Excel BI’s Excel Challenge #319 — solved in R

Defining the Puzzle:

We need to find which from given numbers are “more powerful” than others which means as follow:

List all Powerful numbers.
A powerful is number is that number which is perfectly divisible by square of all its Prime factors.
Ex. 225 — Prime factors are 3 and 5. In turn, 225 is perfectly divided by both 3*3 and 5*5.

Loading Data from Excel:

Lets start loading data and libraries:

library(tidyverse)
library(primes)
library(readxl)
library(data.table) 

input = read_excel(“Powerful Numbers.xlsx”, range = “A1:A10”)
test = read_excel(“Powerful Numbers.xlsx”, range = “B1:B6”)

Approach 1: Tidyverse with purrr

is_powerful = function(number) {
 vec_primes = prime_factors(number)
 count_vec = vec_primes %>% as.data.frame() %>% select(num = 1) %>% group_by(num) %>%
 summarise(a = n())
 check = all(count_vec$a > 1)
 return(check)
}

result = input %>%
 mutate(is_powerful = map(Numbers, is_powerful)) %>%
 filter(is_powerful == TRUE) %>%
 select(Numbers)

Approach 2: Data.table

We will use exactly the same so today, so only method of calling it changes. So today DT version is up.

setDT(input)
input[, is_powerful := sapply(Numbers, is_powerful)]
result <- input[is_powerful == TRUE, .(Numbers)]

Validating Our Solutions:

identical(result$Numbers, test$`Answer Expected`)
# [1] TRUE

identical(test$`Answer Expected`, result$Numbers)
# [1] TRUE

If you like my publications or have your own ways to solve those puzzles in R, Python or whatever tool you choose, let me know.

Powerful Numbers was originally published in Numbers around us on Medium, where people are continuing the conversation by highlighting and responding to this story.