Solution 2: Use Linear Algebra Eigenvalue equation: A.X = λ.X

A =S(x)= $Latex \frac{4}{5}(x-1)$ where x = coconuts

S(x)=λx

Since each iteration of the transformation caused the coconut status ‘unchanged’, which means λ = 1 (see remark below)

$Latex \frac{4}{5}(x-1)=x$

We get

x = – 4

Also by recursive, after the fifth monkey: $Latex S^5 (x)$ = $Latex (\frac{4}{5})^5 (x-1)- (\frac{4}{5})^4-(\frac{4}{5})^3- (\frac{4}{5})^2- \frac{4}{5}$

$Latex S^5 (x)$ = $Latex (\frac{4}{5})^5 (x) – (\frac{4}{5})^5 – (\frac{4}{5})^4 – (\frac{4}{5})^3+(\frac{4}{5})^2 – \frac{4}{5}$

$Latex 5^5$ divides (x)

Minimum positive x= – 4 mod ($Latex 5^{5}$ )= $Latex 5^{5} – 4$= 3,121 [QED]

Note: The meaning of eigenvalue λ in linear transformation is the change by a scalar of λ factor (lengthening or shortening by λ) after the transformation. Here

λ = 1 because “before” and “after” (transformation A) is the SAME status (“divide coconuts by 5 and left 1”).