Question

Difficulty: Very hardPrime Factorization, GCD, and LCM

Let K=25×34×53×112K = 2^5 \times 3^4 \times 5^3 \times 11^2. How many positive integer factors of KK are divisible by 300300 but are not divisible by 900900?

  1. A
    16
  2. 24Answer
  3. C
    36
  4. D
    48
  5. E
    96

Answer

24
The correct answer 24 is obtained by analyzing the prime factorizations 300=22×31×52300 = 2^2 \times 3^1 \times 5^2 and 900=22×32×52900 = 2^2 \times 3^2 \times 5^2. Any factor f=2a×3b×5c×11df = 2^a \times 3^b \times 5^c \times 11^d must satisfy a∈{2,3,4,5}a \in \{2, 3, 4, 5\} (4 choices), b=1b = 1 (1 choice, since b≥1b \ge 1 for 300 but b<2b < 2 for 900), c∈{2,3}c \in \{2, 3\} (2 choices), and d∈{0,1,2}d \in \{0, 1, 2\} (3 choices). Multiplying these choices gives 4×1×2×3=244 \times 1 \times 2 \times 3 = 24.

Step-by-Step Solution

1
Express any factor of KK in terms of prime factor exponent constraints.
Any factor ff of KK has the form f=2a×3b×5c×11df = 2^a \times 3^b \times 5^c \times 11^d, where 0≤a≤50 \le a \le 5, 0≤b≤40 \le b \le 4, 0≤c≤30 \le c \le 3, and 0≤d≤20 \le d \le 2.
The prime factors of ff must be subsets of the prime factors of KK with exponents not exceeding those in KK.
2
Find the prime factorizations of 300300 and 900900.
300=22×31×52300 = 2^2 \times 3^1 \times 5^2 and 900=22×32×52900 = 2^2 \times 3^2 \times 5^2.
Divisibility criteria correspond to minimum exponent requirements for each prime factor.
3
Apply the divisibility conditions to determine constraints on each exponent.
For ff to be divisible by 300300, we need a≥2a \ge 2, b≥1b \ge 1, and c≥2c \ge 2. For ff to NOT be divisible by 900900, we must have b<2b < 2. Thus, b=1b = 1.
Combining b≥1b \ge 1 and b<2b < 2 uniquely restricts bb to 11.
4
Count the number of valid choices for each exponent.
a∈{2,3,4,5}a \in \{2, 3, 4, 5\} (44 choices), b∈{1}b \in \{1\} (11 choice), c∈{2,3}c \in \{2, 3\} (22 choices), and d∈{0,1,2}d \in \{0, 1, 2\} (33 choices).
The exponent dd is unrestricted by 300300 or 900900, so it can take any valid power present in KK.
5
Multiply the number of independent choices using the fundamental counting principle.
4×1×2×3=244 \times 1 \times 2 \times 3 = 24.
Each exponent choice can be paired independently to form a unique factor.

Key Concept

Counting Divisors Using Prime Exponent Constraints
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