MATHS riddle

[J()e]

Another riddle: (btechbah taba3 tawa)

We have 3 bottles the 1st one contains 10L,the 2nd 7L and the 3rd 3L

we have 2 persons....and we should distribute 5L to each one

what we have to do?


Sent Your answers via pm!

[abousoun]

Originally Posted by J()e

It seems enno Ya2asetkon my riddle so here it is
Btw abousoun 7alla 90% menna:P

good one man, i just needed to organize the equation in my head a little bit ...

I'll post a one but it's very difficult ... and the solution isn't in my head right now ...
you have 12 balls same size and shape, one of them has different weight (more or less) ... you have a scale, and you are only allowed to use it 3 times ... figure out the abnormal ball and if it's more or less than the other 11 ones ... remember you have to cover all possible probability with 3 scales only
Hint : enumerate the balls from 1 to 12 to make it easy 4 you
P.S: for this riddle don't PM me ... directly post your answer here

in the meanwhile if any has other riddles ... post it plz


Thank You ...

[enigma]

Originally Posted by abousoun

good one man, i just needed to organize the equation in my head a little bit ...

I'll post a one but it's very difficult ... and the solution isn't in my head right now ...
you have 12 balls same size and shape, one of them has different weight (more or less) ... you have a scale, and you are only allowed to use it 3 times ... figure out the abnormal ball and if it's more or less than the other 11 ones ... remember you have to cover all possible probability with 3 scales only
Hint : enumerate the balls from 1 to 12 to make it easy 4 you
P.S: for this riddle don't PM me ... directly post your answer here

in the meanwhile if any has other riddles ... post it plz


Thank You ...

we put 4 by 4 if they are equal then the diiferent is from the other 4 ones
so we put 2 bye 2 w byerja3 1 bye 1
halla2 2eza mn lawwal ma tol3o equal mn5oud li tabash fiyoun lscale

[Tawa]

Abousoun's One:

First Of All, If The Ball Is Less Weight:
We Put 1,2,3 And 4,5,6 On The Scale. // We Used 1 Time The Scale So Far.
If(They're Equal Weight)
{
// The Ball Is Definitely Between The 7,8,9,10,11 And 12.
We Put 7,8,9 And 10,11,12 On The Scale. // 2 Times The Scale.
If(7,8,9 Weights Less Than 10,11,12)
{
// The Ball Is Between 7,8,9.
We Put 7 And 8 On The Scale. // 3 Times The Scale.
If(They Weight The Same)
The Ball 9 Is The One We're Looking For
Else
The Ball That Weights Less Is The One We're Looking For.
}
Else
{
// The Ball Is Between 10,11,12.
We Put 10 And 11 On The Scale. // 3 Times The Scale.
If(They Weight The Same)
The Ball 12 Is The One We're Looking For
Else
The Ball That Weights Less Is The One We're Looking For.
}
}
Else
{
// The Ball Is Definitely Between The 1,2,3,4,5 And 6.
We Put 1,2,3 And 4,5,6 On The Scale. // 2 Times The Scale.
If(1,2,3 Weights Less Than 4,5,6)
{
// The Ball Is Between 1,2,3.
We Put 1 And 2 On The Scale. // 3 Times The Scale.
If(They Weight The Same)
The Ball 3 Is The One We're Looking For
Else
The Ball That Weights Less Is The One We're Looking For.
}
Else
{
// The Ball Is Between 4,5,6.
We Put 4 And 5 On The Scale. // 3 Times The Scale.
If(They Weight The Same)
The Ball 6 Is The One We're Looking For
Else
The Ball That Weights Less Is The One We're Looking For.
}
}

LOOOL! I Turned It Into A C++ Porgram!
Anyways, If The Ball Weights More, Same Steps, But ... You Know What lol

[J()e]

first min3abbe el 10 liters ....
ba3dein minfadde el 10 bil 3 liters
ba3dein min fadde el 7 elli be2yo bil 7 liters
ba3dein minfadde el 3 liters bil 10 liters
ba3dein minfadde el 7 liters bil 3 liters
sar 3enna 3 (bil 10) 4 (bil 7) w 3 (bil 3)
minfadde el 3 liters bil 10 liters ta yssiro 6 liters
w elli bil 7 liters bottle (4 L) bil 3 liters
heik bissiro 6 (bil 10) 1 bil (7) 3 bil (3)
mnerja3 minzid el 3 liters lal 6 liters bissiro 9 --- > 9 1 0
mnerja3 minzid el 3 liters lal 6 liters bissiro 9 --- > 9 1 0
el 9 liters minfaddihon bl 7 liters bissiro --- > 2 7 1
w elli bil 7 liters minfaddihon bil 3 (already containg 1 )
bissiro 2 liters (bil 10), 5 liters (bil 7) w 3 liters (bil 3)
minzid el 3 3ala el 2
bissiro 5 liters (in 10) and 5 liters (in 7) and 0 liter (in 3)

abousoun's way so that makes him a winner

and enigma good also

[abousoun]

Originally Posted by enigma

we put 3 by 3 if they are equal then the diiferent is from the other 3 ones
so we put 1 bye one of them and we find the different one
halla2 2eza mn lawwal ma tol3o equal mn5oud li tabash fiyoun lscale w mna3moul w me7ott we7de bi we7de

sorry wrong you didn't cover all the probabilities ...

Thank You ...

[enigma]

Originally Posted by abousoun

sorry wrong you didn't cover all the probabilities ...

Thank You ...

it works if you tell me the different one at2al aw a5af

[abousoun]

Originally Posted by Tawa Juniór

Anyways, If The Ball Weights More, Same Steps, But ... You Know What lol

Your thinking is mazbout 100 % and logical only if we consider that the abnormal ball is less or more in weight from the beggining ... you can't take the probability of it's less first and then the probability it's more a second time ... coz you'll be consumming the scale more than 3 times

Though i liked your response and the way of presentation

Thank You ...

[abousoun]

Originally Posted by enigma

it works if you tell me the different one at2al aw a5af

yeah i'm sure it will works ... but here is the trick

Thank You ...

[J()e]

Ok.This is hard and it's Pure maths

A man lost his security code for his pc !! and this code is composed using 5 digits

the 1st number + the fith number=14

the 4th number is bigger then the 2nd number bi 1 number(el ra2em el rabi3 akbar men el ra2em el tene bi ra2em wa7ad)

The 1st number is less than the double of the 2nd bi 1 number(el ra2em el awal 2a2al men do3ef el ra2em el tene bi ra2em wa7ad)

the 2nd # +The 3rd#=10

The addition of all numbers makes=30

what is the code??