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

To find the number of one-to-one functions from set A to the range of the function \( f \), we need to follow the steps outlined below:

1. Identify Set \( A \): The problem states that \( A \) is the set of all prime factors of 2310. To determine these, we perform the prime factorization of 2310.
2. Perform Prime Factorization:
   • 2310 is even, so divide by 2: \( 2310 \div 2 = 1155 \).
   • Sum of digits of 1155 is \( 1+1+5+5=12 \), divisible by 3, so divide by 3: \( 1155 \div 3 = 385 \).
   • 385 ends with 5, so divisible by 5: \( 385 \div 5 = 77 \).
   • 77 is divisible by 7: \( 77 \div 7 = 11 \).
   • 11 is a prime number itself.
   Therefore, the prime factors of 2310 are 2, 3, 5, 7, and 11. Hence, \( A = \{ 2, 3, 5, 7, 11 \} \).
3. Calculate the Range of \( f(x) \): The function is \( f(x) = \left\lfloor \log_2 \left( x^2 + \left\lfloor \frac{x^3}{5} \right\rfloor \right) \right\rfloor \). Evaluate \( f(x) \) for each element in \( A \):
   • For \( x = 2 \):
     • \( x^2 = 4 \), \( \left\lfloor \frac{8}{5} \right\rfloor = 1 \), and \( x^2 + \left\lfloor \frac{x^3}{5} \right\rfloor = 4 + 1 = 5 \).
     • \( \log_2(5) \approx 2.32 \), so \( f(2) = \left\lfloor 2.32 \right\rfloor = 2 \).
   • For \( x = 3 \):
     • \( x^2 = 9 \), \( \left\lfloor \frac{27}{5} \right\rfloor = 5 \), and \( x^2 + \left\lfloor \frac{x^3}{5} \right\rfloor = 9 + 5 = 14 \).
     • \( \log_2(14) \approx 3.81 \), so \( f(3) = \left\lfloor 3.81 \right\rfloor = 3 \).
   • For \( x = 5 \):
     • \( x^2 = 25 \), \( \left\lfloor \frac{125}{5} \right\rfloor = 25 \), and \( x^2 + \left\lfloor \frac{x^3}{5} \right\rfloor = 25 + 25 = 50 \).
     • \( \log_2(50) \approx 5.64 \), so \( f(5) = \left\lfloor 5.64 \right\rfloor = 5 \).
   • For \( x = 7 \):
     • \( x^2 = 49 \), \( \left\lfloor \frac{343}{5} \right\rfloor = 68 \), and \( x^2 + \left\lfloor \frac{x^3}{5} \right\rfloor = 49 + 68 = 117 \).
     • \( \log_2(117) \approx 6.88 \), so \( f(7) = \left\lfloor 6.88 \right\rfloor = 6 \).
   • For \( x = 11 \):
     • \( x^2 = 121 \), \( \left\lfloor \frac{1331}{5} \right\rfloor = 266 \), and \( x^2 + \left\lfloor \frac{x^3}{5} \right\rfloor = 121 + 266 = 387 \).
     • \( \log_2(387) \approx 8.6 \), so \( f(11) = \left\lfloor 8.6 \right\rfloor = 8 \).
4. Thus, the range of \( f(x) \) is \(\{2, 3, 5, 6, 8\}\) which also has 5 distinct elements.
5. Determine the Number of One-to-One Functions: Since both set \( A \) and the range of \( f \) have 5 elements, we need to calculate the number of one-to-one mappings, which is given by \( 5! \) (factorial of 5).
6. Calculate \( 5! = 5 \times 4 \times 3 \times 2 \times 1 = 120 \).

Therefore, the number of one-to-one functions from \( A \) to the range of \( f \) is 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:
\[ \boxed{120.} \]