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Her eyes lit up as the pirates Sparrow, Barbossa and Turner opened the treasure chest and saw the mountain gold coins.

After pirate's tradition, Sparrow first took a coin, then Barbossa two, Turner three, then Sparrow four, Barbossa five, and so on.

So were the gold coins split without rest.

In the end, Barbossa had 20 more coins than Turner, and Sparrow got ...

(A) 117 coins.
(B) 126 coins.
(C) 145 coins.
(D) 187 coins.
(E) 210 coins.

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  • $\begingroup$ Is it to be assumed that there was no remainder and that all pirates always took the correct amount and that there was no left over? If no, then 210 is also a valid answer. $\endgroup$ – LeppyR64 Jul 8 at 13:46
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    $\begingroup$ Yes there was no remainder and all pirates took the right amount. $\endgroup$ – Matti Jul 8 at 13:53
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We see that

Barbossa gets one fewer coin than Turner on each complete turn. This means that, since he was 20 coins up on Turner at the end of the distribution, the final turn was incomplete, and Barbossa had one more turn than Turner. This means that the distribution must have finished with Barbossa.

We then determine

Which Barbossa turn the game ended on. Note that we will add Barbossa’s final turn coins and subtract the number of complete turns there have been, since Turner would have gotten one more each time:
2: Barbossa 2 - Turns 0 = 2
5: Barbossa 5 - Turns 1 = 4
8: Barbossa 8 - Turns 2 = 6
(See a pattern? Looks like 9 turns...) ...
29: Barbossa 29 - Turns 9 = 20

So that means that

It would have gone S1, B2, T3, ... , S28, B29 and then the game would have been over.

So Sparrow got

1 + 4 + 7 + 10 + 13 + 16 + 19 + 22 + 25 + 28 coins, which is 5 x 29 = 145 coins.

The answer is therefore

C: 145.

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I would like to throw in some math this might make it easier to comprehend for some people.

Lets introduce some notation:

  • Agents $a_i,$ in our case $i \in \{1,..,K\}$, where $K$ is the number of agents.
  • $t^n_i = i + K (n - 1)$ is the amount of money $i-$th agent takes during $n-$th turn.
  • $S^n_i = \sum_{k=1}^n t^k_i = ni + K\frac{n(n-1)}{2}$ is the sum an agent has after $n-$th take.
  • $\Delta^n_{i,j} = S^n_j - S^n_i = n(j-i)$.

Here $n \in \{0\cup\mathbf{N}\}$.

So, for the task in hand with $K = 3$ we have

\begin{align*}S^{n+1}_2 - S^n_3 &= 20\\t^{n+1}_2 + S^n_2 - S^n_3 &= 20\\t^{n+1}_2 + \Delta^n_{3,2} &= 20\\2 + 3 n - n &= 20\\n &= 9\end{align*} $S^{n+1}_1 = S^{10}_1 = 10\cdot 1 + 3 \frac{10\cdot 9}{2} = \mathbf{145} $

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