Problem:
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 103.6383346400240996*10^12.
Find the last 8 digits of 1777↑↑1855.
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 103.6383346400240996*10^12.
Find the last 8 digits of 1777↑↑1855.
Solution:
932718654
Code:
The solution may include methods that will be found here: Library.java .
public interface EulerSolution{
public String run();
}
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