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

Answer: (B)

120

Answer 2

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:
$\boxed{120.}$