Among the numbers 99, 101, 176, and 182, which one has the greatest number of positive divisors? Use prime factorization to compare the total count of divisors for each option.

Introduction / Context:
This question tests understanding of how the number of divisors of an integer is related to its prime factorization. Instead of calculating all divisors manually, we use a formula based on the exponents of primes in the factorization. Comparing several numbers in this way is an efficient technique for identifying which has the most divisors.

Given Data / Assumptions:

• Numbers to compare: 99, 101, 176, 182.
• We want to find which number has the greatest number of positive divisors.

Concept / Approach:
If a positive integer n has prime factorization n = p1^a1 * p2^a2 * ... * pk^ak, then the total number of positive divisors of n is (a1 + 1) * (a2 + 1) * ... * (ak + 1). Therefore, we factor each candidate number, compute this divisor count, and then compare the results to see which is largest.

Step-by-Step Solution:
Factor 99: 99 = 3^2 * 11^1. Number of divisors of 99 = (2 + 1) * (1 + 1) = 3 * 2 = 6. Factor 101: 101 is a prime, so 101 = 101^1. Number of divisors of 101 = (1 + 1) = 2. Factor 176: 176 = 2^4 * 11^1. Number of divisors of 176 = (4 + 1) * (1 + 1) = 5 * 2 = 10. Factor 182: 182 = 2^1 * 7^1 * 13^1. Number of divisors of 182 = (1 + 1) * (1 + 1) * (1 + 1) = 2 * 2 * 2 = 8. Compare counts: 6, 2, 10, 8. The maximum is 10 for 176.

Verification / Alternative check:
We can list a few divisors of 176 to see that it has many: 1, 2, 4, 8, 11, 16, 22, 44, 88, 176. This matches the count of 10 divisors computed from the formula. The other numbers have fewer divisors based on the exponent method, so the result is consistent.

Why Other Options Are Wrong:
99 has 6 divisors, 101 has 2 divisors, and 182 has 8 divisors, all of which are fewer than the 10 divisors of 176. 90 is not in the original comparison set and is only included as a distractor.

Common Pitfalls:
Learners sometimes try to list divisors exhaustively for each number, which is time consuming and error prone. Others may incorrectly factor a number and then misapply the formula. Carefully factor each number into primes and use the exponent plus one rule for each prime factor to get the correct divisor count efficiently.

Final Answer:
176