A positive integer n has a prime factorization of the form 2^x \times 3^y \times 7^z, wher… — GRE General Test

A positive integer $n$ has a prime factorization of the form $2^x \times 3^y \times 7^z$ , where $x$ , $y$ , and $z$ are positive integers. The greatest common divisor of $n$ and $420$ is $84$ , and the least common multiple of $n$ and $420$ is $8,820$ . What is the total number of positive integer factors of $n$ ?

A
$8$
B
$12$
C
$18$
$27$ Cevap
E
$36$

Cevap

The total number of positive integer factors of

n

27

By prime factorizing

420

84

, and

8,820

, we find that

420 = 2^2 \times 3^1 \times 5^1 \times 7^1

\gcd(n, 420) = 2^2 \times 3^1 \times 7^1

, and

\text{lcm}(n, 420) = 2^2 \times 3^2 \times 5^1 \times 7^2

. Comparing the minimum and maximum powers for each prime factor shows that

n = 2^2 \times 3^2 \times 7^2

. Applying the divisor counting formula gives

(2+1)(2+1)(2+1) = 27

Adım Adım Çözüm

Express all given values in their prime factorized forms.

420 = 2^2 \times 3^1 \times 5^1 \times 7^1

\gcd(n, 420) = 84 = 2^2 \times 3^1 \times 7^1

, and

\text{lcm}(n, 420) = 8,820 = 2^2 \times 3^2 \times 5^1 \times 7^2

Finding the prime factorizations allows comparison of prime exponent bounds for GCD and LCM.

Determine the exponents

x

y

, and

z

for

n = 2^x \times 3^y \times 7^z

For prime

2

\max(x, 2) = 2

and

\min(x, 2) = 2 \implies x = 2

.
For prime

3

\max(y, 1) = 2

and

\min(y, 1) = 1 \implies y = 2

.
For prime

7

\max(z, 1) = 2

and

\min(z, 1) = 1 \implies z = 2

GCD takes the minimum exponent of each prime factor, while LCM takes the maximum exponent.

Calculate the total number of positive factors of

n = 2^2 \times 3^2 \times 7^2

Total factors

= (x + 1)(y + 1)(z + 1) = (2 + 1)(2 + 1)(2 + 1) = 3 \times 3 \times 3 = 27

The number of positive divisors of

p_1^{a} p_2^{b} \dots p_k^{k}

(a+1)(b+1)\dots(k+1)

Anahtar Kavram

Relationship between prime factorization, greatest common divisor (GCD), least common multiple (LCM), and the number of positive divisors.

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