math problem

[Sphinx]

Keep solving imaginary problems.

 

[Deleted member 13787]

This is it. Though to be fair you should be able to infer this

so if i get their order will i not be able to see the top 3?

 

[Deleted member 13787]

You can only race 5 horses at a time. What is the least number of races to get the top 3 fastest horses out of All 25?

This is not hard, idk what you’re not getting, Am I explaining this wrong or what?

you didnt say i can only race 5 horses at a time, and if its common knowledge i didnt know it

 

[Deleted member 13787]

Oh, youre Right, I forgot to include this in the problem. Do you have enough now?

yeah now i get the problem entirely, its just that without that info it seems like a pointless problem cause i thought i could race all 25 horses at the same time

 

[Sphinx]

Painful to watch @cloUder thinking he's got it but failing each time. also thinking that africa exist

 

[Sphinx]

Oh, youre Right, I forgot to include this in the problem. Do you have enough now?

You motherfucker complete your questions

 

[Deleted member 13787]

Painful to watch @cloUder thinking he's got it but failing each time. also thinking that africa exist

lol i didnt bother actually solving it, just used a calculator which gave me a wrong result
keep being a schizo

 

[Sphinx]

keep being a schizo

I will, doctor said it's for life.

 

[Sphinx]

so funny

Thanks Gunfire.

 

[Sphinx]

I’ll blow ur head off with a 50 cal nigga

So funny

 

[Sphinx]

Nah But I ain’t joking

Funny.

 

[Deleted member 13787]

Are you solving or what?

yeah im thinking of ways to solve it

 

[Deleted member 13787]

alright i figured out one way but i dont think its the best way
you take 5, get the top 3 and race them against another 2, take the top 3 and race them against another 2, keep doing that until youve raced all horses
but thats gonna take 11 races
i think there might be a better way
gimme some time and ill try to figure it out

 

[Harold O'brien]

@cloUder I have another quick problem for you.

You have 25 mechanical horses. You can’t record their exact times, only the order they finish the race. Each mechanical horse finishes the race in the same time in every single race they race in.

How many races would it require to find the 3 fastest horses from the group?

Bonus : Try generalising for all similar problem.

I don't get it

if they always finish in the same time and you know the order surely you know top 3 after one race

 

[Lorsss]

Wolframalpha couldn’t compute this number. I’ll have to get back to you on this one

no programming language on earth can compute a number with one thousand zeros

 

[Deleted member 13787]

I don't get it

if they always finish in the same time and you know the order surely you know top 3 after one race

read his previous comments
you can only let 5 horses race at a time

 

[Deleted member 13787]

no programming language on earth can compute a number with one thousand zeros

it works if you type 10^1334+1

 

[Copeful]

You can only race 5 horses at a time. What is the least number of races to get the top 3 fastest horses out of All 25?

This is not hard, idk what you’re not getting, Am I explaining this wrong or what?

easy af just do 5 races of 5 horses each, pick the ones that finish first out of those, then race them and pick the top three (total of 6 races)

 

[Deleted member 16288]

It is
Doesnt look like it'll be divisible by any except 1 and itself

 

[Lorsss]

it works if you type 10^1334+1

it is a floating point number, it only handles few digits

Floating-point arithmetic - Wikipedia

en.wikipedia.org

 

[Deleted member 13787]

easy af just do 5 races of 5 horses each, pick the ones that finish first out of those, then race them and pick the top three (total of 6 races)

no, that doesnt work...
what if the 2nd and 3rd fastest horse were matched against the fastest one in the first round?

 

[Deleted member 13787]

It is
Doesnt look like it'll be divisible by any except 1 and itself

its divisible by 101

 

[Deleted member 13787]

You’re there dude. I’ll ask you again, how many races do you need?

7

 

[Deleted member 13787]

Solved, describe them for me.

you take 5 diff groups of horses and you race them
the winners race against each other while all thr 4th and fifth places get eliminated
the winner of the 1st place match is known to be the fastest, the bottom 2 of that match get eliminated, along with the 2nd and 3rd place horses of the matchest in which these bottom 2 horses raced in the first round
now we need top 2 and 3 out of the remaining 8 horses. We can also eliminate the top 3 finishes of the 1st round matches except for the top 3 finish in the match of the fastest horse in the first round, as well as the top 2 finish in the first match which the top 3 winners horse won. Which leaves us at 5 horses left to determine which is top 2 and which is top 3, so we just race them in a 7th match and its done
i know my explanation was weird because i didnt provide anything visual but it works

 

[SAR]

its not a prime number
you can divide it by 11
View attachment 1538719

You can't divide by 11, because number of zeros is odd.

 

[Deleted member 13787]

So from race 1, 2nd and 3rd place
From race 2, 1st and 2nd place
From race 3, 1st place

Am I right?

yeah exactly, and its between race 1 2nd place and race 2 1st place for 2nd fastest, and between the other 3 for 3rd fastest

 

[SAR]

View attachment 1538732

Power should be 1334

 

[Deleted member 13787]

Power should be 1334

yeah i know dude we already discussed it

 

[SAR]

Power should be 1334

For example:
101=10^2 + 1
10001=10^4 + 1

 

[SAR]

is 100000000000000000 ... 01 where there are 1333 zeros between the leading and ending 1 a prime number?

why or why not?

@Gunfire
@cloUder

Basically you need to find a prime which satisfies this condition:
10^(1334mod(p-1))mod(p)=p-1
Where p is a prime, and Amod(B) means the reminder of A from dividing it to B.

 

[SAR]

Basically you need to find a prime which satisfies this condition:
10^(1334mod(p-1))mod(p)=p-1
Where p is a prime, and Amod(B) means the reminder of A from dividing it to B.

I checked for small primes, but they are not divisible

 

[SAR]

is 100000000000000000 ... 01 where there are 1333 zeros between the leading and ending 1 a prime number?

why or why not?

@Gunfire
@cloUder

This number is divisible by 101. Check it