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Difficulty: MediumDivisibility, Factors, and Multiples

If k=23⋅3a⋅7bk = 2^3 \cdot 3^a \cdot 7^b, where aa and bb are positive integers, and kk has exactly 36 positive integer divisors, how many positive integer divisors does 3k3k have?

  1. A
    39
  2. B
    40
  3. 48Answer
  4. D
    54
  5. E
    72

Answer

48
The number of positive integer divisors of a prime-factored number p1e1p2e2⋯pmemp_1^{e_1} p_2^{e_2} \cdots p_m^{e_m} is (e1+1)(e2+1)⋯(em+1)(e_1+1)(e_2+1)\cdots(e_m+1). For k=23⋅3a⋅7bk = 2^3 \cdot 3^a \cdot 7^b, the divisor count is 4(a+1)(b+1)=364(a+1)(b+1) = 36, which simplifies to (a+1)(b+1)=9(a+1)(b+1) = 9. Given a,b≥1a, b \ge 1, the only integer solution is a=2a = 2 and b=2b = 2. Multiplying kk by 3 yields 3k=23⋅33⋅723k = 2^3 \cdot 3^3 \cdot 7^2, which has (3+1)(3+1)(2+1)=48(3+1)(3+1)(2+1) = 48 positive divisors.

Step-by-Step Solution

1
Use the formula for the number of positive divisors of an integer.
The number of positive divisors of k=23⋅3a⋅7bk = 2^3 \cdot 3^a \cdot 7^b is given by (3+1)(a+1)(b+1)=4(a+1)(b+1)(3+1)(a+1)(b+1) = 4(a+1)(b+1).
If N=p1e1p2e2⋯pmemN = p_1^{e_1} p_2^{e_2} \cdots p_m^{e_m}, the total number of positive integer divisors is (e1+1)(e2+1)⋯(em+1)(e_1 + 1)(e_2 + 1) \cdots (e_m + 1).
2
Set the divisor count equal to 36 and solve for aa and bb.
4(a+1)(b+1)=36  ⟹  (a+1)(b+1)=94(a+1)(b+1) = 36 \implies (a+1)(b+1) = 9.
Dividing both sides by 4 isolates the term containing aa and bb.
3
Determine the values of aa and bb using the constraint that aa and bb are positive integers.
Since a≥1a \ge 1 and b≥1b \ge 1, we have a+1≥2a+1 \ge 2 and b+1≥2b+1 \ge 2. The only factor pair of 9 where both factors are at least 2 is 3×33 \times 3. Thus, a+1=3  ⟹  a=2a+1 = 3 \implies a = 2 and b+1=3  ⟹  b=2b+1 = 3 \implies b = 2.
Positive integers must be greater than or equal to 1.
4
Write the prime factorization of 3k3k and compute its number of positive divisors.
3k=3⋅(23⋅32⋅72)=23⋅33⋅723k = 3 \cdot (2^3 \cdot 3^2 \cdot 7^2) = 2^3 \cdot 3^3 \cdot 7^2. The number of divisors is (3+1)(3+1)(2+1)=4⋅4⋅3=48(3+1)(3+1)(2+1) = 4 \cdot 4 \cdot 3 = 48.
Multiplying kk by 3 increases the exponent of 3 from 2 to 3.

Key Concept

Prime Factorization and Total Divisors Rule
Estimated Time:1m 30s
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Topics

Divisibility, Factors, and Multiples

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