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Fun Creative Math Problems

elite

elite

Oldie moldie
Josey Wales said:
But you still have no accurate way of telling how long each side of the fuse will burn so regardless of how many sides you use, theres no solution.

You cant lit a fuse from both ends and expect the laws of percentage to take over. It doesnt work like that, one side of the fuse may burn for 40 minutes without moving, the other side might move an inch and burn for 30 minutes in doing so - the problem is not the fuse, but when you light both ends you are trigger two unknown side specific timers that cant be tracked... thus... back to square 1
Click to expand...
If that was true Josey, then I would have said that all of the fuse takes more than 1 hour and 10 minutes. You have valid info which is 1 burn = 1 hour. 2 burns = 0.5 hour. No matter where do the flames meet, but it should take only 1/2 an hour for the fuse to fully burn.
 
Josey Wales

Josey Wales

Evil Poptart
elite said:
If that was true Josey, then I would have said that all of the fuse takes more than 1 hour and 10 minutes. You have valid info which is 1 burn = 1 hour. 2 burns = 0.5 hour. No matter where do the flames meet, but it should take only 1/2 an hour for the fuse to fully burn.
Click to expand...

Ok I see it now, the second fuse will burn for 30 minutes, then when you light the other end that will burn for 15 and the other end 15 as well...

Good one.
 
elite

elite

Oldie moldie
Okay guys. Any solutions for the second one? I am gonna post the solution if none answers.
 
Josey Wales

Josey Wales

Evil Poptart
Yea, you lift them all up and the one that weighs twice as much as the others is the one. Thats pretty simple, you cant fault the way mass feels
 
elite

elite

Oldie moldie
You are practically doing it more than once. You can only use the balance once. I know that here you used your hands, but they work as a balance.
Hint: You can use the screws inside the packs.
 
slicer4ever

slicer4ever

Coding random shit
elite said:
You are practically doing it more than once. You can only use the balance once. I know that here you used your hands, but they work as a balance.
Hint: You can use the screws inside the packs.
Click to expand...

considering that you posted two day's ago, i doubt anyone at this point is ganna step forward to give a correct answer(although i may be wrong), anywho, my solution would be to monitor the output as you put one screw from each pack onto the scale until u get an increase of two gram jump, however this would go against the "you only get one use" rule, so i know it's wrong, and i can't think of any other solution which only requires the one use rule
 
MenaceInc

MenaceInc

Staff Member
We'll either all groan and kick ourselves for missing such a simple solution or call out the solution for being bullshit.
 
psp318player

psp318player

SMOOTH HACKER
For the screw question, just use the 2g on the scale and get a beaker for the 1g. Drop the 1g into the beaker filled with water and record how long it takes for it to drop. The time it takes for it to drop ??
 
elite

elite

Oldie moldie
The ANSWER:

- Place a color marker on every pack. Lift 1 screw from each pack and color every screw with the color matching the source pack. Lift all the screws at once and place them on the digital balance. Remove 1 screw at a time, if the balance status decreases by 1 g, then it's a 1 g screw. If it decreases by 2 g, then you have found the 2 g screw which corresponds to the source pack of 2g screws.

I did use my hand, but not as a scale. However, I used the digital balance only once which is placing all the screws at once.

Third riddle! This is gonna be somewhat easy to release the tension from the second riddle. :p

3 - You bought a box that consists of x numbers of apples. You gave your first friend half of the apples and a half of an apple. Next, you gave your second friend half of the remaining apples and a half of an apple. Then, you gave your third friend half of the remaining apples and half of an apple. At this point, you have no apples. Calculate x. Who solves this algebraically will get +REP. :)
 
Abe Froeman

Abe Froeman

Gamer Dad
Enforcer Team
Your answer involves more than one weigh. The second you pull off one screw you have re-weighed the lot. Grab another screw, another re-weigh.
 
MenaceInc

MenaceInc

Staff Member
elite said:
The ANSWER:

- Place a color marker on every pack. Lift 1 screw from each pack and color every screw with the color matching the source pack. Lift all the screws at once and place them on the digital balance. Remove 1 screw at a time, if the balance status decreases by 1 g, then it's a 1 g screw. If it decreases by 2 g, then you have found the 2 g screw which corresponds to the source pack of 2g screws.

I did use my hand, but not as a scale. However, I used the digital balance only once which is placing all the screws at once.

Third riddle! This is gonna be somewhat easy to release the tension from the second riddle. :p

3 - You bought a box that consists of x numbers of apples. You gave your first friend half of the apples and a half of an apple. Next, you gave your second friend half of the remaining apples and a half of an apple. Then, you gave your third friend half of the remaining apples and half of an apple. At this point, you have no apples. Calculate x. Who solves this algebraically will get +REP. :)
Click to expand...

x = 11

(I swear I was sleep deprived when I thought it was 7 :\


Also, as Abe pointed out, the weighing solution is bollocks.
 
slicer4ever

slicer4ever

Coding random shit
elite said:
- Place a color marker on every pack. Lift 1 screw from each pack and color every screw with the color matching the source pack. Lift all the screws at once and place them on the digital balance. Remove 1 screw at a time, if the balance status decreases by 1 g, then it's a 1 g screw. If it decreases by 2 g, then you have found the 2 g screw which corresponds to the source pack of 2g screws.
Click to expand...

1. i was right!(my method doesn't require marking the screws though)=-)
2. i shouldn't have been right=-(

elite said:
3 - You bought a box that consists of x numbers of apples. You gave your first friend half of the apples and a half of an apple. Next, you gave your second friend half of the remaining apples and a half of an apple. Then, you gave your third friend half of the remaining apples and half of an apple. At this point, you have no apples. Calculate x. Who solves this algebraically will get +REP.
Click to expand...
3.
Friend 3 must have 1 apple(U=x/2-.5 = 0=x/2-.5 = .5 = x/2 = 1=x)
so, Friend 2 must have 3 apples(F3=x/2-.5 = 1=x/2-.5 = 1.5=x/2 = 3=x)
so, Friend 1 must have 7 apples(F2=x/2-.5 = 3=x/2-.5 = 3.5=x/2 = 7=x)
so, you must have 15 apples(F1=x/2-.5 = 7=x/2-.5 = 7.5=x/2 = 15 = x)

so in total, you had 15 apples to start with=-)
 
elite

elite

Oldie moldie
Josey Wales, reread my answer and you'll see:
I did use my hand, but not as a scale.
Click to expand...

Abe, I said that you can only use the balance once. Well tbh I should've really detailed what this means. One balance use = Placing the item/s on the balance and lifting it/them such that no item is currently on the balance. It was really hard to translate that riddle, so excuse me. :/

Anyways, I am really disappointed at your Math calculations guys. Come on slicer, 15? Oh really? xD

Let's say we do have 15:
1st friend: 15 - 15/2 -1/2 = 7
2nd friend: 7 - 7/2 - 1/2 = 3
3rd friend 3 - 3/2 -1/ = 1
You still have one apple. You failed. Calculations are way more complicated.

MenaceInc, let's say we have 11:
1st friend = 11 - 11/2 -1/2 = 5
2nd friend = 5 - 5/2 -1/2 = 2
3rd friend = 2 - 2/2 - 1/2 = 1/2
You also failed. However, your first guess was right (7). Actual number is 7. I just need to see the right algebraic expression.
 
FrozenIpaq

FrozenIpaq

Justin B / Supp. Editor
Enforcer Team
"Using the balance once" generally can be taken as only being able to take one measurement. You took several measurements...using it several times...
 
MenaceInc

MenaceInc

Staff Member
elite said:
Josey Wales, reread my answer and you'll see:


Abe, I said that you can only use the balance once. Well tbh I should've really detailed what this means. One balance use = Placing the item/s on the balance and lifting it/them such that no item is currently on the balance. It was really hard to translate that riddle, so excuse me. :/

Anyways, I am really disappointed at your Math calculations guys. Come on slicer, 15? Oh really? xD

Let's say we do have 15:
1st friend: 15 - 15/2 -1/2 = 7
2nd friend: 7 - 7/2 - 1/2 = 3
3rd friend 3 - 3/2 -1/ = 1
You still have one apple. You failed. Calculations are way more complicated.

MenaceInc, let's say we have 11:
1st friend = 11 - 11/2 -1/2 = 5
2nd friend = 5 - 5/2 -1/2 = 2
3rd friend = 2 - 2/2 - 1/2 = 1/2
You also failed. However, your first guess was right (7). Actual number is 7. I just need to see the right algebraic expression.
Click to expand...

I know where I got 11 from, I added the 1,3 and 7 from each step.
PROTIP, never do math with no sleep.
Also, I fell asleep in the middle of an exam today. Thankfully I had already blasted through and completed it xD
 
R

RoBz

sucker
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
elite said:
At this point, you have no apples.
Click to expand...
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
 
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