Question

Let $[t]$ be the greatest integer less than or equal to t. Let A be the set of all prime factors of 2310 and $f: A \to \mathbb{Z}$ be the function $f(x) = \left[ \log_2 \left( x^2 + \left[ \frac{x^3}{5} \right] \right) \right]$. The number of one-to-one functions from A to the range of f is:

• A. 20
• B. 120
• C. 25
• D. 24

Answer 1

The Correct Option is B

Prime Factorization
The prime factorization of 2310 is:
\[ 2310 = 2 \cdot 3 \cdot 5 \cdot 7 \cdot 11. \]
Thus, \( A = \{2, 3, 5, 7, 11\} \).
Compute \( f(x) \)
For each \( x \in A \), compute:
\[ f(x) = \left\lfloor \log_2\left(x^2 + \frac{x^3}{5}\right) \right\rfloor. \]
For \( x = 2 \):
\[ f(2) = \left\lfloor \log_2\left(2^2 + \frac{2^3}{5}\right) \right\rfloor = \left\lfloor \log_2\left(4 + \frac{8}{5}\right) \right\rfloor = \left\lfloor \log_2\left(\frac{28}{5}\right) \right\rfloor = \left\lfloor \log_2(5.6) \right\rfloor = 2. \]
For \( x = 3 \):
\[ f(3) = \left\lfloor \log_2\left(3^2 + \frac{3^3}{5}\right) \right\rfloor = \left\lfloor \log_2\left(9 + \frac{27}{5}\right) \right\rfloor = \left\lfloor \log_2\left(\frac{72}{5}\right) \right\rfloor = \left\lfloor \log_2(14.4) \right\rfloor = 3. \]
For \( x = 5 \):
\[ f(5) = \left\lfloor \log_2\left(5^2 + \frac{5^3}{5}\right) \right\rfloor = \left\lfloor \log_2\left(25 + 25\right) \right\rfloor = \left\lfloor \log_2(50) \right\rfloor = 5. \]
For \( x = 7 \):
\[ f(7) = \left\lfloor \log_2\left(7^2 + \frac{7^3}{5}\right) \right\rfloor = \left\lfloor \log_2\left(49 + \frac{343}{5}\right) \right\rfloor = \left\lfloor \log_2\left(\frac{588}{5}\right) \right\rfloor = \left\lfloor \log_2(117.6) \right\rfloor = 6. \]
For \( x = 11 \):
\[ f(11) = \left\lfloor \log_2\left(11^2 + \frac{11^3}{5}\right) \right\rfloor = \left\lfloor \log_2\left(121 + \frac{1331}{5}\right) \right\rfloor = \left\lfloor \log_2\left(\frac{1936}{5}\right) \right\rfloor = \left\lfloor \log_2(387.2) \right\rfloor = 8. \]
Range of \( f \)
The range of \( f \) is:
\[ \text{Range of } f = \{2, 3, 5, 6, 8\}. \]
One-to-One Functions
The number of one-to-one functions from \( A \) to the range of \( f \) is:
\[ 5! = 120. \]
Final Answer:
\[ \boxed{120.} \]