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  3. Project Euler #30: Digit Nth powers
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Project Euler #30: Digit Nth powers

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  • Technophobe01
     + 1 comment

    My sense was to do the following.

    MIN = 2
MAX = 1000000


def compute(value):
    ans = sum(i for i in range(MIN, MAX)
              if i == power_digit_sum(value, i))
    return str(ans)


def power_digit_sum(value, n):
    return sum(int(c) ** value for c in str(n))


if __name__ == "__main__":
    value = int(input())
    print(compute(value))

  • kumar_mahendra
     + 1 comment

    My Approach - First find all possible numbers that a K-digit number could have for example I treated [1,2,3],[2,1,3],[1,3,2],[2,3,1],[3,1,2],[3,2,1] all as same and I find only one of them and not repeated numbers with same digits (using recursive appraoch) since what we are interested in only digits which constitute K-digit number . this way my algorithm is very fast and will also work if N (input) is even more than 6 .

  • leduckhai
     + 0 comments

    You can still use brute-force. Just do some experiments to determine lower and upper limit for i, then you will pass all cases.

    def Digit_Nth_Powers(n,lower,upper):
  sumList = 0

  for i in range(lower,upper):
    digits = list(map(int,str(i)))
    tempSum = 0

    for j in digits:
      tempSum += j**n
    
    if i == tempSum:
      sumList += i
        
  return sumList

if __name__ == '__main__':
  n = int(input())
  
  # Lower and upper limit is determined by experiments for faster code execution
  if n == 3:
    lower,upper = [1*(10**2), 1*(10**3)]
  elif n == 4:
    lower,upper = [1*(10**3), 1*(10**4)]  
  elif n == 5:
    lower,upper = [2*(10**3), 2*(10**5)]
  elif n == 6:
    lower,upper = [1*(10**5), 6*(10**5)]
    
  sumList = Digit_Nth_Powers(n,lower,upper)
  print(sumList)

  • George_Emmanuel
     + 0 comments
    N = int(input())
c = []
for i in range(2, 10000000):
    cnt = 0
    for j in str(i):
        cnt += int(j) ** N
    if cnt == i:
        c.append(cnt)
print(c)

    if you want to check for all numbers you can check the numbers in the list then you could simply check their sum for any N value. Just change the line "Print(c)" to print(sum(c)) Be sure to check this in your own ide as here it would give a TLE

  • priyanshu_706811
     + 0 comments

    Try this

    n=int(input())
print(sum([i for i in range(n**2,10**6) if i==sum(int(z)**n for z in str(i))]))
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