Examkin

Question

Difficulty: HardPrime Numbers and Prime Factorization

If n=3x⋅5y⋅7zn = 3^x \cdot 5^y \cdot 7^z, where xx, yy, and zz are positive integers, and nn has exactly 2424 positive divisors, what is the minimum possible value of x+y+zx + y + z?

  1. A
    5
  2. 6Answer
  3. C
    7
  4. D
    9
  5. E
    21

Answer

6
The total number of positive divisors of n=3x⋅5y⋅7zn = 3^x \cdot 5^y \cdot 7^z is (x+1)(y+1)(z+1)=24(x+1)(y+1)(z+1) = 24. Because x,y,zx, y, z are positive integers, each factor term must be at least 2. Factoring 24 into three integer factors each ≥2\ge 2 yields two possibilities: (2,2,6)(2, 2, 6) and (2,3,4)(2, 3, 4). The sum of the exponents x+y+zx+y+z equals (x+1)+(y+1)+(z+1)−3(x+1)+(y+1)+(z+1) - 3. For (2,2,6)(2, 2, 6), the sum is 2+2+6−3=72+2+6-3 = 7. For (2,3,4)(2, 3, 4), the sum is 2+3+4−3=62+3+4-3 = 6. The minimum possible value is 6.

Step-by-Step Solution

1
Express the number of divisors using the prime factorization formula.
For n=3x⋅5y⋅7zn = 3^x \cdot 5^y \cdot 7^z, the number of positive divisors is (x+1)(y+1)(z+1)=24(x+1)(y+1)(z+1) = 24.
The total number of divisors of a prime-factored integer is found by taking the product of each prime factor's exponent plus one.
2
Determine the constraints on the factor terms.
Since x,y,zx, y, z are positive integers (x,y,z≥1x, y, z \ge 1), each term (x+1),(y+1),(z+1)≥2(x+1), (y+1), (z+1) \ge 2.
Exponents must be at least 1, so each factor in the product must be at least 2.
3
Find all valid sets of 3 integer factors of 24 that are all ≥2\ge 2.
The possible triples (a,b,c)(a, b, c) such that a⋅b⋅c=24a \cdot b \cdot c = 24 and a,b,c≥2a, b, c \ge 2 are (2,2,6)(2, 2, 6) and (2,3,4)(2, 3, 4).
We factor 24 into three integer components, each at least 2.
4
Calculate x+y+zx+y+z for each factor triple to find the minimum.
For (2,2,6)(2, 2, 6): x+1=2,y+1=2,z+1=6  ⟹  x+y+z=1+1+5=7x+1=2, y+1=2, z+1=6 \implies x+y+z = 1 + 1 + 5 = 7.
For (2,3,4)(2, 3, 4): x+1=2,y+1=3,z+1=4  ⟹  x+y+z=1+2+3=6x+1=2, y+1=3, z+1=4 \implies x+y+z = 1 + 2 + 3 = 6.
The minimum value is 6.
Comparing the sums of exponents shows that the factor triple (2,3,4)(2, 3, 4) minimizes x+y+zx+y+z.

Key Concept

Divisors from Prime Factorization
Estimated Time:2m 0s
Rate this question

Topics

Prime Numbers and Prime Factorization

More questions from this topic