Euler Problem 29: Distinct Powers

By Peter Prevos

(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 

The post Euler Problem 29: Distinct Powers appeared first on The Devil is in the Data.

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