Math Problems

[mithrand1r]

Originally posted by Radagast The Brown
27 days.

Btw, I still didn't succeed completing the 6x6 magic square.

Here is one soln. to 6x6 magic square:


'''''''''''''''''''''''''111
001 004 013 030 031 032| 111
035 034 008 023 009 002| 111
018 015 017 026 016 019| 111
027 014 028 003 029 010| 111
025 011 021 022 020 012| 111
005 033 024 007 006 036| 111
========================
111 111 111 111 111 111''111


Interesting feature (at least to me) of this 6x6 magic square is;
If you take the center of the square (4x4) and add the rows, columns and diagonals you get the folllowing:


'''''''''''''74
34 08 23 09 |74
15 17 26 16 |74
14 28 03 29 |74
11 21 22 20 |74
============
74 74 74 74''74


and if you broke this 4x4 block into four 2x2 blocks:

34+08+15+17 = 74
23+09+26+16 = 74
14+28+11+21 = 74
03+29+22+20 = 74

[mithrand1r]

Problem with Coins

There are 12 coins. One of them is false; it weights differently. It is not known, if the false coin is heavier or lighter than the right coins.

How can you find the false coin by three weighs on a simple scale (the scale is similar to a see saw in appearence. It does not return a numerical value for the weight of objects) ?

[mithrand1r]

There are 2 boxes with labels 'A' and 'B'. On each box is also a sign.

The sign on box A reads: 'The sign on box B is true and the gold is in box A'

The sign on box B reads: 'The sign on box A is false and the gold is in box A'

In exactly one of theses two boxes is some gold. In which one?

[Radagast The Brown]

Solution to the coins:

It took me a while, (help from my sister), and is very complicated especially at 12:20 AM. but:

First weight:
4 _____ 4

If = then
2 (from the 8) ______ 2 (untouched)

If = then
1 (untouched) ______ 1 (from the 8)

If = then it's the other untouched, if not it's the untouched that was weighted in the third weight.

If not equal then
1 (from the right side of weight 2) ______ 1 (from the 8)

The same as the one before.

If not equal (back to the first weight... now this is comlicated)
Let's call 4 'heavy', 4 'light', and 4 'regular' (even if not all the 4 are heavy/light)

then:
1 heavy, 1 light and 1 regular ______ 2 heavy and 1 lgiht

if = then we weight the two lights we didn't touch in the 2nd weight.

if the left is heavier, we weight the heavy in the left side with regular one.

if the right is heavier, we weight the two 'heavy's from the right side.

That's all. Not as bad as I thought.

[Count Comfect]

The gold must be in box B.
If the statement on A (hereafter called A) is true, then both (B true) and (gold is in box A)
Then since B is true, (A is false) - paradox
So A is false.
If B is true then (A is false) and (gold is in box A)
However, we then have a paradox since A becomes true (gold
was in A, and B was true) but A was false.
So B is false
The only way for both A and B to be false is for the gold to be in box B - then there is no paradox since neither statement is fully true.

[Radagast The Brown]

But if the gold is in box B, both of the statements are wrong - and it means that the statement on box A is true. And from this that the statement on box B is also true. And there is another paradox.

[mithrand1r]

Originally posted by Radagast The Brown
Solution to the coins:

It took me a while, (help from my sister), and is very complicated especially at 12:20 AM. but:

First weight:
4 _____ 4

If = then
2 (from the 8) ______ 2 (untouched)

If = then
1 (untouched) ______ 1 (from the 8)

If = then it's the other untouched, if not it's the untouched that was weighted in the third weight.

If not equal then
1 (from the right side of weight 2) ______ 1 (from the 8)

The same as the one before.

If not equal (back to the first weight... now this is comlicated)
Let's call 4 'heavy', 4 'light', and 4 'regular' (even if not all the 4 are heavy/light)

then:
1 heavy, 1 light and 1 regular ______ 2 heavy and 1 lgiht

if = then we weight the two lights we didn't touch in the 2nd weight.

if the left is heavier, we weight the heavy in the left side with regular one.

if the right is heavier, we weight the two 'heavy's from the right side.

That's all. Not as bad as I thought.

Good job.

It takes a bit of thinking, but if you write everything out as you did the problem can be worked out.

[mithrand1r]

Originally posted by Count Comfect
The gold must be in box B.
If the statement on A (hereafter called A) is true, then both (B true) and (gold is in box A)
Then since B is true, (A is false) - paradox
So A is false.
If B is true then (A is false) and (gold is in box A)
However, we then have a paradox since A becomes true (gold
was in A, and B was true) but A was false.
So B is false
The only way for both A and B to be false is for the gold to be in box B - then there is no paradox since neither statement is fully true.

The gold is in Box A!

The puzzle never said that the labels on the Boxes are either true or false.

They have nothing to do with the puzzle whatsoever.

You can consider it he labels to be a red herring.

[Count Comfect]

mithrand1r - in that case, the gold could just as well be in Box B, because if we ignore the labels then no information is given save that there is gold in one box. So no conclusion could be drawn.
Radagast - if the gold is in box B, statement A is not true - it is still false because the gold is not in Box A, and statement A was that B was true and the gold was in A.

[mithrand1r]

There are 4 men who want to cross a bridge. They all begin on the same side. You have 17 minutes to get all of them across to the other side. It is night and it is pitch black out (i.e. no moon light available). There is one flashlight. A maximum of two people can cross at one time. Any party who crosses, either 1 or 2 people, must have the flashlight with them. The flashlight must be walked back and forth, it cannot be thrown, etc. Each man walks at a different speed. A pair must walk together at the rate of the slower man's pace.

* Man 1: 1 minute to cross
* Man 2: 2 minutes to cross
* Man 3: 5 minutes to cross
* Man 4: 10 minutes to cross

For example if Man 1 and Man 4 walk across first, 10 Minutes have elapsed when they get to the other side of the bridge. If Man 4 returns with the flashlight, a total of 20 minutes have passed, and you have failed the mission.

Can you get them all across?

[mithrand1r]

Originally posted by Count Comfect
mithrand1r - in that case, the gold could just as well be in Box B, because if we ignore the labels then no information is given save that there is gold in one box. So no conclusion could be drawn.
Radagast - if the gold is in box B, statement A is not true - it is still false because the gold is not in Box A, and statement A was that B was true and the gold was in A.

True.

However, since I wrote this puzzle, I can decide which box to put the gold in so there.

(I probably should have used a puzzle that has more basis in mathematics than semantic word play.)

post 400

[Janny]

Originally posted by mithrand1r
Can you get them all across?

I presume one can, as you ask. But I can't. Um... okay, at some point man 4 must cross, so that's 10 mins. It makes sense if man 1 goes and comes back to reduce the time of one walking back.
However way you do it there will be 10 + 5 + 2 mins going and 3 * 1 mins coming back... 20 mins?

There must be another way tho... wait!
If man 3 goes with man 4 you save 4 mins (5 - 1)... which you would then lose when he comes back...
No, I reckon it's 20 mins at least.

[mithrand1r]

Originally posted by Janny
I presume one can, as you ask. But I can't. Um... okay, at some point man 4 must cross, so that's 10 mins. It makes sense if man 1 goes and comes back to reduce the time of one walking back.
However way you do it there will be 10 + 5 + 2 mins going and 3 * 1 mins coming back... 20 mins?

There must be another way tho... wait!
If man 3 goes with man 4 you save 4 mins (5 - 1)... which you would then lose when he comes back...
No, I reckon it's 20 mins at least.

I'll give a better hint out tomorrow or Wednesday, but for now:

The 10 minute man is the key. Think of the best way to bring him across.

[Janny]

Gottit!!! Gotta take the 5 man with the 10 man, but not at the start...
1 and 2 =>
<= 1
5 and 10 =>
<= 2
1 and 2 =>

[mithrand1r]

Originally posted by Janny
Gottit!!! Gotta take the 5 man with the 10 man, but not at the start...

1 and 2 =>
<= 1
5 and 10 =>
<= 2
1 and 2 =>

Correct. There may be another soln. out there, but yours works well.

[Janny]

You guys are pure mathematicians, right?
Ever been tempted to do stuff like Fermat's Theorum based on the notion that he wouldn't have much more knowledge or skills than a modern student?
(Tell me if I'm being a little too precocious!)

[mithrand1r]

Originally posted by Janny
You guys are pure mathematicians, right?
Ever been tempted to do stuff like Fermat's Theorum based on the notion that he wouldn't have much more knowledge or skills than a modern student?
(Tell me if I'm being a little too precocious!)

Tempted and interested? Yes.

But this quickly went over my head.

I have briefly looked through the proof, by some professor at Princeton University and much of the proof went over my head.

[Janny]

Well, the point was th guy who proved it, Andrew Wiles*, began at the age of about 12 with the assmption that Fermat knew only a little more than Wiles when Fermat made his undisclosed proof. For example, I think I can prove there are no solutions which can fit in the equation x^n + y^n = z^n, when z is the hypothenuse and x and y are sides at 90 degrees to one another. Having said this, I think most likely there will be some kind of obvious flaw in it...
x^2 + y^2 = z^2 is true, stated by Pythagorous.
Multiply through by z^(n-2)
x^2 z^(n-2) + y^2 z^(n-2) = z^n
If solution is true then z^n = x^n + y^n
Therefore x^n + y^n = x^2 z^(n-2) + y^2 z^(2-n)
And I think... therefore z^(n-2) must = x and must = y
But if x = y = z, then:
z^n + z^n = z^n
So there can be no solutions where z =/= 0.
Go on, find the glaringly obvious flaw!
If there isn't one, then it's mine, MINE!!!

If this were right, I would sugest moving to the equation:
z^2 = x^2 + y^2 - 2xy cos angle
Which would sort out angles not 90 degrees.

But hey, I may be completely barking up the wrong tree!

[Mercutio]

Originally posted by Janny
For example, I think I can prove there are no solutions which can fit in the equation x^n + y^n = z^n, when z is the hypothenuse and x and y are sides at 90 degrees to one another. Having said this, I think most likely there will be some kind of obvious flaw in it...
x^2 + y^2 = z^2 is true, stated by Pythagorous.
Multiply through by z^(n-2)
x^2 z^(n-2) + y^2 z^(n-2) = z^n
If solution is true then z^n = x^n + y^n
Therefore x^n + y^n = x^2 z^(n-2) + y^2 z^(2-n)
And I think... therefore z^(n-2) must = x and must = y
But if x = y = z, then:
z^n + z^n = z^n
So there can be no solutions where z =/= 0.
Go on, find the glaringly obvious flaw!
If there isn't one, then it's mine, MINE!!!
If this were right, I would sugest moving to the equation:
z^2 = x^2 + y^2 - 2xy cos angle
Which would sort out angles not 90 degrees.

*stands awestruck with mouth gaping open*

[Janny]

No, it's a game... hunt the absurdity.