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I found this puzzle in my Logic and Reason book from 2000's. The topic is ordering information. From the looks of it seems to be an adaptation from a reprinted version of Martin Gardner's 70's book of Puzzle Carnival.

The puzzle is as follows:

Marina, Sakura, and Hina were finalists at an Idol Athletics competition. They take part in the final, which has three trials: archery, rhythmic gymnastics and a 100-m sprint. In each test, the one who ends up first gets $a$ points, the second gets $b$ points and the third gets $c$ points. We know that $a$, $b$ and $c$ are positive integers such as $a>b>c$ there are no draws. In total Marina got 20 points, Sakura 10 points and Hina 9 points. We know that Marina ended up second place in the rhythmic gymnastics trial. Who ended up third in the archery trial and second place in 100-m sprint trial?

The choices given by my book are as follows:

  1. Hina and Marina
  2. Sakura and Hina
  3. Marina and Sakura
  4. Hina

I'm confused on how to arrange this information in a logical manner. My approach was to make a table. So far I have this table:

Sport Marina Sakura Hina
Archery x y z
Rhythmic gymnastics v u w
100-m sprint d e f

Where do I go from here?

It doesn't say that $a$, $b$, and $c$ must be contiguous, but in order to have them to add up for 20, Marina's first-place score, she must have ended up either third or first for either Archery or 100-m sprint. The same for the other two finalists, Hina and Sakura. How can this information be arranged more simply?

I attempted to break down the numbers to get 20, 10 and 9 and these are:

20 = 1+19, 2+18, 3+17, 4+16, 5+15, 6+14, 7+13, 8+12, 9+11, 10+10

But this didn't help much. How can this puzzle be solved? Is there a trick or a method of simplification?

Should any sort of equation be used? Please include a diagram or sketch explaining how to approach this situation. Placing these people in order is very confusing for me, I don't get what logic should be used.

The puzzle doesn't specify the order the trials were conducted in. Would that affect the method of solution or does it not matter?

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  • $\begingroup$ Please stop using MathJax to format lists - there are much better ways, such as simply writing "1.", "2.", etc. Also don't use MathJax to format non-equation things such as "100-m sprint". Finally, your titles keep having poor grammar - you can see my rephrasing edits for examples of better titles. I'll keep editing if this continues, but it would be nice to not have to. $\endgroup$ – bobble Oct 27 '20 at 22:21
  • $\begingroup$ @bobble I'm so sorry for the poor grammar on how this question was stated. I did it in a rush. I'll improve this in future questions. Regarding the use of Mathjax. I got used to it and I felt that it displays better lists. But if there's a rule in this community for the discouragement to use them outside of the necessity of mathematics then I'll reduce its usage. Regards. $\endgroup$ – Chris Steinbeck Bell Oct 28 '20 at 2:55
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Because the total score for all three contestants over three events

is equal to $39$, there are $3$ events, and each event awards $a$ points for first, $b$ points for second, and $c$ points for third, $3a + 3b + 3c = 39$ is the total number of points in the event. Dividing this by the number of disciplines gives $a+b+c = 13$.

The highest possible $b$ is

$5$, because if $b=6$, then $a \geq 7$ and $a+b+c > 13$ because $a, b, c$ are all $> 0$.

Since Marina placed second in rhythmic gymnastics,

we know she must have gained at least 15 points from the other two disciplines; $15 \div 2 > 5$, so she must have placed first in both other disciplines, having scored $2a + b = 20$ points. This further constrains $b$; it cannot be 3 or 5, because $2a$ cannot be odd due to all three values being integers. Thus, $b=2\lor b=4$.

If $b=2$, then $a=9$ (because $2a+b=20$) and $c=1$ (because $b>c>0$), and this leads to a contradiction as $2+9+1 \neq 13$. Therefore, $b = 4$, $a = 8$, and $c = 1$.

With these values, there is only one way to get Sakura and Hina's scores of 10 and 9 respectively;

Sakura must have placed first in rhythmic gymnastics (the one discipline Marina didn't place first in) and third in both other disciplines, while Hina must have placed second in archery and 100m sprint and third in rhythmic gymnastics.

The answer to the question is B: Sakura placed third in archery, and Hina placed second in the 100m sprint.

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  • $\begingroup$ I'm trying very hard to understand the way how you solve it, but I can't because I'm stuck at the very beginning. I got the part that the sum of all three contestants is 39, but why did you divide by $3$ this quantity and why did you come to the conclusion that such number is equal to $a+b+c$?. The second part, seems logical, as the highest number must be $5$, as you mentioned if $6$ is the highest would be equal and higher than $7$ hence making the equation $a+b+c>13$. $\endgroup$ – Chris Steinbeck Bell Oct 28 '20 at 3:43
  • $\begingroup$ But Again why $a+b+c=13$ is the sum of all points?. Splitting 15 between the other two disciplines makes sense. The rest of the constraining $b$ makes sense as well. The last part also makes sense as $c=1$ is the only logical conclusion if b=2 or 4 are used. $\endgroup$ – Chris Steinbeck Bell Oct 28 '20 at 3:44
  • $\begingroup$ Finally from this I can understand, but as a domino pieces falling down. This all lands in I'm not getting why $a+b+c=13$? Why should I assume that the total score of the three divided by three accounts for the sum of each points given by the performance in each discipline?. Is it because there's a condition which says draws aren't allowed?. I'm not getting that idea. Can you help me with that please?. $\endgroup$ – Chris Steinbeck Bell Oct 28 '20 at 3:45
  • $\begingroup$ The total number of points all contestants have is 39. Since "In each test, the one who ends up first gets $a$ points, the second gets $b$ points and the third gets $c$ points." and "there are no draws", the total number of points over the entire event is $3a + 3b + 3c = 39$, and dividing by 3 will show that one test's total point value is $a+b+c=13$. $\endgroup$ – Braegh Oct 28 '20 at 10:52
  • $\begingroup$ Okay I got the part of why the total number of points over the entire event is $3a+3b+3c$ which in words would meant someone scored 1st in all events, other 2nd in all events and the last one 3rd in all events. But, why is this quantity equal to 39?. We cannot say for sure that this sort of combination happened to the contestants in the mentioned compeition. Or is it that it doesn't matter in the end?. Will it yield the same?. In order to clear out this, I made a table and figured out that by the way how it is mentioned that aren't draws it will fill in all the table. $\endgroup$ – Chris Steinbeck Bell Oct 28 '20 at 23:43
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If we add up all the scores from the three rounds, we get 20 + 10 + 9 = 39.

We divide 39 by 3 to get the sum of a, b and c: a + b + c = 13.

With this, we can find that the maximum possible value for a is 10, and the minimum possible value for a is 3.

@Braegh has the best answer, but for those who like brute force strategies, here is my python code to find the answer:

from itertools import permutations

for a in range(3, 11): # For every possible value for a
    for b in range(2, a): # For every possible value for b
        v = b # We already know that v = b
        for c in range(1, b): # For every possible for c
            if a + b + c <= 13: # If a + b + c is less or equal to 13
                for p in permutations([a, b, c, a, c, a, b, c]): # For every permutation for 3 as, 2 bs and 3 cs
                    x, y, z, u, w, d, e, f = p # Set the values to the corresponding variables
                    if x + v + d == 20 and y + u + e == 10 and z + w + f == 9: # If the conditions meet
                        archery_winners = {x: 'Maria', y: 'Sakura', z: 'Hina'} # Dictionary to determine who won which place in the archery trial
                        sprint_winners = {d: 'Maria', e: 'Sakura', f: 'Hina'} # Dictionary to determine who won which place in the sprint trial
                        print(f'{archery_winners[c]} won thrid place for the archery trial')
                        print(f'{sprint_winners[b]} won second place for the sprint trial')
                        break

Output:

Sakura won third place for the archery trial
Hina won second place for the sprint trial

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  • $\begingroup$ The logical solution, presented in the other answer, is much nicer and is actual puzzle-solving. You just brute-forced with a computer. That isn't interesting or in the spirit of the puzzle. $\endgroup$ – bobble Oct 27 '20 at 23:38
  • $\begingroup$ @bobble I know, I'm just presenting a brute force method :) $\endgroup$ – risky mysteries Oct 28 '20 at 0:12
  • $\begingroup$ @riskymysteries Actually I think it is interesting to see that a computer can also handle this sort of question, but as bobble mentioned, the purpose was to solve it without any machine aid. If you don't mind, can you add a flowchart for your solution?. Its easier to understand the logic of the program this way rather than reading python lines (to which I'm unfamiliar with), perhaps could you add a LibertyBasic equivalent (to which I'm better familiar with).? $\endgroup$ – Chris Steinbeck Bell Oct 28 '20 at 3:02

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