Problem Description

The hyperexponentiation or tetration of a number a by a positive integer b, denoted by a↑↑b or ^ba, is recursively defined by:

a↑↑1 = a,
a↑↑(k+1) = a^(a↑↑k).

Thus we have e.g. 3↑↑2 = 3³ = 27, hence 3↑↑3 = 3²⁷ = 7625597484987 and 3↑↑4 is roughly 10^{3.6383346400240996*10^12}.

Find the last 8 digits of 1777↑↑1855.

Analysis

Using the little known optional third parameter of Python’s pow(x, y [,z]) function that provides x to the power y, modulo z (computed more efficiently than pow(x, y) % z) we were able to simply loop through the hyperexponentiation of a↑↑b.

This is similar to problem 97.

Solution

Runs < 1 second in Python.

def hyper_exp(a, b, m):
    temp = 1
    while b:
        temp = pow(a, temp, 10**m)
        b -= 1
    return temp
 
print "Answer to PE188 = ",hyper_exp(1777, 1855, 8)

Comments

Note: The last eight digits remain constant by the eighth iteration.
See problem 48 & problem 97