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The archery match math puzzle

The archery match math puzzle

An archery match was held between Sen, Moitra and Koul. The point score for the bull's eye of the target was 40 and for the rest five increasingly larger concentric rings 39, 24, 23, 17 and 16 respectively. The target shown as below.

archery target

Each of the three men shot six arrows. The total scores of Sen was 110, Moitra 100 and Koul 120. Every arrow scored and the bull's eye was hit only once.

Can you say what were the six hits of the three competitors? How many such combinations can you find?

Recommended time to solve: 30 minutes.

Solution to the archery match math puzzle

We'll work on two parameters balancing each as we go forward,

  • Grand total scores left, and
  • Number of arrow hits left.

At the start these are 330 and 18. Isn't it?

Bull's eye of value 40 having hit only once, you are left with grand total score left as 290 and number of arrow hits left 17.

These two numbers ring a bell in your mind as you recall $17\times{17}=289$, very near to the grand total score left.

This is the most important pattern you discover.

The main strategy of forming combinations of arrow hit scores for the three players must then be to include as many numbers of 16 and 17 as possible in the combinations. This can only keep the grand score left to the minimum of 290 from 17 hits.

The second important conclusion can also be made with confidence,

The point score of 39 cannot be considered. The corresponding ring area was not hit at all. The point score values to be considered are 16, 17, 23 and 24.

This is assessment, but an assessment made with 100% certainty.

Let us now jot down the score combinations of the three players in cases of each hitting the bull's eye.

For Sen hitting bull's eye of value 40: Rest scores are Sen 70, Moitra 100, Koul 120.

For Moitra hitting bull's eye of value 40: Rest scores are: Sen 110, Moitra 60, Koul 120.

For Koul hitting bull's eye of value 40: Rest scores are: Sen 110, Moitra 100, Koul 80.

This again is the second instance when the figure of 80 as 5 times 16 catches attention.

This is the second key pattern discovery.

At this point we can only say, $5\times{16}=80$ for Koul satisfies one of the three possible score combinations as well as satisfies the guiding principle of considering as many of 16 and 17 as possible. It is more or less a certainty that this way will be the solution.

Let us go ahead with this score of Koul as $5\times{16}=80$ and see whether it leads actually to the solution.

With this score of Koul, the two scores left are Sen 110 and Moitra 100 with arrow hits left as 12.

Can we compile 110 or 100 as a sum of 16s and 17s only?

With focus narrowed down, answer is immediate. Yes, $4\times{17}+2\times{16}=100$. Six arrow hits left with Sen's 110 to compile yet.

It is clear (again by an assessment, but a certain assessment) that with only 16s and 17s 110 cannot be compiled. The next larger point score of 23 has to be considered. And rightly so. Answer is immediate.

$2\times{23}+4\times{16}=110$. Six hits left are also consumed.

Final solution:

Sen: 4 hits of 16 and 2 hits of 23.

Moitra: 2 hits of 16 and 4 hits of 17.

Koul: 5 hits of 16 and 1 hit of 40.

Any other possible combination?

Now we can say with confidence that this is the only possible combination in which number of 16s and 17s are used in maximum numbers. This confidence hinges on the single point of $5\times{16}=80$ for Koul. This is the only feasible score of 80 among the three combinations that enables use of maximum numbers of 16s and 17s.

The step by step methodical approach ensures this uniqueness.

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