If n = 3^x \cdot 5^y \cdot 7^z, where x, y, and z are positive integers, and n has exactly… — GMAT

If $n = 3^x \cdot 5^y \cdot 7^z$ , where $x$ , $y$ , and $z$ are positive integers, and $n$ has exactly $24$ positive divisors, what is the minimum possible value of $x + y + z$ ?

A
5
6Cevap
C
7
D
9
E
21

Cevap

The total number of positive divisors of

n = 3^x \cdot 5^y \cdot 7^z

(x+1)(y+1)(z+1) = 24

. Because

x, y, z

are positive integers, each factor term must be at least 2. Factoring 24 into three integer factors each

\ge 2

yields two possibilities:

(2, 2, 6)

and

(2, 3, 4)

. The sum of the exponents

x+y+z

equals

(x+1)+(y+1)+(z+1) - 3

. For

(2, 2, 6)

, the sum is

2+2+6-3 = 7

. For

(2, 3, 4)

, the sum is

2+3+4-3 = 6

. The minimum possible value is 6.

Adım Adım Çözüm

Express the number of divisors using the prime factorization formula.

For

n = 3^x \cdot 5^y \cdot 7^z

, the number of positive divisors is

(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.

Determine the constraints on the factor terms.

Since

x, y, z

are positive integers (

x, y, z \ge 1

), each term

(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.

Find all valid sets of 3 integer factors of 24 that are all

\ge 2

The possible triples

(a, b, c)

such that

a \cdot b \cdot c = 24

and

a, b, c \ge 2

are

(2, 2, 6)

and

(2, 3, 4)

We factor 24 into three integer components, each at least 2.

Calculate

x+y+z

for each factor triple to find the minimum.

For

(2, 2, 6)

x+1=2, y+1=2, z+1=6 \implies x+y+z = 1 + 1 + 5 = 7

.
For

(2, 3, 4)

x+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)

minimizes

x+y+z

Anahtar Kavram

Divisors from Prime Factorization

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