Solution 3 (Modelling): Monkeys & Coconuts

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’ = \frac {1}{5} X

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

b= \frac{1}{5} .4a’ = \frac{4}{25}X

… Left 4b= \frac {16}{25}X
[b][b][b][b] => [c][c][c][c][c]
c=\frac {1}{5} .4b= \frac {16}{125}X
… Left 4c= \frac {64}{125}X
[c][c][c][c] => [d][d][d]d][d]
d= \frac {1}{5}.4c= \frac {64}{625}X
… Left 4d= \frac {256}{625}X
[d][d][d]d] => [e][e][e][e][e]
e= \frac {1}{5}.4d= 256.(X/3125)
Since e is integer
=> X = 3125 or multiples of 3125
Minimum X=3125
x+4 = 3125
x= 3121 = minimum Coconut initially.

Note: This solution used the Singapore Modelling Math taught in all Primary Schools for 11-year-old pupils.

2 thoughts on “Solution 3 (Modelling): Monkeys & Coconuts

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