To solve this problem, we use the principle of inclusion and exclusion to count the integers divisible by exactly two of the given divisors $3, 5, 7, 11$. Here’s how:

Step 1: Define the sets

Let:

• $A_3$: Set of integers divisible by $3$,
• $A_5$: Set of integers divisible by $5$,
• $A_7$: Set of integers divisible by $7$,
• $A_{11}$: Set of integers divisible by $11$.

We want to count the integers divisible by exactly two of these divisors. This means we count numbers divisible by the pairwise products of the divisors, but not by the product of three or all four divisors.

Step 2: Count numbers divisible by two divisors

The integers divisible by exactly two divisors $p$ and $q$ are: $\text{Count}(p, q) = \left\lfloor \frac{10000}{p \cdot q} \right\rfloor - \text{overlaps with three divisors}$

For each pair $(p, q)$, calculate the contribution.

Pairwise contributions:

1. Numbers divisible by $3 \times 5 = 15$: $\left\lfloor \frac{10000}{15} \right\rfloor = 666$
2. Numbers divisible by $3 \times 7 = 21$: $\left\lfloor \frac{10000}{21} \right\rfloor = 476$
3. Numbers divisible by $3 \times 11 = 33$: $\left\lfloor \frac{10000}{33} \right\rfloor = 303$
4. Numbers divisible by $5 \times 7 = 35$: $\left\lfloor \frac{10000}{35} \right\rfloor = 285$
5. Numbers divisible by $5 \times 11 = 55$: $\left\lfloor \frac{10000}{55} \right\rfloor = 181$
6. Numbers divisible by $7 \times 11 = 77$: $\left\lfloor \frac{10000}{77} \right\rfloor = 129$

Subtract overlaps with three divisors:

For numbers divisible by three divisors:

1. Numbers divisible by $3 \times 5 \times 7 = 105$: $\left\lfloor \frac{10000}{105} \right\rfloor = 95$
2. Numbers divisible by $3 \times 5 \times 11 = 165$: $\left\lfloor \frac{10000}{165} \right\rfloor = 60$
3. Numbers divisible by $3 \times 7 \times 11 = 231$: $\left\lfloor \frac{10000}{231} \right\rfloor = 43$
4. Numbers divisible by $5 \times 7 \times 11 = 385$: $\left\lfloor \frac{10000}{385} \right\rfloor = 25$

Add overlaps with all four divisors:

For numbers divisible by $3 \times 5 \times 7 \times 11 = 1155$: $\left\lfloor \frac{10000}{1155} \right\rfloor = 8$

Step 3: Total count for exactly two divisors

Using inclusion-exclusion, the total count is: $\text{Total} = \sum_{\text{pairs}} \text{Count}(p, q) - \sum_{\text{triplets}} \text{Count}(p, q, r) + \text{Count}(p, q, r, s)$

Substitute the values: $\text{Total} = (666 + 476 + 303 + 285 + 181 + 129) - (95 + 60 + 43 + 25) + 8$

$\text{Total} = 2040 - 223 + 8 = 1825$

Final Answer:

The number of integers between 1 and 10,000 divisible by exactly two of $3, 5, 7, 11$ is: $\boxed{1825}$

Would you like a more detailed breakdown or clarification of any steps?

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Related Questions:

1. How can this method be extended to find integers divisible by exactly three divisors?
2. What would change if the range were from 1 to 20,000 instead of 10,000?
3. Can this method be applied to a different set of divisors, such as $2, 3, 5, 7$?
4. How does the principle of inclusion and exclusion work in general?
5. What is the computational complexity of this method for larger ranges?

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Tip:

Always carefully track overlaps when using inclusion-exclusion to avoid double-counting.