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The question asks for the probability that a randomly chosen positive divisor of \(10^{23}\) is an integral multiple of \(10^{16}\).

### Step-by-step solution:

1. **Prime Factorization of Powers of 10:**
   \[
   10^{23} = 2^{23} \times 5^{23}
   \]
   \[
   10^{16} = 2^{16} \times 5^{16}
   \]
   So, the divisors of \(10^{23}\) are of the form \(2^a \times 5^b\), where \(0 \leq a \leq 23\) and \(0 \leq b \leq 23\).

2. **Number of Divisors of \(10^{23}\):**
   The total number of divisors of \(10^{23}\) can be found by using the formula for the number of divisors:
   \[
   \text{Number of divisors of } 10^{23} = (23+1)(23+1) = 24 \times 24 = 576
   \]
   So, there are 576 divisors of \(10^{23}\).

3. **Condition for Being a Multiple of \(10^{16}\):**
   For a divisor of \(10^{23}\) to be a multiple of \(10^{16}\), the divisor must be of the form \(2^a \times 5^b\), where:
   \[
   a \geq 16 \quad \text{and} \quad b \geq 16
   \]
   This is because any multiple of \(10^{16}\) must contain at least \(2^{16} \times 5^{16}\).

4. **Number of Divisors That Are Multiples of \(10^{16}\):**
   The values of \(a\) and \(b\) can range from 16 to 23, inclusive. So, both \(a\) and \(b\) have 8 possible values (16, 17, ..., 23).

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

5. **Probability Calculation:**
   The probability is the ratio of favorable outcomes (divisors that are multiples of \(10^{16}\)) to the total outcomes (all divisors of \(10^{23}\)):
   \[
   \text{Probability} = \frac{64}{576} = \frac{1}{9}
   \]

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

### Answer: \(\boxed{\frac{1}{9}}\)

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Would you like further clarification on any of these steps?

Here are 5 related questions for deeper understanding:
1. How do you find the number of divisors of a number?
2. What is the significance of prime factorization in solving this type of problem?
3. How would the probability change if the powers in the problem were different, say \(10^{30}\) and \(10^{20}\)?
4. How do you apply combinatorics to divisor-related probability problems?
5. What is the formula for the number of divisors of a number given its prime factorization?

### Tip: 
When solving divisor-related probability questions, always break the problem into prime factorizations and count the ranges of exponents carefully.

What is the probability that a randomly chosen positive divisor of 10^23 is an integral multiple of 10^16?

Explore the probability that a randomly chosen divisor of 10^23 is a multiple of 10^16. This problem uses prime factorization and the number of divisors formula to find the solution. Suitable for high school students in grades 9-12.

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