There are 4 classrooms. Each classroom can accommodate in prime numbers but each classroom can't have the same number of students. There are 140 boys and n girls who have to be accommodated in these rooms. The ratio of boys and girls must be as close to 1 as possible in every single class and also the total amount. What is the lowest possible difference between the highest and lowest capacity of the classrooms?

To solve this problem, we need to distribute the 140 boys and \( n \) girls into 4 classrooms, with each classroom accommodating a prime number of students and no two classrooms having the same number of students. Additionally, the number of boys and girls in each classroom should be as close as possible to a 1:1 ratio, and the total number of boys and girls in each class must follow this ratio as closely as possible.

### Steps:

1. **Identify prime numbers:** 
   Let's start by listing some prime numbers as candidates for the capacities of the classrooms. For example, the first few primes are:
   \[
   2, 3, 5, 7, 11, 13, 17, 19, 23, 29, \dots
   \]
   However, the prime numbers must also allow us to distribute the students fairly evenly.

2. **Determine the total number of students:** 
   Let the total number of students be \( S = 140 + n \), where \( n \) is the number of girls.

3. **Select prime numbers for classroom capacities:** 
   We need to select 4 distinct prime numbers such that the sum of these primes is as close as possible to the total number of students \( S \). The difference between the largest and smallest prime numbers should be minimized to ensure the lowest possible difference in capacities.

4. **Balance the boys and girls:** 
   The ratio of boys to girls in each classroom must be as close to 1:1 as possible. This means that for each classroom with a prime number of students \( p_i \), about half should be boys and half should be girls.

5. **Minimize the difference in capacities:** 
   We need to minimize the difference between the largest and smallest prime numbers selected for the classroom capacities.

Let’s calculate and consider possible configurations.

### Trial 1: Pick 4 distinct primes close to \( S = 140 + n \)

We try different sets of prime numbers that sum up to approximately \( 140 + n \).

- Set 1: Prime numbers \( 31, 29, 23, 17 \)
  \[
  31 + 29 + 23 + 17 = 100
  \]
  This total is smaller than 140, so this set will not work unless \( n \) is very large.

- Set 2: Prime numbers \( 47, 41, 37, 31 \)
  \[
  47 + 41 + 37 + 31 = 156
  \]
  This is more than 140, so we can adjust by determining the right number of girls \( n \).

### Conclusion:
The lowest possible difference between the highest and lowest capacities can be minimized using prime numbers that are close to each other but still distinct. For example, choosing the primes \( 47, 41, 37, \) and \( 31 \), the difference between the highest and lowest capacity is \( 47 - 31 = 16 \).

Thus, the lowest possible difference between the highest and lowest capacity of the classrooms is \( \boxed{16} \).

Would you like to explore more details or verify with different configurations?

Here are 5 related questions for further exploration:

1. How does changing the number of girls \( n \) affect the prime number selection?
2. Can you find a different set of 4 primes with a smaller difference?
3. What happens if we relax the condition that no two classrooms can have the same number of students?
4. How can we distribute boys and girls evenly if the number of total students isn't perfectly divisible?
5. Can this problem be generalized to more than 4 classrooms?

Tip: When solving problems with prime constraints, balancing different factors like total sum and individual limits can often be done by testing ranges of prime numbers.

Explore how to minimize the difference between the highest and lowest classroom capacities when distributing 140 boys and n girls across 4 classrooms. Learn the strategic use of prime numbers and ratio balancing to achieve optimal classroom setups.

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