Brain Teaser

1

>this shit again
at least try something not for brainlets

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how is everyone getting 1

9

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1 and you call Vergens retard. Blame on you

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1.000000090000001

its 9 idiots

6

6?

1

8 dumbasses

The answer is 9.

9

7

Im 50% in TNB should i kill myself biz?

Doesn't matter because it's not an oracle problem

no

Make 5 races, pick the fastest horse in each group, and make a sixth race to establish the top three.
Am I hired?

9

7. Because if you devide 3 by a third you end up with 3÷1 which is 3. Pemdas. 9-3+1

9-3/1/3+1
Negative 3 divided by 1 is negative three.
9-3/3+1
Negative three divided by three is negative 1.
9-1+1
It's 9.

Wrong. What if two or three of the top three are in a single group?

4. Group 3 races of 8+8+9 horses, top 2 of each go in final race.

11

When the other 60% is not XVG
>YES

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Oh, up to 5. So yeah 6 right? Horse racing is subjective though as differnet horses have better races and worse races, so who knows you'd have to do some complicated probabilistic formula right?

PEMDAS you faggots - my fuck, it’s negative 1

wrong
can only race 5 at a time

12?

9

How is this achieved, by cross section?

How would you introduce the subjectiveness and other factors however, a probabilistic formula of some sorts? I've been to horse races and the fastest horses can have off runs, so you'd need to do more to account for that right?

This shouldn't surprise me but most of you have not been taught simple math properly. It's 1.

PEMDAS would mean the answer is 9, not -1, Kenneth.

Run off races for the top three of each race, til you have a final three.

assume the horses are mechanical and in working condition

Brainlets
9 – 3 ÷ (1/3) + 1
3 divided by a third is 9
9-9+1=1

this

Oops I fucked up. The answer is 1.

wrong

5

use your smartphone to time them

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shit it gets complicated into 1v1 round robin after 10, 19 would be enough

The problem is that the 2nd fastest and 3rd fastest horses could've been "eliminated" if they were in the same group as the fastest horse.

So after your 6 races (5 elimination rounds + 1 final round) to get the fastest horse:
you need to do another race of these 5 horses:
- 2nd fastest horse in the final race + 4 horses from the elimination round the top 1 horse came from
This will give you the 2nd fastest horse.

THEN, you have 2 possible outcomes:
a. the winner is from the horses that were in the "elimination round" with horse #1, or
b. the winner is from the final round

In case of (a), the final race you need to do is:
3 remaining horses from the "elimination round" horse #1 and #2 came from + 2nd fastest horse in the "final round" earlier that determined horse #1

In case of (b), it's possible that the 3rd fastest horse was eliminated in the elimination rounds by horse #1 or horse #2.
So you'll need 2 races more:
- 4 horses from the elimination round of horse #1 + the 3rd horse in the "final round" earlier
- 4 horses from the elimination round of horse #2 + the 3rd horse in the "final round" earlier

So you can do it in 8 or 9 races.

no

wrong

11.. You have to race 5 at a time, each time taking the top 3 and racing them against another pair

25 horses divides into 4 groups of 4 and 1 group of 5
race the group of 5 first, then take the winner and race it against a group of 4, take the winner and race it against the next group of 4, do this until all horses have raced
5 races total

the final 6 horses have to battle to the death. 22 races MUST be more than enough

US maths? Damn boy, its hard to sit next to some Niggers.

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not hired due to poor reading comprehension

Why the hell would I trouble myself with horse races?

it's asking for the minimum

Who needs math when you have digits.

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PEMDAS
parentheses - none
exponents - none
multiplication and division - expression becomes 9 - 9 + 1
addition and subtraction - expression becomes 1

1. 25/5=5
5x3=15
2.15/5=3
3x3=9
3.Group into 5 and 4
6 remaining
4.Race 4 and have the two remaining in a final race with the top three.
Rounds:5+3+ 2+2=12

Multiplication or division goes first, then addition and subtraction, moving from left to right. 3 divided by 1/3 = 3*3 =9, 9-9=0, +1 =1.

Some of us made it past basic 7th grade arithmetic.

6

My God.

nope

>t. does not work at Google
Lol. Yeah it is. You can't figure out the top three in less than 12 and mine would

Add before subtract faggots, that means -1

nope
solution:
youtube.com/watch?v=i-xqRDwpilM

6 races faggot

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13?

9-3=6
6 /1/3 --flip the denominator--> 6*3/1=18/1=18
18+1=19

wrong

Oh, I get it... I was including ALL the horses in each group.

So I guess after the 6 races to get the winner, you can do one more race:

- 2nd and 3rd horse from the elimination round horse #1 came from
- 2nd and 3rd horse from the final round
- 2nd horse from the elimination round that final round's horse #2 came from (the only horse that was possibly eliminated but could've been 3rd fastest)

>implying I come here to read and didnt just look at the picture
>not purposely getting the answer wrong so you dont have to work for jewgle in a $9k/month rent cuckshack

Lmao. Brainlets these days

Damn

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It's left to right. Type it in a calculator. 9-9+1=1.

How am I not rich yet trading against you brainlets?

6 races. 5 races for each group of 5, and then race 6 would for the 5 winners of each group. You would then be able to discover 1st, 2nd, and 3rd place horses.

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Because biz is for brainlets and semibrainlets.
Youre just one semibrainlet

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what if the 3 fastest are all in one race?

The fastest of a group weeds out the rest among the 5 groups. The remaining 5 fastest horses of each 5 groups compete for 3 titles while the other 2 go back to their stables. If I'm wrong then I'm sorry because I don't even deserve to be considered for the position.

this is fucking gay just race all 25 at once you cunts the first three to the finish are fastest

That's the problem, in the initial 5 "elimination rounds", once a horse loses, it's eliminated.

So if the 2nd fastest or the 3rd fastest (or both) are in an elimination round with the fastest horse, they don't get to compete in the "finals", they're eliminated early.

That will only get you the top three of winners from each race. Not the top 3 of the whole 25, you fucking faggot.

You are thinking like most championship brackets but that isn't always fair. Imagine if one of those first five races were stacked with elites and the other four races were chumps. The 2nd place of the elite race would smoke any of the "winners" of the other four but it doesn't get to race in your 6th race. Try again, chump

Race 5 sets of 5 and eliminate the bottom 2 of each race. This leaves 15. Do the same thing with 3 sets of 5 to leave 9. Race one set (a) of 5 and then a second (b) with the 3rd place getter and the remaining 4. Run the last race with the top 2 of a and the top 3 of b. 11 races total.

Numbers aren't real, only approximations.

Simple - the answer is: 19.000000000000000000000000000018

> 9 minus 3 = 6
> 1 divided by 3 = 0.33333333333333333333333333333333
> 6 divided by 0.33333333333333333333333333333333
> = 18.000000000000000000000000000018
> + 1
> = 19.000000000000000000000000000018

My method is correct except after the last part that comprises race 6. The answer is 7.

Here's a view after labeling of the last race (7). Makes sense when you remove the slowest of the winners.

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underrated

Is the american educational system so poor?

Split horses in 5 groups and have them race (5 races). Take top three horses from each group and make 5 new groups, take 1st horse from each and race them, repeat 3 times (3 races). The winner of each group is the fastest horse. Therefore, 8 races.

replies like this make my day

> My method is correct
Everyone and their mom figured out that method.
That's the entire point of the question, to get beyond that step, and realize that the 2nd and 3rd fastest could've been prematurely eliminated. Hence the 7th race.

3

race 5 groups of 5 - record rankings of 1st to 3rd of each group. (5 races)
Race the winners. Get a top 3 of the winners. Remember the groups of each winner - denoted 1f, 2f, 3f (1 race)
take 2nd (1f2) and 3rd (1f3) place from original group of winner and 2nd (2f2) place of the bracket from second finalist
race 1f2, 1f3, 2f1, 2f2, 3f1 (1 race)
The two winners there are second and third place overall.

Total 7 races.

This is faster than some strats posted above by avoiding repeat races with the 2nd and 3rd runners from dumpster brackets. but I think it should still guarantee an accurate top 3.

The way you put it is as if I'm patting myself on the back. Yes, I understand I was stating what was obvious to being with. My intention wasn't to sound pompous, but to highlight that I wasn't that far off.

You got a problem with that?

-1