Count prime factors with multiplicity How many prime factors (counting multiplicity) are there in (6)^10 * (7)^17 * (11)^27?

Introduction / Context:This question checks understanding of “number of prime factors with multiplicity.” For a prime power p^k, there are exactly k prime factors counted with multiplicity. For a product of prime powers, add the exponents.

Given Data / Assumptions:

• (6)^10 * (7)^17 * (11)^27.
• 6 is not prime, but its prime factorization is 2 * 3. However, the expression presents 6 as a base; we are asked to count prime factors of the whole product with multiplicity.

Concept / Approach:Rewrite (6)^10 as (2 * 3)^10 = 2^10 * 3^10. Then the whole product becomes 2^10 * 3^10 * 7^17 * 11^27. The total number of prime factors with multiplicity is the sum of all exponents.

Step-by-Step Solution:(6)^10 contributes 10 twos and 10 threes.(7)^17 contributes 17 sevens.(11)^27 contributes 27 elevens.Total = 10 + 10 + 17 + 27 = 54.

Verification / Alternative check:You can explicitly state the prime-power breakdown: 2^10, 3^10, 7^17, 11^27. Summing exponents is the standard method and is unambiguous.

Why Other Options Are Wrong:64, 71, 81, and 44 are simple mis-sums or include an error from misinterpreting 6 as a prime power without expanding it.

Common Pitfalls:Forgetting to expand 6 = 2 * 3, or mistakenly adding 10 twice to get 20 but then also miscounting the other exponents.

Final Answer:54