Monkeys & Coconuts Problem

Solution 1 : iteration problem => Use sequence

$Latex U_{j} =\frac {4}{5} U_{j- 1} -1 $

(initial coconuts)

$Latex U_0 =k$

Let

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

$Latex U_1 =f(U_0)=f(k)= \frac{4}{5}(k+4)-4$

$Latex U_2 =f(U_1)=f(\frac{4}{5}(k+4)-4)= \frac{4}{5}((\frac{4}{5}(k+4)-4+4)-4$

$Latex U_2=(\frac{4}{5})^2 (k+4)-4$

$Latex U_3=(\frac{4}{5})^3 (k+4)-4$

$Latex U_4=(\frac{4}{5})^4 (k+4)-4$

$Latex U_5=(\frac{4}{5})^5 (k+4)-4$

Since

$Latex U_5$ is integer ,

$Latex 5^5 divides (k+4)$

k+4 ≡ 0 mod($Latex 5^5$)

k≡-4 mod($Latex 5^5$)

Minimum {k} = $Latex 5^5 -4$= **3121** [QED]

Note: *The solution was given by Paul Richard Halmos (March 3, 1916 – October 2, 2006)*