The Case of the Pilfered Coconuts

Story:

Leo, known for his stealthy coconut thieving, went one night with two of his friends to raid a coconut grove. After they had gathered their loot and were climbing down the tree, the estate owner's voice startled them, and a flashlight beam cut through the darkness. In a hurry, they hid a pile of coconuts, and all three fled.

The very next morning, one of Leo's friends came back to count the coconuts and tried to divide them fairly into three equal shares. However, after dividing them, there was one coconut left over. Leo's friend took one of the three shares plus the leftover coconut, and left.

Later, the second friend of Leo came to the spot. He counted the remaining coconuts and again divided them into three equal shares. Once more, there was one coconut left over. He also took one of the three shares plus the leftover coconut and went his way.

Finally, Leo returned. He wasn't aware that his friends had already taken some coconuts. He took the pile of coconuts that was left and divided them equally among himself and his two friends. This time, the division was perfect, with no coconuts left over.

The Question:

How many coconuts did Leo initially hide?

Solution to The Case of the Pilfered Coconuts

Let's work backward to solve this problem:

• Final Division (Leo's Return): When Leo returned, the remaining number of coconuts must have been perfectly divisible by 3 (for him and his two friends). Let's say the number of coconuts at this stage was 3z, where z is a whole number.

• Second Friend's Action: Before the second friend arrived, there were $3y+1$ coconuts (since dividing by 3 left a remainder of 1). After taking his share (y) and the remainder (1), there were 2y coconuts left. This 2y must be equal to the 3z coconuts Leo found. So, $2y=3z$. The smallest whole number values that satisfy this are $y=3$ and $z=2$. This means the second friend found $3\times 3+1=10$ coconuts, took $3+1=4$, leaving 6 ($3\times 2$).

• First Friend's Action: Before the first friend arrived, there were $3x+1$ coconuts. After taking his share (x) and the remainder (1), there were 2x coconuts left. This 2x must be equal to the 10 coconuts the second friend found. So, $2x=10$, which means $x=5$. Therefore, the first friend found $3\times 5+1=16$ coconuts, took $5+1=6$, leaving 10.

Therefore, Leo initially hid 16 coconuts.

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Answer: Leo hid 16 coconuts.