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Zorluk: ZorFactors, Multiples, and Prime Factorization

Let NN be a positive integer with the prime factorization N=24×32×53N = 2^4 \times 3^2 \times 5^3. Which of the following statements regarding the positive factors of NN are correct?

  1. The total number of positive factors of NN that are perfect squares is 1212.Cevap
  2. The number of positive factors of NN that are divisible by 1515 is 3030.Cevap
  3. C
    The total number of positive even factors of NN is 5050.
  4. D
    The number of positive odd factors of NN is 1515.

Cevap

The correct statements are that the total number of positive factors of NN that are perfect squares is 1212, and the number of positive factors of NN that are divisible by 1515 is 3030.
The statement asserting that NN has 1212 factors that are perfect squares is correct because choosing even powers (0,2,40, 2, 4 for 22; 0,20, 2 for 33; and 0,20, 2 for 55) yields 3×2×2=123 \times 2 \times 2 = 12 factors. The statement asserting that 3030 factors are divisible by 1515 is also correct because forcing exponents of 33 and 55 to be at least 11 yields 5×2×3=305 \times 2 \times 3 = 30 factors.

Adım Adım Çözüm

1
Analyze the prime factorization N=24×32×53N = 2^4 \times 3^2 \times 5^3
Any positive factor of NN is of the form 2a×3b×5c2^a \times 3^b \times 5^c, where 0a40 \le a \le 4, 0b20 \le b \le 2, and 0c30 \le c \le 3.
By the Fundamental Theorem of Arithmetic, factors inherit prime bases with constrained exponents.
2
Count the number of factors that are perfect squares
For a perfect square, a{0,2,4}a \in \{0, 2, 4\} (3 choices), b{0,2}b \in \{0, 2\} (2 choices), and c{0,2}c \in \{0, 2\} (2 choices). Total square factors =3×2×2=12= 3 \times 2 \times 2 = 12.
All prime exponents in a perfect square must be even non-negative integers.
3
Count the number of factors divisible by 1515
Since 15=31×5115 = 3^1 \times 5^1, a{0,1,2,3,4}a \in \{0, 1, 2, 3, 4\} (5 choices), b{1,2}b \in \{1, 2\} (2 choices), and c{1,2,3}c \in \{1, 2, 3\} (3 choices). Total factors divisible by 15=5×2×3=3015 = 5 \times 2 \times 3 = 30.
Divisibility by 1515 mandates at least one factor of 33 and at least one factor of 55.
4
Verify even and odd factor counts to evaluate the remaining claims
Even factors require a1a \ge 1, giving 4×3×4=484 \times 3 \times 4 = 48 even factors. Odd factors require a=0a = 0, giving 1×3×4=121 \times 3 \times 4 = 12 odd factors.
Evaluating a1a \ge 1 vs a=0a = 0 confirms that claims specifying 5050 even factors and 1515 odd factors are incorrect.

Anahtar Kavram

Counting factors with constrained exponents in prime factorizations
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