Is there an unbeatable Cross-Country score?

Spiralina wrote:

math glitch? wrote:

why have two people now thought that 22+23 = 55?

Well to be fair, I think one of them lives in Houston. Educational standards arent high in the South.

My kids went to public schools in suburban Dallas- and let me tell you brother- those big schools had a ton of academic talent, mostly strong teachers, and prepared the kids really well for college.

Other people say Texas isn’t “the South”. Idk- but Texas has some really strong school systems.

And yeah, 27 is the answer obviously.

i understand that, but the point is moot. in that specific iteration of 35 points the score is unbeatable. However there are numerous iterations of 35 that are beatable. At best this is simply a semi-interesting caveat to the question initially posed.

UTEP's 17 points in the 1981 NCAAs.

27 is the correct response by the way, anything lower automatically wins.

But had a meet recently where the last finisher of a race 35minute+ 5k impacted the team results and the team champion, how you ask? Well the slow runner, was a team's 5th runner and they had 2-3 individuals in the top 10. So this meant when he finished, the finishers went from not counting in the team score, to then displacing certain teams more than others and it changed the team scores by a few points changing the overall winner. A unique element of XC where even the last finisher can impact the scoring a meet. Everyone counts

Now that this has been solved, what is the best score that guarantees last place?

My guess is (Number of teams * 35) - 40

Or what is the worst score that can still win?

My guess is (Number of teams * 14) + 1

this is not an addition problem or something you can solve without knowing the overall numbers of a race.

This could be solved using a recursive function. Just remember that in recursion you get stacks for free.

math nerd wrote:

Now that this has been solved, what is the best score that guarantees last place?

My guess is (Number of teams * 35) - 40

Or what is the worst score that can still win?

My guess is (Number of teams * 14) + 1

This would be a more math intensive question. But lets go of off NCAA DI championships- 31 teams are listed on the results from 2017 so for the TEAM competition with individuals thrown out, thats 217 places, 155 of which are counted (top 5).

What is the worst possible score that could win? Without putting too much thought into it..................I'm drawing a blank, I have no idea how to do this and have to get back to work

Skeeple wrote:

math nerd wrote:

Now that this has been solved, what is the best score that guarantees last place?

My guess is (Number of teams * 35) - 40

Or what is the worst score that can still win?

My guess is (Number of teams * 14) + 1

This would be a more math intensive question. But lets go of off NCAA DI championships- 31 teams are listed on the results from 2017 so for the TEAM competition with individuals thrown out, thats 217 places, 155 of which are counted (top 5).

What is the worst possible score that could win? Without putting too much thought into it..................I'm drawing a blank, I have no idea how to do this and have to get back to work

Oh geez again... the worst score that CAN win and the best score that CANNOT win? This obviously depends on the number of teams... and I imagine, but am not positive, that displacements would matter... but holy cow that makes my head hurt.

math nerd wrote:

Now that this has been solved, what is the best score that guarantees last place?

My guess is (Number of teams * 35) - 40

Or what is the worst score that can still win?

My guess is (Number of teams * 14) + 1

...

2*35=70

70-40=30

Can't win a dual meet with 30 points. Ok.

2*14=28

28-1=27

27 wins any meet.

Alright, checks out for a dual meet, does it work for any meet? If so, wow, super impressive.

Skeeple wrote:

math nerd wrote:

Now that this has been solved, what is the best score that guarantees last place?

My guess is (Number of teams * 35) - 40

Or what is the worst score that can still win?

My guess is (Number of teams * 14) + 1

This would be a more math intensive question. But lets go of off NCAA DI championships- 31 teams are listed on the results from 2017 so for the TEAM competition with individuals thrown out, thats 217 places, 155 of which are counted (top 5).

What is the worst possible score that could win? Without putting too much thought into it..................I'm drawing a blank, I have no idea how to do this and have to get back to work

Ok, this is the new question. At NCAA's What is the highest possible score that could win,

tried something that I know is completely wrong but just to start getting a ballpark: 1+32+63+94+125 = 315.

Supertramp wrote:

math nerd wrote:

Now that this has been solved, what is the best score that guarantees last place?

My guess is (Number of teams * 35) - 40

Or what is the worst score that can still win?

My guess is (Number of teams * 14) + 1

...

2*35=70

70-40=30

Can't win a dual meet with 30 points. Ok.

2*14=28

28-1=27

27 wins any meet.

Alright, checks out for a dual meet, does it work for any meet? If so, wow, super impressive.

His formula was actually +1, not -1. For the most points but winning, I think the answer is generally floor(25n/2 + 5/2) - 1.

I get that with an average score per team of 25n/2+5/2 (5n/2(5n+1) total points in the meet). In this case (win with as many points as possible), teams will all be within 1 point of this probably non-integer number. Since some will be above and some below, you have to take the lower possibility (floor function), then subtract 1 to break any ties. This makes some assumptions I don't want to get into and actually goes too low in the dual meet case because by being below the average (27.5), the other team has to be above so the -1 is unnecessary for breaking ties. I think a real answer would have to include some number theory or something and get complicated, but I'd like to be proven wrong there.

For worst possible score that could win, my theory is that if a team went 1,2,3 and then the last 4 available places in the race (214, 215, 216, 217) the rest of the places could be spread out to the remaining teams by giving the next team the next 3 places available and last 4 places available.

Scoring top is: 1+2+3+(7* numberOfTeams - 2) +(7* numberOfTeams - 3)

reduces to: (Number of teams * 14) + 1

For the case of NCAA with 31 teams, this would give team scores ranging from 435 to 465

I was wrong though, because my theory doesn't have any teams with the same score.

For example, once you have 5 teams, it is possible to get scores of 72, 73, 73, 73 and 74. Which is 1 point higher than my prediction.

So for NCAA 435 is possible, but I'm thinking there could be a way to prove up to 449.

One team gets 449, next 29 all tie with 450 and then 1 teams loses with 451.

math nerd wrote:

Now that this has been solved, what is the best score that guarantees last place?

The worst case would be all members of the two worst scoring teams bringing up the rear. This makes the competition for last essentially a dual meet between the last two places.

The first finisher in this "dual meet" would score after the runner in place (#teams - 2) * (5+2) where so we can just add that value to each runner in the "dual meet". The (5+2) represents the 5 scorers and 2 pushers for all but the last two teams.

Thus a team scoring 5*(#teams -2) *7 + 28 would always be last.

Magic Number 20? wrote:

Magic Number 34 wrote:

How about this.

Team A goes 1-4 which totals 10 points.

Team B finishes with places 5-9, which totals 35 points. Then runners 6 and 7 come in at 10th and 11th.

Team A can still win as long as their #5 runner finishes no higher than 24th place. A 25th place finish would be 35 but lose since Team B’s 6th and 7th already finished breaking the tie.

So 34 is the magic number

Or is it 20?

Team A

1, 7,8,9,10=35

Team B

2,3,4,5,6= 20

You clearly do not understand the question. Did you take something magic today?

Citizen Runner wrote:

math nerd wrote:

Now that this has been solved, what is the best score that guarantees last place?

The worst case would be all members of the two worst scoring teams bringing up the rear. This makes the competition for last essentially a dual meet between the last two places.

The first finisher in this "dual meet" would score after the runner in place (#teams - 2) * (5+2) where so we can just add that value to each runner in the "dual meet". The (5+2) represents the 5 scorers and 2 pushers for all but the last two teams.

Thus a team scoring 5*(#teams -2) *7 + 28 would always be last.

I had the same logic except I added 30 instead of 28.

Then reduced: 5*(#teams -2) *7 +30 = 35*(#teams -2) + 30 = 35*#teams -70 +30 = 35*#teams -40

malmo wrote:

What the hell does 1-4 plus 24 have to do with anything? It's not relevant to the question being ssked.

Just listen for 2 minutes. If a team in XC finishes in the following positions.

1,2,3,4,24 to score 34

What positions would another team have to finish in to beat them?

The answer is no team can thus making 34 the highest unbeatable score

Magic Number 34 wrote:

malmo wrote:

Your first sentence answers the question. The rest has nothing to do with the question.

15 is perfect and can never be beaten

27 can’t be beat

34 is the highest a team can score without getting beat if winning team goes 1-4, 24 thus making this the correct answer because no other combo can score lower.

A team can score 28 and lose

A team can go 1-4, 25 and score 35 and win

A team can go 5-9 and score 35 and lose

Agree to disagree

Fun thread, OP.

Well, my team scored 34 by going 1,5,6,7,15, but another team went 2,3,4,8,16 and got 33. But, hey, there must be something wrong in their addition. I'm going to go to the officials and point them to this thread. We scored 34. We should be guaranteed the win!! Hooray for us.

numskull wrote:

Magic Number 34 wrote:

15 is perfect and can never be beaten

27 can’t be beat

34 is the highest a team can score without getting beat if winning team goes 1-4, 24 thus making this the correct answer because no other combo can score lower.

A team can score 28 and lose

A team can go 1-4, 25 and score 35 and win

A team can go 5-9 and score 35 and lose

Agree to disagree

Fun thread, OP.

Well, my team scored 34 by going 1,5,6,7,15, but another team went 2,3,4,8,16 and got 33. But, hey, there must be something wrong in their addition. I'm going to go to the officials and point them to this thread. We scored 34. We should be guaranteed the win!! Hooray for us.

I think the better way to phrase magic number 34 would be "if you were to go 1-4, you would need your 5th man to be 24th or higher to guarantee the win, because the next highest possible score would be 5-6-7-8-9 = 35 points." your 1-5-6-7-15 vs 2-3-4-8-16 scenario does not apply to that because nobody is going 1-4.

Retrace wrote:

malmo wrote:

What the hell does 1-4 plus 24 have to do with anything? It's not relevant to the question being ssked.

Just listen for 2 minutes. If a team in XC finishes in the following positions.

1,2,3,4,24 to score 34

What positions would another team have to finish in to beat them?

The answer is no team can thus making 34 the highest unbeatable score

You can't just change the question as you see fit. The question did you state that you get to add a condition to the race. If my team places x, y, z. Why not just add a second or third or more conditions. If we score 1,2,3,4, 26 and no other team has consecutive runners in the top 20. Well, then 36 guarantees a win. Heck, under those conditions 44 guarantees a win.

Lets add another condition, and I can get the score even higher;->