Question

Difficulty: HardFactors, Multiples, and Prime Factorization

Let N=25×34×52N = 2^5 \times 3^4 \times 5^2. How many positive integer factors of NN are divisible by 1212 but not divisible by 7272?

Answer: 21

Answer

21
The correct answer is 21. By analyzing the exponent bounds of 2a×3b×5c2^a \times 3^b \times 5^c, there are 48 factors of NN divisible by 1212 (a2,b1a \ge 2, b \ge 1). Among these, 27 factors are also divisible by 7272 (a3,b2a \ge 3, b \ge 2). Subtracting the two counts gives 4827=2148 - 27 = 21 factors divisible by 12 but not by 72.

Step-by-Step Solution

1
Represent the general form of a factor of NN
Every positive factor of N=25×34×52N = 2^5 \times 3^4 \times 5^2 can be uniquely written as f=2a×3b×5cf = 2^a \times 3^b \times 5^c, where a{0,1,2,3,4,5}a \in \{0, 1, 2, 3, 4, 5\}, b{0,1,2,3,4}b \in \{0, 1, 2, 3, 4\}, and c{0,1,2}c \in \{0, 1, 2\}.
By the Fundamental Theorem of Arithmetic, any factor of a number in prime-factored form takes exponents bounded by the original number's exponents.
2
Identify exponent constraints for divisibility by 12
Since 12=22×3112 = 2^2 \times 3^1, for ff to be divisible by 1212, we must have a2a \ge 2 and b1b \ge 1.
Divisibility requires prime factor exponents in the factor to be at least as large as those in the divisor.
3
Calculate the total number of factors divisible by 12
Number of choices for aa: 52+1=45 - 2 + 1 = 4 (values 2,3,4,52, 3, 4, 5). Number of choices for bb: 41+1=44 - 1 + 1 = 4 (values 1,2,3,41, 2, 3, 4). Number of choices for cc: 20+1=32 - 0 + 1 = 3 (values 0,1,20, 1, 2). Total count = 4×4×3=484 \times 4 \times 3 = 48.
By the multiplication principle of counting, independent choices for each exponent are multiplied.
4
Identify exponent constraints and count factors divisible by 72
Since 72=23×3272 = 2^3 \times 3^2, a factor is divisible by 7272 if a3a \ge 3 and b2b \ge 2. Number of choices: a{3,4,5}a \in \{3, 4, 5\} (3 choices), b{2,3,4}b \in \{2, 3, 4\} (3 choices), c{0,1,2}c \in \{0, 1, 2\} (3 choices). Total count = 3×3×3=273 \times 3 \times 3 = 27.
Every factor of NN that is divisible by 72 is inherently also a factor divisible by 12, forming a strict subset.
5
Apply the principle of subtraction to find the required count
Required count = (Factors divisible by 12) - (Factors divisible by 72) = 4827=2148 - 27 = 21.
The set of factors divisible by 12 but not by 72 is the set difference between factors divisible by 12 and factors divisible by 72.

Key Concept

Counting Factors with Divisibility Constraints using Prime Factorization
Estimated Time:2m 0s
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