Solution 3 (Modelling): Monkeys & Coconuts

Math Online Tom Circle

Let x the min number of coconuts initially.

1st monkey took “a” coconuts away, 2nd monkey “b” coconuts….5th monkey took “e” coconuts.

[a][a][a][a][a] + 1 =x

Loan 4 coconuts to the initial pool of x coconuts to divide by 5 evenly at each monkey.

[a][a][a][a][a] + 1 + 4 = [a][a][a][a][a] + 5 = x+4 = X (inflated x by 4 )
1st Monkey: [a’][a’][a’][a’][a’] = X
a’ = $Latex \frac {1}{5} X$

… Left 4a’= $Latex \frac {4}{5} X$
[a’][a’][a’][a’] => [b][b][b][b][b]

b= $latex \frac{1}{5} $ .4a’ = $latex \frac{4}{25}X$

… Left 4b= $Latex \frac {16}{25}X$
[b][b][b][b] => [c][c][c][c][c]
c=$Latex \frac {1}{5}$ .4b= $Latex \frac {16}{125}X$
… Left 4c= $Latex \frac {64}{125}X$
[c][c][c][c] => [d][d][d]d][d]
d= $Latex \frac {1}{5}$.4c= $Latex \frac {64}{625}X$
… Left 4d= $Latex \frac {256}{625}X$
[d][d][d]d] => [e][e][e][e][e]
e= $Latex \frac {1}{5}$.4d= 256.(X/3125)
Since e is integer
=> X = 3125 or multiples of…

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