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Fun Creative Math Problems

ilyace

Member
friend 1: x/2 + 1/2
friend 2: (x/2 - 1/2)/2 + 1/2
friend 3: ((x/2 - 1/2)/2 - 1/2)/2 + 1/2

therefore ((x/2 - 1/2)/2 - 1/2)/2 = 1/2
=> (x/2 - 1/2)/2 - 1/2 = 1
=> (x/2 - 1/2)/2 = 3/2
=> x/2 - 1/2 = 3
=> x - 1 = 6
=> x = 7
hurrrrrrrrrrrr

awww shit robz in da house
 

slicer4ever

Coding random shit
ah, i see what my problem was, i took it 1 step too far, and came up with 15, when i needed to stop at 7, ok, i failed that one=-(
 

elite

Oldie moldie
Nice RoBz. +rep

Now for a brain smasher. ;)

4 - Four friends were in the house. One of them went out and bought 5 hats, 3 of the hats are red and the rest is white. The friend came back. He showed them the hats then placed blindfolds on each one (which means 3 of them). After that, he placed 1 red hat for each of his friends (himself is not counted). The man burnt the 2 remaining white hats away from the house.

- He opened the blindfold from the first man and asked him:"What is the color of the hat you are wearing". The man answered:"I have either red or white".
- He opened the blindfold from the second man and asked him:"What is the color of the hat you are wearing". The man answered:"I have either red or white".
- He opened the blindfold from the third man and asked him:"What is the color of the hat you are wearing". The man answered:"I absolutely have a red hat".

How did the third guy know that he is absolutely wearing a red hat?
 

Josey Wales

Evil Poptart
The third man said he had a red hat because the odds of him having a red hat over a white hat were greater than of a white hat - This is because the first two men think they might have white hats on, since there was only 2 white hats, and 2 men think they have the 2 hats then it would only make sense for the third man to pick a red hat based on the probability.

But if youre looking for some sort of scientific algebraic equation i dont have one
 

FrozenIpaq

Justin B / Supp. Editor
Enforcer Team
That's actually a common probability problem. Just work out a chart (3x7 chart I think) and the only remaining posibilities are a red hat. Although the problem generally states that the people are standing in a line or in a circle and can see the other people's hats clearly once theirs are off (unless they are in front of a person then they can't see what the person behind them has and must base their decision off what the person behind them said)
 

elite

Oldie moldie
The third man said he had a red hat because the odds of him having a red hat over a white hat were greater than of a white hat - This is because the first two men think they might have white hats on, since there was only 2 white hats, and 2 men think they have the 2 hats then it would only make sense for the third man to pick a red hat based on the probability.

But if youre looking for some sort of scientific algebraic equation i dont have one

If that was true, then he can't 100% know that he is wearing a red hat.
That's actually a common probability problem. Just work out a chart (3x7 chart I think) and the only remaining posibilities are a red hat. Although the problem generally states that the people are standing in a line or in a circle and can see the other people's hats clearly once theirs are off (unless they are in front of a person then they can't see what the person behind them has and must base their decision off what the person behind them said)
This and with Josey's answer make up the answer. Josey figured out the probability and FrozenIpaq figured out the use of sight. Good job! :)

I'll post my next riddle soon.
 

slicer4ever

Coding random shit
If that was true, then he can't 100% know that he is wearing a red hat.

This and with Josey's answer make up the answer. Josey figured out the probability and FrozenIpaq figured out the use of sight. Good job! :)

I'll post my next riddle soon.

if he can see the first two are wearing red hat's, wouldn't probability dictate a higher percentage that he's wearing a white hat then a red hat?
 

FrozenIpaq

Justin B / Supp. Editor
Enforcer Team
if he can see the first two are wearing red hat's, wouldn't probability dictate a higher percentage that he's wearing a white hat then a red hat?

Yes but he can't be certain what hat he is wearing. There's still a chance he is wearing a red hat
 

slicer4ever

Coding random shit
Make a chart of the possibilities and you will see how it is possible once you narrow the possible situations down

hmm, i guess this one must be above my understanding then, i'll ask you to point out where i'm wrong in how i'm understanding this:

first person's blindfold is removed, his chance's of having a red hat is: 60% Red/40% White
second person's blindfold is removed, after seeing the first person has a red hat, his chance of having a red hat is: 50% Red/50% White
Third person's blindfold is removed, after seeing the first+second person has a red hat, his chance of having a red hat is: 33% Red/66% White

i'm not exactly certain where a chart comes into play, isn't a chart, in this instance, just a way to visualize the possible outcome's?
also, i'm sorry if i'm dragging this out, i just don't understand how the person is absolutely certain he has a red hat.
 

Josey Wales

Evil Poptart
If you have someone blindfolded, put a hat on their head, and then unblind fold them - they can never know to a true 100% certainty what hat they have on.

The only solution is to use probability and deduction.
 

elite

Oldie moldie
Okay, next riddle! :)

2eevupw.png


You have 16 sticks arranged together to form 5 squares/quadrilaterals (actually any quadrilateral is compatible with the riddle) as shown in the above figure. You can only move 3 sticks and you should form a shape of 4 quadrilaterals. What can you do?
 

elite

Oldie moldie
Oh, well that's true. Now try to do it while maintaining the same geometric figure in all quadrilaterals (all squares).
 

MenaceInc

Staff Member
Oh, well that's true. Now try to do it while maintaining the same geometric figure in all quadrilaterals (all squares).

There was no mention of them having to stay as squares, just as quadrilaterals :p


You have 16 sticks arranged together to form 5 squares/quadrilaterals (actually any quadrilateral is compatible with the riddle) as shown in the above figure. You can only move 3 sticks and you should form a shape of 4 quadrilaterals. What can you do?



Just got home, I'll have another look at it to see about keeping squares in it.
 
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