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  1. All Contests
  2. ProjectEuler+
  3. Project Euler #21: Amicable numbers
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Project Euler #21: Amicable numbers

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65 Discussions

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  • hamblinglen
     + 0 comments

    The amicable numbers are a well-studied sequecence, and a list of several of them can be found here. Technically, you could just use this list to find your result, but I encourage you to use them only for checking and testing your code. Otherwise, what's the point?

  • mansityagi472
     + 1 comment
    def sumof(n):
    some=0
    for i in range(1,n):
        if n%i==0:
            some+=i
    return some
        


def is_amicable(a,b):
    return  (sumof(a)==b) and ((sumof(b))==a) and a!=b
        




t = int(input())
for _ in range(t):
    n = int(input())
    you=0
    for i in range(1,n+1):
        for j in range(i+1,n+1):
            if (is_amicable(i,j)):
                you+=(i+j)
    print(you)

  • thummaganti_rav1
     + 0 comments

    // C# using System; using System.Collections.Generic;

    class Program { static int S(int n) { int sum = 1;

        for (int x = 2; x <= Math.Sqrt(n); x++)
    {
        if (n % x == 0)
        {
            sum += x + n / x;
        }
    }

    return sum;
}

static void Main()
{
    int numTestCases = int.Parse(Console.ReadLine());
    List<int> testCases = new List<int>();

    for (int i = 0; i < numTestCases; i++)
    {
        testCases.Add(int.Parse(Console.ReadLine()));
    }

    HashSet<int> result = new HashSet<int>();

    for (int x = 0; x <= testCases.Max(); x++)
    {
        if (x == S(S(x)) && x != S(x))
        {
            result.Add(x);
        }
    }

    foreach (int testCase in testCases)
    {
        Console.WriteLine(result.Where(x => x <= testCase).Sum());
    }
}

    }

  • blackspring
     + 0 comments
    #python
def s(n):
    s=1
    for x in range(2, int(n**.5)+1):
        if n %x==0:
            s+=x+n//x
    return s
n=[int(input()) for x in range(int(input()))]
res=set()
for x in range(max(n)+1):
    if x==s(s(x)) and x!=s(x):
        res.add(x)
for _ in n:
    print(sum(x for x in res if x<=_))

  • vidush_rk11
     + 0 comments

    easy sol

    def divi_sum(n):
    r_sum=0
    for i in range(1,int(n**0.5)+1):
        if n%i==0:
            r_sum+=i+n//i
    if int(n**0.5)==n**0.5:
        r_sum+=-int(n**0.5)
    r_sum+=-n
    return r_sum

req_numbers={}
our_sum=0
for i in range(1,10**5):
    req_numbers[i]=our_sum
    x=divi_sum(i)
    if x!=i and i==divi_sum(x):
        our_sum+=i
        
for _ in range(int(input())):
    no=int(input())
    print(req_numbers[no])