime factors of \(24!\) are

\[ 3,\;5,\;7,\;11,\;13,\;17,\;19,\;23. \]

Their exponents are

\[ \begin{aligned} 3 & amp;: \left\lfloor\frac{24}{3}\right\rfloor+\left\lfloor\frac{24}{9}\right\rfloor =8+2=10,\\ 5 & amp;: \left\lfloor\frac{24}{5}\right\rfloor=4,\\ 7 & amp;: \left\lfloor\frac{24}{7}\right\rfloor=3,\\ 11 & amp;: \left\lfloor\frac{24}{11}\right\rfloor=2,\\ 13 & amp;: \left\lfloor\frac{24}{13}\right\rfloor=1,\\ 17 & amp;: \left\lfloor\frac{24}{17}\right\rfloor=1,\\ 19 & amp;: \left\lfloor\frac{24}{19}\right\rfloor=1,\\ 23 & amp;: \left\lfloor\frac{24}{23}\right\rfloor=1. \end{aligned} \]

Therefore,

\[ 24! = 2^{22} \cdot 3^{10} \cdot 5^4 \cdot 7^3 \cdot 11^2 \cdot 13 \cdot 17 \cdot 19 \cdot 23. \]

Step 2: Find the number of odd factors.

An odd factor cannot contain the prime factor \(2\).

Hence, the number of odd factors is

\[ (10+1)(4+1)(3+1)(2+1)(1+1)^4. \] \[ = 11\times5\times4\times3\times2^4. \] \[ = 11\times5\times4\times3\times16. \] \[ = 10560. \]

Therefore,

\[ \boxed{10560.} \]

Hence, the correct option is

\[ \boxed{(C)}. \]