This is one of my favorite problems. It is hard to get the wrong answer if you use Einstein's Principle of Equivalence, providing a beautiful example of how intellectual depth can make the solution to a difficult problem obvious.

http://www.youtube.com/watch?v=XXpU...

The only thing you need to understand what's going on is to remember at the invisible air is definitely experiencing the forces as one would expect. Slamming the brakes makes the air rush to the front of the cabin. Accelerating makes it rush to the back. And so forth.

A 3x3 magic square is a square array of nine numbers arranged so that they add up to the same total horizontally, vertically, and diagonally. For example, the rows, columns, and diagonals of

6 1 8
7 5 3
2 9 4

all add up to 15. Of course an array obtained by rotating this one, e.g.,

2 7 6
9 5 1
4 3 8

or by inverting it, e.g.,

8 1 6
3 5 7
4 9 2

or by a sequence of rotations and inversions is equivalent to the original one.

Adding the same number to all nine entries (an "additive transformation") yields another equivalent magic square. For example, adding -5 to all nine entries changes it to

1 -4 3
2 0 -2
-3 4 -1

where now all the sums equal 0. Likewise, multiplying all nine entries in a magic square by the same number (a "multiplicative transformation") yields another equivalent magic square. For example, multiplying all nine entries of the original square by 2 changes it to

12 2 16
14 10 6
4 18 8

where now all the sums equal 30.

Here's the puzzle: Is the first square equivalent to the one in clue #54 of the contest,

17 89 71
113 59 5
47 29 101

Sort all numbers in the squares. Consider the effects of the transformation on that sequence.

Rotation, inversion ("mirroring" would be a better word), addition, and multiplication by positive numbers do not change the ordering in the sequence. Multiplication by negative numbers reverses the sequence. This implies that if the transformation was posible, you'd map original numbers to final numbers either thus, after even number of multiplications by negative numbers:

1 -> 5
...
9 -> 113

or thus, after odd number of negative multiplications:

1 -> 113
...
9 -> 5

Additive and multiplicative transformations apply identical linear transformation to all numbers, and their aggregate effect can be always simplified to one multiplication (by x) and one addition (of y).

If the squares can be transformed, then either this system of equations:

1x + y = 5
2x + y = 17
3x + y = 29
4x + y = 47
5x + y = 59
6x + y = 71
7x + y = 89
8x + y = 101
9x + y = 113

or this one:

1x + y = 113
2x + y = 101
3x + y = 89
4x + y = 71
5x + y = 59
6x + y = 47
7x + y = 29
8x + y = 17
9x + y = 5

would have to have a solution. But it is easily shown that they are inconsistent, therefore the transformation is impossible, Q.E.D.

Note that adding two magic squares together element-wise also produces a magic square; let's call this an "aggregation" of squares.

The numbers 1,2,...,9, are consecutive, so all the first differences are the same (1). The equivalence transformations either leave the 9 numbers unchanged or change them to new numbers that are again equidistant. But if you arrange the numbers in the final square in ascending order, viz.,

5, 17, 29, 47, 59, 71, 89, 101, 113,
and take first differences, you get

12, 12, 18, 12, 12, 18, 12, 12.

QED.

199, 409, 619, 829, 1039, 1249, 1459, 1669, 1879.

"Primes in arithmetic progression"

[ http://en.wikipedia.org/wiki/Primes... ]

Another solution to your original problem is to set up a canonical form for each magic square, so that rotations and reflections achieve

NW corner <= NE corner <= SW corner with NW corner <= SE corner

(Secondary rules on other elements are required to break the ties.)

One can then apply a linear transformation so that

NW corner = 1

NE corner = 2 (or some secondary rule if NW corner = NE corner).

I have what seems to be an answer to your question.

First, some notation. I will write A ≡ B to indicate that the corresponding elements of the two 3x3 magic squares A and B (which I'll just call “squares” from now on) are identical and use I to denote the “unit square”

1 1 1
1 1 1
1 1 1.

I will use ++ and -- to indicate that one square is added to or subtracted from another element by element. If I add the same number c to each element of a square A, then the result can be written

A' ≡ A ++ cI,

i.e., as an aggregation. Multiplying each element of a square A by 2 can also be regarded as an aggregation:

A' ≡ 2A ≡ A ++ A.

Likewise for multiplying by 3, 4, or any other integer:

A" ≡ 3A ≡ A ++ A ++ A, etc.

If a, b, and c are the numbers
in the center cells of A, B, and C, respectively, let

A' ≡ A -- aI; B' ≡ A -- bI; C' ≡ C -- cI.

Then if A ≡ B ++ C, it follows that A' ≡ B' ++ C'.

Hence I can restrict consideration to squares having 0 in the center cell.

Now let Z denote the original 1,2,...,9 square

6 1 8
7 5 3
2 9 4,

so Z' ≡ Z -- 5I ≡

1 -4 3
2 0 -2
-3 4 -1.

Let P denote the prime square

71 5 101
89 59 29
17 113 47

and P' ≡ P -- 59I ≡

12 -54 42
30 0 -30
-42 54 -12.

Evidently P' -- 6Z' ≡ 6Q, where Q denotes

1 -5 4
3 0 -3
-4 5 -1.

Q can also be written as Z' -- Y, where Y denotes the rather trivial magic square

0 1 -1
-1 0 1
1 -1 0.

But Y can also be expressed in terms of squares that are equivalent to Z. Let Z§ denote Z' rotated 90 degrees in the clockwise direction, i.e.,

-3 2 1
4 0 -4
-1 -2 3

and let Z> denote Z' reflected (mirrored, inverted) about a vertical axis, i.e.,

3 -4 1
-2 0 2
-1 4 -3.

Then -2Y ≡ Z§ ++ Z>.

Putting them all together, we have P' ≡ 6Z' ++ 6Q ≡ 12 Z' -- 6Y ≡ 12Z' + 3Z§ ++ 3Z>,

=== Simple Math ===

You have a bunch of bags of nearly pure gold coins. How many bags is up to you. Choose a number.

Each bag has a certain number of coins, which can vary from bag to bag. Choose a number, it's up to you.

Now AFTER you have chosen both those numbers, let me present the problem:

ONE and only one of the bags has coins that are fake---lead clad in gold. They are only about half the weight as gold coins. You need to determine which bag has the fake coins.

BUT: Here is the kicker: All you have available for a scale is a penny weight machine (it will read up to 450kg/1000lbs), and only one penny.

That's right, you need to figure out which bag has the false coins, with just one weighing/reading.

HOW TO DO THAT?

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

In the episode, Mrs Columbo is apparently the one who came up with the correct solution (according to the detective).

Are you as smart as Mrs. Columbo?

Kate Mulgrew was Mrs. Columbo.>

You are my hero. Good going. Took the practical approach, gave it a shot of reality, made a working model of the problem, and got the correct answer!

Or else remembered the show.

For all the people with problems about this problem, can you explain why <BS> is right?

"Let's just pretend it does!"

<Sneaky: <Each bag has a certain number of coins, which can vary from bag to bag. Choose a number, it's up to you.> I don't think that is the proper statement of the puzzle. Expressed in that fashion I could say I have one million bags, each with one coin each. I'm quite sure you won't find the fake coin with one weighing.>

So when you present this problem to your friends, modify it as follows:

"...your choice, from three to a dozen bags, and any number of coins, so long as each bag contains at least the number of coins as there are bags. And oh, the penny weight machine is deadly accurate---one of a kind!"

An Alien, a sinister creature from another world, has slipped inside the life support dome of the Moon Base. If not detected within 24 hours it will radio a message to its fellow extra-terrestrials in their space ship lurking out by the asteroid belt. But the Alien has assumed an appearance anatomically indistinguishable from humans; only a microscopic examination of its blood will give it away. The authorities have taken blood samples from each of the 1000 individuals occupying the base, but there is only one apparatus suitable for carrying out the appropriate tests, and each test requires two hours.