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Zorluk: OrtaDivisibility, Factors, and Multiples

Let nn be a positive integer with exactly two distinct prime factors. If nn is a multiple of 1212, is not a multiple of 88, and has exactly 1212 positive factors, what is the number of positive factors of n2n^2?

  1. A
    24
  2. B
    33
  3. 35Cevap
  4. D
    45
  5. E
    144

Cevap

The number of positive factors of n2n^2 is 35.
Since nn is a multiple of 12=22⋅3112 = 2^2 \cdot 3^1 and has exactly two distinct prime factors, its prime factorization must be n=2a⋅3bn = 2^a \cdot 3^b. The condition that nn is not a multiple of 8=238 = 2^3 forces a=2a = 2. Using the factor count formula (a+1)(b+1)=12(a+1)(b+1) = 12, we get (2+1)(b+1)=12(2+1)(b+1) = 12, which gives b=3b = 3. Therefore, n2=(22⋅33)2=24⋅36n^2 = (2^2 \cdot 3^3)^2 = 2^4 \cdot 3^6. The total number of positive factors of n2n^2 is (4+1)(6+1)=35(4+1)(6+1) = 35.

Adım Adım Çözüm

1
Determine the prime factors of nn
n=2a⋅3bn = 2^a \cdot 3^b, where a≥2a \geq 2 and b≥1b \geq 1
Since nn has exactly two distinct prime factors and 12=22⋅3112 = 2^2 \cdot 3^1 divides nn, the only prime factors of nn are 22 and 33.
2
Apply the divisibility constraints on the exponent aa
a=2a = 2
Since 1212 divides nn, a≥2a \geq 2. Since nn is not a multiple of 8=238 = 2^3, a<3a < 3. Thus aa must equal 22.
3
Use the factor count formula to find bb
b=3b = 3, so n=22⋅33n = 2^2 \cdot 3^3
The number of positive factors of nn is (a+1)(b+1)=(2+1)(b+1)=12  ⟹  3(b+1)=12  ⟹  b=3(a+1)(b+1) = (2+1)(b+1) = 12 \implies 3(b+1) = 12 \implies b = 3.
4
Calculate the number of positive factors of n2n^2
(4+1)(6+1)=35(4+1)(6+1) = 35
Squaring nn gives n2=(22⋅33)2=24⋅36n^2 = (2^2 \cdot 3^3)^2 = 2^4 \cdot 3^6. By the factor count formula, n2n^2 has (4+1)(6+1)=35(4+1)(6+1) = 35 positive factors.

Anahtar Kavram

Factor Count Formula & Prime Factorization Constraints
Tahmini Süre:1m 45s
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Konular

Divisibility, Factors, and Multiples

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