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 b1b \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 0a50 \le a \le 5, 0b40 \le b \le 4, 0c30 \le c \le 3, and 0d20 \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 a2a \ge 2, b1b \ge 1, and c2c \ge 2. For ff to NOT be divisible by 900900, we must have b<2b < 2. Thus, b=1b = 1.
Combining b1b \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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