Question

Difficulty: Very hardFactors, Multiples, and Prime Factorization

Consider the positive integer N=24×33×52N = 2^4 \times 3^3 \times 5^2. Which of the following statements regarding the properties of the factors and multiples of NN is/are correct?

  1. The number of positive factors of NN that are perfect squares is equal to 12.Answer
  2. The sum of the reciprocals of all positive factors of NN is equal to σ(N)N\frac{\sigma(N)}{N}, where σ(N)\sigma(N) is the sum of all positive factors of NN.Answer
  3. The number of ordered pairs of positive integers (x,y)(x, y) such that lcm(x,y)=N\text{lcm}(x, y) = N is equal to 315.Answer
  4. D
    The product of all positive factors of NN is equal to N60N^{60}.

Answer

The statements asserting that the number of perfect square factors is 12, the sum of factor reciprocals equals σ(N)N\frac{\sigma(N)}{N}, and the number of ordered pairs (x,y)(x, y) such that lcm(x,y)=N\text{lcm}(x, y) = N is 315 are all correct.
The statements asserting that NN has 12 perfect square factors, that the sum of reciprocals of all factors is σ(N)N\frac{\sigma(N)}{N}, and that there are 315 ordered pairs (x,y)(x, y) with lcm(x,y)=N\text{lcm}(x, y) = N are all mathematically correct applications of prime factorization principles.

Step-by-Step Solution

1
Determine the number of perfect square factors of N=24×33×52N = 2^4 \times 3^3 \times 5^2.
Any factor d=2a×3b×5cd = 2^a \times 3^b \times 5^c is a perfect square if a,b,ca, b, c are even. The allowed values are a{0,2,4}a \in \{0, 2, 4\} (3 options), b{0,2}b \in \{0, 2\} (2 options), and c{0,2}c \in \{0, 2\} (2 options). Total perfect square factors = 3×2×2=123 \times 2 \times 2 = 12.
Perfect square factors require non-negative even exponents for each prime factor.
2
Evaluate the sum of the reciprocals of all factors of NN.
As dd ranges over all divisors of NN, Nd\frac{N}{d} also ranges over all divisors of NN. Therefore, dN1d=dNN/dN=1NdNd=σ(N)N\sum_{d|N} \frac{1}{d} = \sum_{d|N} \frac{N/d}{N} = \frac{1}{N} \sum_{d'|N} d' = \frac{\sigma(N)}{N}.
Pairing each factor dd with its complementary factor N/dN/d converts the reciprocal sum into the total factor sum divided by NN.
3
Calculate the number of ordered pairs (x,y)(x, y) such that lcm(x,y)=N\text{lcm}(x, y) = N.
For each prime pieip_i^{e_i}, the maximum exponent in xx and yy must be eie_i. The number of choices for (ai,bi)(a_i, b_i) such that max(ai,bi)=ei\max(a_i, b_i) = e_i is 2ei+12e_i + 1. For N=24×33×52N = 2^4 \times 3^3 \times 5^2, total ordered pairs = (2(4)+1)(2(3)+1)(2(2)+1)=9×7×5=315(2(4)+1)(2(3)+1)(2(2)+1) = 9 \times 7 \times 5 = 315.
The exponent of each prime factor in lcm(x,y)\text{lcm}(x, y) is determined independently by max(ai,bi)=ei\max(a_i, b_i) = e_i.
4
Compute the product of all positive factors of NN.
Total number of factors D(N)=(4+1)(3+1)(2+1)=60D(N) = (4+1)(3+1)(2+1) = 60. The product of all factors is ND(N)/2=N60/2=N30N^{D(N)/2} = N^{60/2} = N^{30}.
Factors can be grouped into D(N)/2D(N)/2 pairs, each multiplying to NN.

Key Concept

Advanced properties of prime factorization including perfect square factor enumeration, reciprocal factor sum identity, LCM pair counting formula, and factor product relations.
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