The Olympic rock climbing competition has 20 climbers. Each climber competes in 3 separate events, where they rank from 1st to 20th. The final score of a climber is the product of their rankings from the 3 events, where lower is better. For example, if a climber finishes 5th, 3rd and 10th then their final score will be $5∗3∗10=150$.

Now I have two questions. You can answer one or both:

  1. What is the most number of climbers that can tie on the same score?

  2. What is the longest sequence of consecutive scores possible? A consecutive sequence is one where each successive term increases by 1, such as $x, x+1, x+2, \ldots, x+N$.

  • $\begingroup$ Hmmm looks like people don't like this puzzle? The solutions are actually quite interesting. $\endgroup$ Aug 7, 2021 at 14:34

3 Answers 3


What is the most number of climbers that can tie on the same score?

12, with a score of 720:

Three climbers with score of 3 * 15 * 16
Three climbers with score of 4 * 9 * 20
Three climbers with score of 5 * 8 * 18
Three climbers with score of 6 * 10 * 12

Proof that this is maximal:

The product of all players' scores is $20!^3=2^{54}\cdot3^{24}\cdot5^{12}\cdot7^6\cdot11^3\cdot13^3\cdot17^3\cdot19^3$

Any score with a prime factor greater than five can occur at most six times.

Scores with a factor of five can occur at most 12 times, as in the solution shown.

Scores that only have prime factors less than five must be the product of three numbers in $\{1,2,3,4,6,8,9,12,16,18\}$ and as there are only ten of these, there cannot be more than ten climbers with such scores. (The limit in this case is actually 9 climbers with a score of 144)


Question 1

12 people is the maximum, which can only be reached with scores of 240 360 or 720.

The small prime factor counts of all the scores, i.e. 20!^3 are:
7: 6
5: 12
3: 24
2: a lot
So if we want to have more than 6 people we cannot use 7 (or higher) Variation in the prime factors is important, so we probably need a factor 5 and 2 factor 3s Taking this into account it makes sense to try 12 people This limit our option since all multiples of 3 and 5 are needed, and in addition the factors of 2 need to be a multiple of 12
numbers used: from each event 3,5,6,9,10,12,15,18,20,2^a,2^b,2^c
using the highest factors of 2 : 4,8,16 exclusively allows a tying score of 720
half that score may also be possible using 111444888, 112244888 or something similar as factors of 2

option 720
Distributing the multiples of 5 and some trying easily give the following options
A: 5-9-16,10-4-18,15-6-8,20-3-12
B: 5-8-18,10-6-12,15-3-16,20-4-9
Rotating A or B over the events gives 2 solutions
I did not find more solution

option 360
Distributing the multiples of 5 and some trying easily give the following options without 2s
A: 5-8-9,10-3-12,15-4-6,20-1-18
B: 5-6-12,10-4-9,15-3-8,20-1-18
Rotating A or B over the events gives 2 solutions
However, there are more solutions e.g.
C: 5-6-12,10-2-18,15-3-8,20-2-9 ,combined with 2 rotated B

If we do not use factor 5 , we cannot use 15 and thus have less than 24 factor 3s Thus we have only one factor 3 per person available if aiming for at least 12 persons, meaning each person scores at least 2 times a power of 2 There are only 15 of those, less than the 24 needed. This leaves us with only one other option-group 2^n*15:
Then: 12 spots are used for multiples of 5
9 additional spots are used for multiples of 3: 3, 6 and 12 since 9 and 18 cannot be used
This means all 15 powers of 2 must be used
Using all of {1,2,3,4,5,6,8,10,12,15,16,20} 3 times turns out to yield valid results:
option 240
A: 5-6-8, 10-2-12, 15-1-16, 20-3-4
B: 5-3-16, 10-4-6, 15-2-8, 20-1-12
C: 5-4-12, 10-3-8, 15-1-16, 20-2-6
Rotating A, B or C over the events gives 3 solutions

  • 1
    $\begingroup$ Nice work. After finding and posting the my solution (your 720-B), I wrote some code to find other solutions. There are a total of 8 solutions for 720. There are 13 solutions for 240. And there are 23 solutions for 360, including one in which all triplets are distinct. $\endgroup$ Aug 7, 2021 at 21:07
  • $\begingroup$ Well done! This is the correct answer for part 1. $\endgroup$ Aug 8, 2021 at 1:17
  • $\begingroup$ @DanielMathias Wow, thx, that is more than I expected. I did add my answer to show that there are more optimal solutions than one / than yours, but clearly I am far from complete. Even if mine had been complete, I would find your to the point answer better than my wall of text though. $\endgroup$
    – Retudin
    Aug 8, 2021 at 6:58

Possible answer to #1

I think the answer to the first question might be:

9 people.

If three people come 5th, 6th and 12th, three come 4th, 9th and 10th, and three come first, 18th and 20th, 9 people are tied at a score of 360.

This is in theory - After a lot of trial-and-error I can't come up with anything better.

Update: According to the OP it's possible to do better.

Proven answer to #2


There are a couple of candidates for ranges of numbers with length >4: - 1-22
- 24-28
- 32-36
- 48-52
The range 24-28 can be created using the following: - 24 - 6 * 4 * 1
- 25 - 5 * 5 * 1
- 26 - 13 * 2 * 1
- 27 - 3 * 3 * 3
- 28 - 2 * 2 * 7
And there are no possivle ranges larger than length 5 aside from within 1-22.
But note that we can only use 3 1s. This means we can only have at most one prime, and at most three semiprimes, one if we have a prime.
We can express numbers with >2 prime factors for free, as long as we have enough 2s.
So if we want a range of length 6, it must have either three semiprimes and three numbers with >2 prime factors, or a prime, a semiprime, and four numbers with >2 prime factors.
There are only four numbers with >2 prime factors within 1-22: 12, 16, 18, and 20.
Therefore, the second is impossible, as these are not within a range of 6.
16, 18 and 20 are within a range of 6, but any range containing them contains the two primes 17 and 19, which we can't have.
Therefore, the most consecutive scores you can get is 5.

99% sure this is correct, let me know if I've made a mistake with my proof here.

  • $\begingroup$ Please hide your answer. You can do better for #1. $\endgroup$ Aug 7, 2021 at 9:41
  • $\begingroup$ You got the correct answer for #2, but there are more solutions $\endgroup$ Aug 7, 2021 at 11:19

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.