halla n si A=28^n x 50 tiene 150 divisores

To determine the value of \( n \) such that the number of divisors of \( A = 28^n \times 50 \) equals 150, let's break it down.

### Step 1: Prime Factorization of \( 28 \) and \( 50 \)
- \( 28 = 2^2 \times 7 \)
- \( 50 = 2 \times 5^2 \)

Now, substitute these factorizations into the expression for \( A \):
\[
A = (28)^n \times 50 = (2^2 \times 7)^n \times (2 \times 5^2)
\]

This simplifies to:
\[
A = 2^{2n} \times 7^n \times 2 \times 5^2
\]
\[
A = 2^{2n+1} \times 7^n \times 5^2
\]

### Step 2: Formula for the Number of Divisors
The formula for the number of divisors of a number given its prime factorization is:
\[
\text{Divisors} = (e_1 + 1)(e_2 + 1)(e_3 + 1) \dots
\]
where \( e_1, e_2, \dots \) are the exponents of the prime factors.

From the prime factorization of \( A \):
\[
A = 2^{2n+1} \times 7^n \times 5^2
\]
the exponents are:
- \( 2^{2n+1} \) has exponent \( 2n+1 \)
- \( 7^n \) has exponent \( n \)
- \( 5^2 \) has exponent \( 2 \)

Thus, the number of divisors \( D(A) \) is:
\[
D(A) = (2n+1 + 1)(n + 1)(2 + 1)
\]
\[
D(A) = (2n+2)(n+1)(3)
\]

We are given that the number of divisors is 150:
\[
(2n+2)(n+1)(3) = 150
\]
\[
(2n+2)(n+1) = 50
\]

### Step 3: Solve the Equation
Expand the left-hand side:
\[
(2n+2)(n+1) = 2n^2 + 3n + 2
\]
Now, set the equation equal to 50:
\[
2n^2 + 3n + 2 = 50
\]
\[
2n^2 + 3n - 48 = 0
\]
Divide by 2 to simplify:
\[
n^2 + \frac{3}{2}n - 24 = 0
\]


Discover how to solve for n in the equation A = 28^n x 50, given that the number of divisors of A is 150. This solution covers prime factorization, divisor formulas, and solving quadratic equations, suitable for high school students in Grades 9-12.

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