Game: Math skill

reply to post by rival


Hmmm, I just realized that my answer doesn't deduce whether the "odd ball"
is lighter or heavier...it only finds the "odd ball".

Be back in a second...

I hate throwing in the towel, but I cannot find an answer that
finds the odd ball and tells whether it is heavier or lighter.

Finding the odd ball is easy, but finding the lighter or heavier
(as well) stumps me.

No matter how many balls (per side) I start with it always boils
down to four balls left, one side heavier (or lighter) with only
one try remaining.

So the question could be stated simpler (unless I am missing
something)...

You have four balls on the scale, two on each side. The scale
is tilted to one side. Three of the balls weigh the same, and
one ball is either heavier or lighter. Find the odd ball and
tell if it is heavier or lighter in one try.

I have to assume I AM missing something because I can't
solve from this scenario in one try.

NullVoid
reply to post by ignorant_ape

 

Very nice indeed, took me 15 minutes.
Since the answer revealed, how about the algebraic way ?
Some people never able to solve it, probably they are using algebra/formula/whatever instead of basic arithmetic ?

6x + 3y + 0.10z = 100 = x + y + z

Yes, I think its associative algebra (whatever that is), see R modules in that Wikipedia.
Assuming we know z. Is it possible to get x and y ? I'm really no idea about this.
x + y + 70 = 6x + 3y + 70 = 100

Yes, if you know the value of one variable, like x=1, then you have 2 unknowns with 2 equations - can solve algebraically.
1+y+z=100 so y=99-z
6+3y+0.1z=100 so also y=(94-0.1z)/3
so 99-z=(94-0.1z)/3 and z=70 and then y=99-z=29

You will get different answers for each value of x you choose, however the answers may not be nice!

reply to post by rival


There is an answer and you can tell its lighter or heavier.
I did remember the first step, but kinda forgot next step, but, done it and solved it.
Took 3 trial and know the weight.

reply to post by saneguy


If x = something else , of course, it will break the 2nd equation.
Your algebra have to in step down a bit so others can follow what and where, it took me few minutes to see what is happening there.

So, its proven, we just need 1 var to be exposed to solve it. I did try exposing it with papamath and that other website, didnt work. I'm wondering if it can be solve without exposing any var ? because we do have a value there: 100 plus 3 other numerators. Possible ?

Okay, I think I've got it...there's no math here, just deduction


There are two possible solutions starting with three balls per side.

First solution

1. Start with three balls per side. If these are even, then...

2. Weigh one of the remaining balls against any of the control balls
on the scale. If they are even, then...

3. Weigh the remaining ball against the control ball
to find if it is heavier or lighter---solved.

----------

Second solution

Part A

1. Start with three balls per side. If these are odd, then...

2. (a) Remove the three balls from the heavier side and set aside.
(b) Take one of the three balls from the lighter side
and place it on the other side of the scale with one of the
two remaining control balls that were originally unweighed.

If these four balls are even then you can logically deduce that
one the three balls from the heavy side (of the original weigh) is
in fact "heavier"...so...

3. (a) Place any two of the three original heavy balls
on the scale. If one of those two balls is heavier, then that
ball is the "heavy" and "odd" ball.

(b) if the scale shows them even, then the remaining ball
is the "heavy" and "odd" ball.

-------------------------------


Second solution

Part B

1. Start with three balls per side. If these are odd, then...

2. (a) Remove the three balls from the heavier side and set aside.
(b) Take one of the three balls from the lighter side
and place it on the other side of the scale with one of the
two remaining control balls that were originally unweighed.

If these four balls are uneven you can now deduce that you
are searching for a "lighter" ball.

3. Clear the two balls from the heavier side of the scale and
weigh the two balls from the lighter side against each other to
find the "lighter" and "odd" ball.


....whew

That took alot of work. Sad thing is, it really isn't that complicated
an answer. But it sure seemed like it...

Anway, thx for the puzzle. It was good one

You wont be able to solve for three variable unless you have more than two equations to work with. You can find two variables for this equation 6x + 3y + 0.10z = 100 because x+y+z=100 which is a second equation. You wont be able to solve for a third variable without a third equation. If you are given one of the variable (example, they bought at least 3 6$ items), then you would substitute that number in to your x value and solve for your other two variables. So given one other variable you can solve for the other two, the answers are fluid there is multiple answers for each variable that satisfy the equation

If you had three you could use linear algebra to solve using gaussian elimination (also known as row echelon form) to get your xyz values.

If you need help with any math though check out khanacademy.org, its wonderful.

reply to post by rival


Yes, its correct, just by looking at "three", its entertaining and at same time push our gray cells to work a bit.
And its much easier than the x+ y+ z = 100 problem. The algebra problem, seems can only be solve by brute force.

I got few others but we can always find these question on the net, right.

For the prize, heres something for you to go experimenting, a picture that was taken on a very dark night. A 5 minute exposure picture reveal almost all, fun way to create lots of blurry UFOs.



Original picture

reply to post by NullVoid


So the top one i just a long exposue at night? Thats actually really cool.

reply to post by AzureSky


Yep, and in the original exposed picture, I can see faint stars streak.
I fail that, I thought its 6pm, never cross my mind its that dark