Consider the number 2^222 * 3^333 * 5^555. How many prime factors does it have when counted with multiplicity (i.e., total prime-power exponents summed)?

Introduction / Context:
When a number is written in its prime factorization form, such as p1^a * p2^b * p3^c, the total number of prime factors with multiplicity is a + b + c. This is distinct from the number of distinct prime factors, which would simply be the count of different primes involved (here, three primes).

Given Data / Assumptions:

• Number: 2^222 * 3^333 * 5^555.
• We are asked for the total count of prime factors with multiplicity.

Concept / Approach:
For N = 2^222 * 3^333 * 5^555, the prime-power exponents are 222, 333, and 555, respectively. The required count equals the sum of these exponents.

Step-by-Step Solution:

Verification / Alternative check:
“With multiplicity” means each prime factor is counted as many times as its exponent indicates. Thus, the result is the exponent sum, not the number of distinct primes (which would be 3).

Why Other Options Are Wrong:

• 3: Counts only distinct primes, not with multiplicity.
• 1107, 1113, 1272: Incorrect sums or distractors; none equal 222 + 333 + 555.

Common Pitfalls:

• Confusing “distinct prime factors” with “prime factors counted with multiplicity.”
• Arithmetic slips when adding large exponents.

Final Answer:
1110