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 determine the count of one-to-one functions mapping from set A to the codomain derived from function \( f \), follow these procedures:

1. Define Set \( A \): Set \( A \) comprises the unique prime factors of 2310. Prime factorization of 2310 is required to identify these factors.
2. Prime Factorization Process:
   • 2310 divided by 2 yields 1155.
   • The sum of digits for 1155 (1+1+5+5=12) is divisible by 3, resulting in 1155 divided by 3 equaling 385.
   • 385, ending in 5, is divisible by 5, producing 385 divided by 5 as 77.
   • 77 is divisible by 7, resulting in 77 divided by 7 equaling 11.
   • 11 is a prime number.
   Consequently, the prime factors of 2310 are 2, 3, 5, 7, and 11. Therefore, \( A = \{ 2, 3, 5, 7, 11 \} \).
3. Compute \( f(x) \) Range: Given \( f(x) = \left\lfloor \log_2 \left( x^2 + \left\lfloor \frac{x^3}{5} \right\rfloor \right) \right\rfloor \), evaluate \( f(x) \) for each member of \( A \):
   • For \( x = 2 \):
     • \( x^2 = 4 \). \( \left\lfloor \frac{8}{5} \right\rfloor = 1 \). Sum: \( 4 + 1 = 5 \).
     • \( \log_2(5) \approx 2.32 \). Thus, \( f(2) = \left\lfloor 2.32 \right\rfloor = 2 \).
   • For \( x = 3 \):
     • \( x^2 = 9 \). \( \left\lfloor \frac{27}{5} \right\rfloor = 5 \). Sum: \( 9 + 5 = 14 \).
     • \( \log_2(14) \approx 3.81 \). Thus, \( f(3) = \left\lfloor 3.81 \right\rfloor = 3 \).
   • For \( x = 5 \):
     • \( x^2 = 25 \). \( \left\lfloor \frac{125}{5} \right\rfloor = 25 \). Sum: \( 25 + 25 = 50 \).
     • \( \log_2(50) \approx 5.64 \). Thus, \( f(5) = \left\lfloor 5.64 \right\rfloor = 5 \).
   • For \( x = 7 \):
     • \( x^2 = 49 \). \( \left\lfloor \frac{343}{5} \right\rfloor = 68 \). Sum: \( 49 + 68 = 117 \).
     • \( \log_2(117) \approx 6.88 \). Thus, \( f(7) = \left\lfloor 6.88 \right\rfloor = 6 \).
   • For \( x = 11 \):
     • \( x^2 = 121 \). \( \left\lfloor \frac{1331}{5} \right\rfloor = 266 \). Sum: \( 121 + 266 = 387 \).
     • \( \log_2(387) \approx 8.6 \). Thus, \( f(11) = \left\lfloor 8.6 \right\rfloor = 8 \).
4. The resultant range of \( f(x) \) is \(\{2, 3, 5, 6, 8\}\), containing 5 distinct values.
5. Calculate One-to-One Functions: As both set \( A \) and the range of \( f \) have cardinality 5, the number of one-to-one mappings is \( 5! \).
6. \( 5! = 5 \times 4 \times 3 \times 2 \times 1 = 120 \).

The total number of one-to-one functions from \( A \) to the range of \( f \) is 120.