Euler Problem 29: Distinct Powers

Euler Problem 29 is another permutation problem that is quite easy to solve using brute force. The MathBlog site by Kristian Edlund has a nice solution using only pen and paper.

Raising number to a power can have interesting results. The video below explains why this pandigital formula approximates e to billions of decimals:

(1 + 9^{-4^{6 \times 7}})^{3^{2^{85}}} \approx e

Euler Problem 29 Definition

Consider all integer combinations of: a^b for 2 \leq a \leq 5 and \leq b \leq 5 .

2^2=4, \quad 2^3 = 8,\quad 2^4 = 16,\quad 2^5 = 32

3^2 = 9,\quad 3^3 = 27,\quad 3^4 = 81,\quad 3^5 = 243

4^2 = 16,\quad 4^3 = 64,\quad 4^4 = 256, \quad 4^5 = 1024

5^2 = 25,\quad 5^3 = 125,\quad 5^4 = 625,\quad 5^5 = 3125

If they are then placed in numerical order, with any repeats removed, we get the following sequence of 15 distinct terms:

4, \ 8, \ 9, \ 16, \ 25, \ 27, \ 32, \ 64, \ 81, \ 125, \ 243, \ 256,\ 625, \ 1024, \ 3125

How many distinct terms are in the sequence generated by a^b for 2 \leq a \leq 100 and 2 \leq b \leq 100 ?

Brute Force Solution

This code simply calculates all powers from 2^2 to 2^{1000} and determines the number of unique values. Since we are only interested in their uniqueness and not the precise value, there is no need to use Multiple Precision Arithmetic.

# Initialisation
target <- 100
terms <- vector()
i <- 1
# Loop through values of a and b and store powers in vector
for (a in 2:target) {
   for (b in 2:target) {
     terms[i] <- a^b
     i <- i + 1
   }
}
# Determine the number of distinct powers
answer <- length(unique(terms))
print(answer)

View the latest version of this code on GitHub.