Math Problem Statement
p and q are prime numbers. Find the values of p and q such that 784 × p/q is a perfect cube
Solution
We are tasked with finding the values of and , where both are prime numbers, such that is a perfect cube.
Step 1: Prime factorization of 784
First, we perform the prime factorization of 784:
392 \div 2 = 196 \\ 196 \div 2 = 98 \\ 98 \div 2 = 49 \\ 49 \div 7 = 7 \\ 7 \div 7 = 1$$ Thus, the prime factorization of 784 is: $$784 = 2^4 \times 7^2$$ ### Step 2: Setting up the equation We are given that $$784 \times \frac{p}{q}$$ must be a perfect cube. Substituting the prime factorization of 784: $$784 \times \frac{p}{q} = 2^4 \times 7^2 \times \frac{p}{q}$$ We need this expression to be a perfect cube. The prime factorization of a perfect cube must have all exponents divisible by 3. ### Step 3: Balancing the exponents In $$2^4 \times 7^2 \times \frac{p}{q}$$, the current exponents of 2 and 7 are not divisible by 3: - For $$2^4$$, we need at least 2 more factors of 2 to make the exponent divisible by 3. - For $$7^2$$, we need 1 more factor of 7 to make the exponent divisible by 3. Let’s now adjust the expression using $$p$$ and $$q$$ such that the exponents become divisible by 3. ### Step 4: Testing different values of $$p$$ and $$q$$ To balance the powers of 2 and 7: - We multiply by $$p = 2^2$$ (i.e., set $$p = 2$$) to increase the power of 2 to 6. - We divide by $$q = 7$$ to reduce the power of 7 by 1, resulting in an exponent of 3 for 7. Thus, we have: $$784 \times \frac{2}{7} = 2^4 \times 7^2 \times \frac{2}{7} = 2^5 \times 7^1$$ This is still not a perfect cube. Adjust \( p = 8 \ or qAsk a new question for Free
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Math Problem Analysis
Mathematical Concepts
Prime Factorization
Perfect Cubes
Exponentiation
Formulas
Prime Factorization
Exponent Rules for Perfect Powers
Theorems
Fundamental Theorem of Arithmetic
Perfect Cube Theorem
Suitable Grade Level
Grades 9-12
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