1
$\begingroup$

The problem is as follows:

A group of archaeologists found an artifact to which they believe tells an account of the results of a race in an ancient Hippodrome. After days of studying the document they were able to decode the message which is summarized below.

The race was comprised of six horses each one wearing an insignia with a number on it. The judges declare that there was no tie at the end and this was reassured by the spectators. We also noted:

I. The order of arrival does not match with the numbers used by the horses.

II. The sum of the numbers of those who went in the last two places is 7.

III. The horse which wears the badge with the number 3 on it arrived on the second place.

IV. The horse which arrived in third place has an odd number written in its identification badge.

Find the sum of the numbers on horses badges which arrived in 4th and 5th places.

The alternatives given in my book are:

  • 7
  • 6
  • 5
  • 9
  • 8

In my attempt to solve this puzzle and going in circles for several minutes I concluded the following using a grid as I will explain in the following lines.

The first clue that it was clear to me is III.

This was a tricky thing since the number on the badges can be confused with the number that states the order of arrival from the horses so to avoid that confusion I put a circle outside the number to represent the badges used by the horses. Since it was mentioned that the one which arrived in second place wears number 3 then this cancels the other choices as seen below:

First grid

Now for the second clue that I used was IV which states that the horse who went 3rd had an odd number insignia so this meant that it cannot be neither 2, 4 or 6. This is seen below.

Second grid

Now that was kind of easy, here is where it all became into guessing. And for this part I used the second clue which states that the sum of the horses who went in the last two places is 7. From those numbers the only choices that can make 7 are 3+4, 1+6 or 2+5. 3 and 4 can't be as it is known 3 is in second place and by extension four cannot be in the last two places. Seen below:

Third grid

So the remaining choices can be 1 and 6 or 2 and 5 for the places of the remaining horses. To add up with the first clue which meant that no horse arrived with the same number written on its badge I tried with 1 being the 6th place so that 6 being the one which ended 5th place. This is shown below:

Fourth grid

By this time, all cascaded into plugging checks with remaining choices so that they do not share the same number:

Fifth grid

So finally that's what I've found and since what it is being asked is the sum of the numbers on the badges from the horses which arrived in 4th and 5th place, from the table can be concluded:

$$2+6=8$$

And this appears within the alternatives. To which I hope that is correct. But this whole process took me a while to complete and needless to say time that is kind of limited in an exam. Moreover what If i had started filling the table using 2 and 5?. It doesn't seem that it may had got me tangled with trial and error.

So, Is there a better method or alternative that can be used to speed up this process?

$\endgroup$
2
$\begingroup$

Second grid

Now that was kind of easy, here is where it all became into guessing.

...

To add up with the first clue which meant that no horse arrived with the same number written on its badge ...

It was not at all obvious to me that that's what the first clue means, but assuming that you've interpreted it correctly we don't need to guess yet. We cross out the diagonal:

    1 2 3 4 5 6

(1) x x . . . .
(2) . x x . . .
(3) x o x x x x
(4) . x x x . .
(5) . x . . x .
(6) . x x . . x

Applying the same reasoning that prevents (4) from being in position 5 or 6 because that would require (3) to be in the other position, we have

    1 2 3 4 5 6

(1) x x . . . .
(2) . x x . . .
(3) x o x x x x
(4) . x x x x x
(5) . x . . x .
(6) . x x . . x

and then (4) is forced to position 1, ruling out all other possibilities for position 1:

    1 2 3 4 5 6

(1) x x . . . .
(2) x x x . . .
(3) x o x x x x
(4) o x x x x x
(5) x x . . x .
(6) x x x . . x

At this point (6) is definitely one of the numbers in the sum. If (6) is in 5th place then (1) is in 6th place and (2) can only go in 4th place, giving result 8. Alternatively (6) is in 4th place, meaning that (2) and (5) take 5th and 6th places, so (5) is in 6th place and (2) in 5th place, again giving result 8. QED


Alternatively, from the last table, we apply clue II to say that (2) cannot be in 6th place since that would put (5) in 5th place: then we have

    1 2 3 4 5 6

(1) x x . . . .
(2) x x x . . x
(3) x o x x x x
(4) o x x x x x
(5) x x . . x .
(6) x x x . . x

at which point it's obvious that since (2) and (6) are both restricted to 4th and 5th place1, the sum of 4th and 5th place is 8.

This is about as much as we can say for certain: if we update the table to show this last point we get

    1 2 3 4 5 6

(1) x x . x x .
(2) x x x . . x
(3) x o x x x x
(4) o x x x x x
(5) x x . x x .
(6) x x x . . x

but each of the remaining .s can be o in one valid solution and x in another:

    1 2 3 4 5 6             1 2 3 4 5 6

(1) x x o x x x            (1)  x x x x x o
(2) x x x x o x            (2)  x x x o x x
(3) x o x x x x            (3)  x o x x x x
(4) o x x x x x            (4)  o x x x x x
(5) x x x x x o            (5)  x x o x x x
(6) x x x o x x            (6)  x x x x o x

1 In sudoku circles this is called a naked pair.

$\endgroup$
  • $\begingroup$ Interesting observation. It is nice to verify that all my guesses were in the right road but I'm still confused at what is the logic of crossing the diagonal?. What argument did I overlooked or not known to support this procedure?. I must say that doing this helps a lot. The following two tables that you used reassure what I deduced. The last paragraph where you mention a naked pair, I don't get it. I'm not fond of sudoku. What is that term and how does it applies in this problem? Can you develop this concept a little bit more?. Btw Can you complete the final table with the remaining circles?. $\endgroup$ – Chris Steinbeck Bell Sep 4 '18 at 19:12
  • $\begingroup$ "no horse arrived with the same number written on its badge ..." means that the diagonal can be crossed out. $\endgroup$ – Peter Taylor Sep 4 '18 at 19:14
  • $\begingroup$ "Naked pair": since there are two horses down to the same two places, no other horse can take those places because that would leave one of the original two without a place. $\endgroup$ – Peter Taylor Sep 4 '18 at 19:15
  • $\begingroup$ And I can't complete the table, because there are two possibilities. $\endgroup$ – Peter Taylor Sep 4 '18 at 19:16
  • $\begingroup$ Sorry about the late reply. Yes I noticed that we can make a diagonal from that statement. Initially I overlooked at, but late in that evening I realized that. Thanks for the explanation about "naked pair". Perhaps could you recommend a website where I can read more about it with some examples?. That would help me a lot!. I don't understand very well what do you mean about two possibilities, can you tell which of them are they?. Thanks in advance!. $\endgroup$ – Chris Steinbeck Bell Sep 10 '18 at 15:12

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.