Python - obtaining the sum of digits of cube numbers

Question

I want to write a code in python to print The sum of the digits of all whole cube numbers in a range that received from the user. for example if he chose [1,100], 1,8,9,10 are displayed that are related to (1,8,27,64).

Answer 1

Generally, it would be good practice to show what you've tried to solve the problem...and, if you allow me to give a personal opinion, you'll learn much faster if you at least try something.

EDIT: thanks to @Python learner, I misinterpreted the question. This should go:

def main():
    low, up = 1, 100


    # list of all cubes in selected range
    cubes = []
    # go through numbers in selected range, check if they are cubes, append them to list if they are
    for i in range(low, up):
         if is_cube(i):
             cubes.append(i)

    # alternative way, using list comprehension
    #cubes = [n for n in range(low, up) if is_cube(n)]

    # once cubes have been found, go through list
    for cube in cubes:
        # convert each cube to a str, so you can iterate over the single digits, convert them back to int, and append them to a list
        cube_str = str(cube)
        digits_list = []
        for digit in cube_str:
            digits_list.append(int(digit))
        print(f'{cube} -> {sum(digits_list)}')


def is_cube(n):
    n = abs(n)  # if n is negative
    return round(n ** (1 / 3)) ** 3 == n


if __name__ == '__main__':
    main()

Answer 2

Use a generator and list comprehension:

import numpy as np
from math import ceil
from typing import Generator


def cubes(start: int, stop: int) -> Generator[int]:
    for i in range(start, stop+1):
        cbrti = np.cbrt(i)
        if cbrti == ceil(cbrti):
            yield i


def sum_digits(n: int) -> int:
    s = 0
    while n:
        s += n % 10
        n //= 10
    return s


result = [sum_digits(i) for i in cubes(1, 100)]
print(result)

Output:

[1, 8, 9, 10]

Answer 3

Instead of checking one by one that each number in the range is a cube or not, Which is quite a waste of time, you can rather modify the bounds to know all the numbers such as raising them to the power of 3 will create numbers in the range.

The advantage of this method is that you don't have to compute the cube root of each number, after changing the bounds, you know directly which numbers are valid.

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Explanation

• If we calculate the cubic root of the start and the end of the range, the new range will contain all numbers with a cube between start and end.
• If start and end are not perfect cubes, we take either the integer above if we are at the beginning, or the integer below if we are at the end.

Example: the range is [1,100], By calculating the cube root, we get [1, 4]. Thus 1, 2, 3 and 4 cubed are in the range.

Note: This is only possible because f(x) = x^3 is strictly increasing.

---

import math

start = 1
stop = 100

cubes = []
for i in range(math.ceil(start**(1/3)), math.floor(stop**(1/3)) + 1):
    cubes.append(i**3)
print(cubes)  # [1, 8, 27, 64]

• math.ceil() is used to round upward to the nearest integer
• math.floor() is used to round a downward to the nearest integer.
• **(1/3) is used to compute the cubic root of the number (raised to the power 1/3).

---

Once we have calculated all the cubes, we have to sum them up. There are many ways to do this, either by converting the number to a string and then looking at each character, or by successively dividing the number by 10.

We can for example use a recursive function, which adds each remainder of the division by 10 recursively, until the quotient is zero.

import math

def sum_digits(num):
    q, r = divmod(num, 10)
    if not q:
        return r
    return sum_digits(q) + r

start = 1
stop = 100

cubes = []
for i in range(math.ceil(start**(1/3)), math.floor(stop**(1/3)) + 1):
    cubes.append(sum_digits(i**3))
print(cubes)  # [1, 8, 9, 10]

---

Note:

If the bounds can be negative:

• The cube root must be calculated in a different way (using numpy or multiplying by -1 the cube if it is negative).
• The absolute value of the cube must be used to calculate the sum of the digits.