Euler Problem 29: Distinct Powers

(This article was first published on The Devil is in the Data, and kindly contributed to R-bloggers)

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 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
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