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Difficulty: HardPrime Numbers and Prime Factorization

For any positive integer nn, let f(n)f(n) denote the product of all distinct prime factors of nn. For example, f(12)=2×3=6f(12) = 2 \times 3 = 6. What is the value of f(26⋅34⋅52−24⋅32⋅54)f(2^6 \cdot 3^4 \cdot 5^2 - 2^4 \cdot 3^2 \cdot 5^4)?

Answer: 330

Answer

330
To find f(26⋅34⋅52−24⋅32⋅54)f(2^6 \cdot 3^4 \cdot 5^2 - 2^4 \cdot 3^2 \cdot 5^4), first factor out the greatest common factor 24⋅32⋅522^4 \cdot 3^2 \cdot 5^2. This yields 24⋅32⋅52(22⋅32−52)=24⋅32⋅52(36−25)=24⋅32⋅52⋅112^4 \cdot 3^2 \cdot 5^2 (2^2 \cdot 3^2 - 5^2) = 2^4 \cdot 3^2 \cdot 5^2 (36 - 25) = 2^4 \cdot 3^2 \cdot 5^2 \cdot 11. The distinct prime factors present in this expression are 2, 3, 5, and 11. Multiplying these distinct prime factors gives 2×3×5×11=3302 \times 3 \times 5 \times 11 = 330.

Step-by-Step Solution

1
Factor out the greatest common term 24⋅32⋅522^4 \cdot 3^2 \cdot 5^2 from 26⋅34⋅52−24⋅32⋅542^6 \cdot 3^4 \cdot 5^2 - 2^4 \cdot 3^2 \cdot 5^4
24⋅32⋅52⋅(22⋅32−52)2^4 \cdot 3^2 \cdot 5^2 \cdot (2^2 \cdot 3^2 - 5^2)
Factoring out common prime powers simplifies the expression and avoids large calculations.
2
Evaluate the arithmetic expression inside the parentheses
22⋅32−52=4⋅9−25=36−25=112^2 \cdot 3^2 - 5^2 = 4 \cdot 9 - 25 = 36 - 25 = 11
11 is itself a prime number.
3
Write the full prime factorization of the overall number
24⋅32⋅52⋅1112^4 \cdot 3^2 \cdot 5^2 \cdot 11^1
All bases (2, 3, 5, 11) are prime numbers, giving the complete prime factorization.
4
Multiply each distinct prime factor together to evaluate f(n)f(n)
2×3×5×11=3302 \times 3 \times 5 \times 11 = 330
The function f(n)f(n) takes the product of each unique prime factor exactly once.

Key Concept

Prime Factorization and Product of Distinct Prime Factors
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Topics

Prime Numbers and Prime Factorization

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