
    Design an algorithm that determines the optimal difficulty level for a class assignment by evaluating whether keeping the current difficulty intact (n) or breaking it down into sub-components (n/2, n/3, n/4, n/5) yields better student performance. The algorithm should recursively analyze these subdivisions to find the maximum achievable performance score, considering that each subdivision represents a different approach to presenting the same material. The optimal difficulty is determined by the formula f(n) = max((f(n/2) + f(n/3) + f(n/4) + f(n/5)), n), where n is the initial difficulty level.
    

    
def get_max_sum(n):
    """
    Design an algorithm that determines the optimal difficulty level for a class assignment by evaluating whether keeping the current difficulty intact (n) or breaking it down into sub-components (n/2, n/3, n/4, n/5) yields better student performance. The algorithm should recursively analyze these subdivisions to find the maximum achievable performance score, considering that each subdivision represents a different approach to presenting the same material. The optimal difficulty is determined by the formula f(n) = max((f(n/2) + f(n/3) + f(n/4) + f(n/5)), n), where n is the initial difficulty level.
    

    """
def get_max_sum(n):
    """
    Design an algorithm that determines the optimal difficulty level for a class assignment by evaluating whether keeping the current difficulty intact (n) or breaking it down into sub-components (n/2, n/3, n/4, n/5) yields better student performance. The algorithm should recursively analyze these subdivisions to find the maximum achievable performance score, considering that each subdivision represents a different approach to presenting the same material. The optimal difficulty is determined by the formula f(n) = max((f(n/2) + f(n/3) + f(n/4) + f(n/5)), n), where n is the initial difficulty level.
    

    """
    res = list()
    res.append(0)
    res.append(1)
    i = 2
    while i < n + 1:
        res.append(max(i, res[int(i / 2)] + res[int(i / 3)] + res[int(i / 4)] + res[int(i / 5)]))
        i = i + 1
    return res[n]