To find the maximum 4-digit positive integer that leaves a remainder of 3 when divided by 5, 7, and 13, we follow these steps:

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Step 1: Find the least common multiple (LCM)

The divisors are $5$, $7$, and $13$. Since these are prime numbers, their LCM is their product:

$\text{LCM}(5, 7, 13) = 5 \times 7 \times 13 = 455.$

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Step 2: General form of the number

A number that leaves a remainder of 3 when divided by these numbers can be expressed as:

$N = 455k + 3,$

where $k$ is a non-negative integer.

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Step 3: Determine the maximum value of $N$ under 4 digits

To ensure $N$ is a 4-digit number, we solve:

$455k + 3 \leq 9999.$

Simplify:

$455k \leq 9996, \quad k \leq \frac{9996}{455} \approx 21.97.$

Thus, $k = 21$ is the largest integer value.

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Step 4: Calculate $N$ for $k = 21$

Substitute $k = 21$ into the formula for $N$:

$N = 455 \times 21 + 3 = 9555 + 3 = 9558.$

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Step 5: Verify the conditions

1. $9558 \div 5 = 1911$ remainder $3$,
2. $9558 \div 7 = 1365$ remainder $3$,
3. $9558 \div 13 = 735$ remainder $3$.

Thus, $9558$ satisfies the conditions.

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Final Answer:

The maximum 4-digit positive integer is:

$\boxed{9558}.$

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Would you like a detailed explanation of any specific step or a related example? Here are some relative questions:

1. How do you calculate the LCM of multiple numbers?
2. What is the significance of modular arithmetic in such problems?
3. Can we extend this method to find numbers with other remainders?
4. How can such problems be solved using programming?
5. How do you check divisibility for larger sets of conditions?

Tip: Always verify your solution by substituting it back into the problem