A function f maps the set of natural numbers to whole numbers such that f(xy) = f(x)f(y) + f(x) + f(y) for all natural numbers x, y, and f(p) = 1 for every prime number p. Then, the value of f(54000) is:
Answer and solution
Correct answer: B
Solution
Answer: B — 1023
1. The given rule becomes much easier if 1 is added to both sides.
2. f(xy) + 1 = f(x)f(y) + f(x) + f(y) + 1
3. So,
4. f(xy) + 1 = [f(x) + 1][f(y) + 1]
5. For every prime number, f(p) = 1. Therefore, each prime factor contributes 2 to the expression on the right.
6. Now, 54000 = 2 × 2 × 2 × 2 × 3 × 3 × 3 × 5 × 5 × 5.
7. There are 10 prime factors when repeated factors are counted.
8. Hence,
9. f(54000) + 1 = 2¹⁰ = 1024
10. So, f(54000) = 1024 − 1 = 1023.
11. Answer: 1023.
Why this method works:
Introduce g(n)=f(n)+1 because the added 1 completes the product (f(x)+1)(f(y)+1). The rule is completely multiplicative, so repeated prime factors each contribute a factor of 2; only counting distinct primes would be incorrect.
Independent check and common trap:
The factorization 2⁴×3³×5³ has 4+3+3=10 prime factors with multiplicity. Therefore g(54000)=1024 and subtracting the introduced 1 gives f=1023. Do not report g as the requested f.
Apply: Add 1: g=f+1 is multiplicative and g(p)=2. 54000=2⁴×3³×5³, so g=2¹⁰ and f=1023.
Check: The factorization 2⁴×3³×5³ has 4+3+3=10 prime factors with multiplicity. Therefore g(54000)=1024 and subtracting the introduced 1 gives f=1023. Do not report g as the requested f.
1. The given rule becomes much easier if 1 is added to both sides.
2. f(xy) + 1 = f(x)f(y) + f(x) + f(y) + 1
3. So,
4. f(xy) + 1 = [f(x) + 1][f(y) + 1]
5. For every prime number, f(p) = 1. Therefore, each prime factor contributes 2 to the expression on the right.
6. Now, 54000 = 2 × 2 × 2 × 2 × 3 × 3 × 3 × 5 × 5 × 5.
7. There are 10 prime factors when repeated factors are counted.
8. Hence,
9. f(54000) + 1 = 2¹⁰ = 1024
10. So, f(54000) = 1024 − 1 = 1023.
11. Answer: 1023.
Why this method works:
Introduce g(n)=f(n)+1 because the added 1 completes the product (f(x)+1)(f(y)+1). The rule is completely multiplicative, so repeated prime factors each contribute a factor of 2; only counting distinct primes would be incorrect.
Independent check and common trap:
The factorization 2⁴×3³×5³ has 4+3+3=10 prime factors with multiplicity. Therefore g(54000)=1024 and subtracting the introduced 1 gives f=1023. Do not report g as the requested f.
G Strategy shortcut
Spot: Algebra – Functions.Apply: Add 1: g=f+1 is multiplicative and g(p)=2. 54000=2⁴×3³×5³, so g=2¹⁰ and f=1023.
Check: The factorization 2⁴×3³×5³ has 4+3+3=10 prime factors with multiplicity. Therefore g(54000)=1024 and subtracting the introduced 1 gives f=1023. Do not report g as the requested f.
G Strategy: Spot: Algebra – Functions. Apply: Add 1: g=f+1 is multiplicative and g(p)=2. 54000=2⁴×3³×5³, so g=2¹⁰ and f=1023. Check: The factorization 2⁴×3³×5³ has 4+3+3=10 prime factors with multiplicity. Therefore g(54000)=1024 and subtracting the introduced 1 gives f=1023. Do not report g as the requested f.