Question

Difficulty: MediumProperties of Integers and Divisibility

If N=24×33×52N = 2^4 \times 3^3 \times 5^2, how many positive integer factors of NN are divisible by 66 but not divisible by 1212?

Answer: 9

Answer

The number of positive integer factors of NN divisible by 6 but not by 12 is 9.
Any factor of N=24×33×52N = 2^4 \times 3^3 \times 5^2 is of the form 2a×3b×5c2^a \times 3^b \times 5^c. For the factor to be divisible by 66, we must have a1a \ge 1 and b1b \ge 1. For it to not be divisible by 12=22×3112 = 2^2 \times 3^1, we must have a<2a < 2. Therefore, aa must equal 11. The possible choices for aa are 11 value (11), for bb are 33 values (1,2,31, 2, 3), and for cc are 33 values (0,1,20, 1, 2). Multiplying these options gives 1×3×3=91 \times 3 \times 3 = 9.

Step-by-Step Solution

1
Express the prime factorization structure of a factor of NN.
Any factor of NN takes the form 2a×3b×5c2^a \times 3^b \times 5^c with bounds 0a40 \le a \le 4, 0b30 \le b \le 3, and 0c20 \le c \le 2.
Divisors of a number are formed by taking prime factors with exponents between zero and their maximum powers in the original number.
2
Determine exponent constraints for divisibility by 6 and non-divisibility by 12.
Divisibility by 66 requires a1a \ge 1 and b1b \ge 1. Non-divisibility by 1212 requires a<2a < 2. Thus, a=1a = 1 exactly.
A factor must contain at least one factor of 2 and one factor of 3 to be a multiple of 6, but containing two or more factors of 2 makes it a multiple of 12.
3
Count combinations of choices for the exponents.
Exponent aa has 11 option (a=1a = 1), bb has 33 options (b{1,2,3}b \in \{1, 2, 3\}), and cc has 33 options (c{0,1,2}c \in \{0, 1, 2\}). Total =1×3×3=9= 1 \times 3 \times 3 = 9.
Applying the fundamental counting principle by multiplying the number of choices for independent prime factor exponents.

Key Concept

Counting Divisors with Prime Factorization and Divisibility Constraints
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