What is the probability that a randomly chosen positive divisor of 10^23 is an integral multiple of 10^16 options are 1/9,2/9,1/3,4/15

We are tasked with finding the probability that a randomly chosen positive divisor of \( 10^{23} \) is an integral multiple of \( 10^{16} \).

### Step 1: Prime Factorization of \( 10^{23} \)
The number \( 10^{23} \) can be written as:
\[
10^{23} = (2 \cdot 5)^{23} = 2^{23} \cdot 5^{23}
\]
So, the divisors of \( 10^{23} \) are of the form \( 2^a \cdot 5^b \), where \( 0 \leq a \leq 23 \) and \( 0 \leq b \leq 23 \).

### Step 2: Total Number of Divisors of \( 10^{23} \)
The total number of divisors of \( 10^{23} \) is given by the formula for the number of divisors of a number \( n = p^a \cdot q^b \), which is \( (a+1)(b+1) \). In this case, the total number of divisors is:
\[
(23+1)(23+1) = 24 \times 24 = 576
\]

### Step 3: Condition for Divisors to be Multiples of \( 10^{16} \)
A divisor of \( 10^{23} \) is an integral multiple of \( 10^{16} = 2^{16} \cdot 5^{16} \). Thus, we need the divisor to be of the form \( 2^a \cdot 5^b \) where \( a \geq 16 \) and \( b \geq 16 \).

- For \( a \), the possible values are \( a = 16, 17, \dots, 23 \), which gives \( 23 - 16 + 1 = 8 \) options.
- Similarly, for \( b \), the possible values are \( b = 16, 17, \dots, 23 \), which also gives \( 8 \) options.

Thus, the number of divisors that are multiples of \( 10^{16} \) is:
\[
8 \times 8 = 64
\]

### Step 4: Probability Calculation
The probability is the ratio of the number of favorable outcomes (divisors that are multiples of \( 10^{16} \)) to the total number of divisors:
\[
\text{Probability} = \frac{64}{576} = \frac{1}{9}
\]

Thus, the probability that a randomly chosen divisor of \( 10^{23} \) is an integral multiple of \( 10^{16} \) is \( \boxed{\frac{1}{9}} \).

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Do you want more details or have any questions?

Here are 5 related questions to explore:
1. What is the prime factorization of \( 10^{16} \)?
2. How do you calculate the number of divisors of a number?
3. What is the probability that a randomly chosen divisor of \( 10^{23} \) is an integral multiple of \( 2^5 \)?
4. How does the probability change if we consider \( 10^{25} \) instead of \( 10^{23} \)?
5. Can the method used here be applied to other prime factorizations?

**Tip:** To calculate the number of divisors of a number, express it in its prime factorized form and use the divisor formula.

What is the probability that a randomly chosen positive divisor of 10^23 is an integral multiple of 10^16? Options are 1/9, 2/9, 1/3, 4/15.

Discover how to calculate the probability that a randomly chosen divisor of 10^23 is an integral multiple of 10^16. This problem uses prime factorization, the divisor formula, and probability concepts, suitable for students in Grades 10-12.

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