Excel BI - Excel Challenge 659

excel-challenges

excel-formulas

🔰 Find those 2 digits to 5 digits numbers whose cube results into numbers where frequency of appearing digits are same.

Published

March 24, 2026

Illustration for Excel BI - Excel Challenge 659

Challenge Description

🔰 Find those 2 digits to 5 digits numbers whose cube results into numbers where frequency of appearing digits are same.

Solutions

library(tidyverse)
library(readxl)

path = "Excel/659 Cube Frequency of All Digits Same.xlsx"
test  = read_excel(path, range = "A1:A81")

find_numbers <- function(min_n = 10, max_n = 99999) {
  nums <- min_n:max_n
  keep(nums, ~ length(unique(table(str_split(as.character(.x^3), "", simplify = TRUE)))) == 1) %>%
    enframe(name = NULL, value = "Answer Expected") 
}

result <- find_numbers()

all.equal(result, test)
#> [1] TRUE

Logic: Read the workbook ranges needed for the challenge; Parse the packed text or string structure.
Strengths: The code maps the workbook rule into a compact, reproducible pipeline.
Areas for Improvement: The solution assumes the workbook layout and selected ranges remain stable, so any structural change in the sheet would require small adjustments.
Gem: The elegant part is how little code is needed once the correct intermediate representation is chosen.

import pandas as pd

path = "659 Cube Frequency of All Digits Same.xlsx"
test = pd.read_excel(path, usecols="A", nrows=81)

def find_numbers(min_n=10, max_n=99999):
    return pd.DataFrame(
        [num for num in range(min_n, max_n + 1) 
         if len(set(str(num ** 3).count(digit) for digit in set(str(num ** 3)))) == 1], 
        columns=["Answer Expected"]
    )

result = find_numbers()

print(result.equals(test)) # True

The Python version keeps the algorithm explicit, which helps when the challenge depends on a greedy or iterative rule.

Difficulty Level

Easy / Medium

The business rule is clear, though the workbook still needs a few transformation steps to reach the expected output.