Math puzzles

[flOvermind]

Now the interesting question: Is it optimal? How to prove that? Or can anyone do better?

Yes it is optimal.
First we observe that it only makes sense to balance two equal number of coins against each other.
Next we observe that as long as the scale didn't tilt yet a next scaling gives only two discrimating results neutral or tilting. Whether it tilts to the left or right doesn't help us to eliminate suspects. ( N , T )
After the scale has tilted for the first time a next scaling can give three discriminating results neutral, equal tilting or opposite tilting. ( N, E, O )

We can start sensibly only with two equal piles A and B balancing against each other leaving out a pile C.
If this balance result is neutral and C has more than 5 coins we are stuck.
Why? If the next balancing act again gives neutral the final one has only two outcomes. But if the next balancing act tilts the scale the third one has three outcomes. So together only 5 outcomes possible after a first neutral scaling. We can discrimate between 5 suspects at most.

If the scale tilts from the start the second and third scaling gives three outcomes each so at most 9 outcomes. A and B has the same number of coins but together not more than 9. So 4 coins each at most. That gives 4 + 4 + 5 at most.

Nice.

The next question: can we decide between 14 suspects given a 15th that is genuine.

Or perhaps more? Because once you introduce a genuine coin, you have three possible outcomes on every trial involving the genuine coin, even if it is the first one: Balanced, and tilted with the genuine coin down or up. The question is: can we extract useful information from that? I haven't found a way yet, but also no argument why that isn't possible

[Redundant]

This thread demands additional problems!

This one's from a Putnam from a couple years ago. Given an empty 2010 by 2010 matrix, Alice and Bob alternate filling entries with real numbers. After all entries are filled, Alice wins if the determinant is nonzero, while Bob wins if the determinant is zero. Who has a winning strategy?

[cyclops]

This thread demands additional problems!

This one's from a Putnam from a couple years ago. Given an empty 2010 by 2010 matrix, Alice and Bob alternate filling entries with real numbers. After all entries are filled, Alice wins if the determinant is nonzero, while Bob wins if the determinant is zero. Who has a winning strategy?

Bob wins on a 2X2 matrix. If he starts he enters a zero anywhere and next makes a row or a column of zeros. If he is second he enters a zero diagonally opposite Alice's entry and next makes a row or a column of zeros.

So my guess is: Bob.

[Solomon]

Bah these problems are too complicated, so here's a simple one: abcd = a^a + b^b + c^c + d^d. What are the values of a, b, c, and d? (note: abcd is not a*b*c*d, but a 4-digit number where a is in the thousands' place, b is in the hundreds' place, etc.; e.g: if abcd = 4819, then a = 4, b = 8, c = 1, d = 9).

[Redundant]

Araban's problem

3435 = 3*3 + 4^4 + 3^3 + 5^5

By brute force. I wonder if there's some more elegant way to derive this.

[Harleqin]

abcd

If we assume that 0^0 = 1, then there is another solution: 0030.

[cyclops]

This one's from a Putnam from a couple years ago. Given an empty 2010 by 2010 matrix, Alice and Bob alternate filling entries with real numbers. After all entries are filled, Alice wins if the determinant is nonzero, while Bob wins if the determinant is zero. Who has a winning strategy?

Pls. Solve it, redundant.

[Redundant]

Solution:

Whenever Alice places an element in an even indexed row, Bob places the same number in the spot immediately above it.

Whenever Alice places an element in an odd row, Bob places the same number immediately below it.

This way we're guaranteed to have two rows which are identical, so the determinant will be zero. Bob wins.

Note that the size of the matrix is even, so we won't run into any cases where this strategy fails.

[CSamurai]

Solution:

Whenever Alice places an element in an even indexed row, Bob places the same number in the spot immediately above it.

Whenever Alice places an element in an odd row, Bob places the same number immediately below it.

This way we're guaranteed to have two rows which are identical, so the determinant will be zero. Bob wins.

Note that the size of the matrix is even, so we won't run into any cases where this strategy fails.

Assuming that Alice goes first. If Bob goes first, Alice wins.

[cyclops]

Assuming that Alice goes first. If Bob goes first, Alice wins.

Can you prove that?

[Magicwand]

abcd

it took 30 minute for me to write a script by brute force!!
wow..i suck at programming.
i have not written any program for past 5 years and forget all my syntex!!!

[unkx80]

Bah these problems are too complicated, so here's a simple one: abcd = a^a + b^b + c^c + d^d. What are the values of a, b, c, and d? (note: abcd is not a*b*c*d, but a 4-digit number where a is in the thousands' place, b is in the hundreds' place, etc.; e.g: if abcd = 4819, then a = 4, b = 8, c = 1, d = 9).

Since 0^0 is undefined, assume none of the four digits is 0.

1^1 = 1
2^2 = 4
3^3 = 27
4^4 = 256
5^5 = 3125
6^6 = 46656
...

No digits can be 6 or above.

4^4 + 4^4 + 4^4 + 3^3 = 795 < 1000
4^4 + 4^4 + 4^4 + 4^4 = 1024
These are not solutions, so there is at least one digit 5.

5^5 + 5^5 + x^x + x^x > 6250
This number is either at least five digits or a four-digit number whose first digit is at least 6. Therefore, at most one digit 5.

5^5 + 1^1 + 1^1 + 1^1 = 3128
5^5 + 4^4 + 4^4 + 4^4 = 3893
3128 < 5^5 + x^x + x^x + x^x < 3893
First digit is always a 3.

With only sixteen choices for the remaining two digits, easiest to just brute force.

[Magicwand]

Bah these problems are too complicated, so here's a simple one: abcd = a^a + b^b + c^c + d^d. What are the values of a, b, c, and d? (note: abcd is not a*b*c*d, but a 4-digit number where a is in the thousands' place, b is in the hundreds' place, etc.; e.g: if abcd = 4819, then a = 4, b = 8, c = 1, d = 9).

Since 0^0 is undefined, assume none of the four digits is 0.

1^1 = 1
2^2 = 4
3^3 = 27
4^4 = 256
5^5 = 3125
6^6 = 46656
...

No digits can be 6 or above.

4^4 + 4^4 + 4^4 + 3^3 = 795 < 1000
4^4 + 4^4 + 4^4 + 4^4 = 1024
These are not solutions, so there is at least one digit 5.

5^5 + 5^5 + x^x + x^x > 6250
This number is either at least five digits or a four-digit number whose first digit is at least 6. Therefore, at most one digit 5.

5^5 + 1^1 + 1^1 + 1^1 = 3128
5^5 + 4^4 + 4^4 + 4^4 = 3893
3128 < 5^5 + x^x + x^x + x^x < 3893
First digit is always a 3.

With only sixteen choices for the remaining two digits, easiest to just brute force.

What is 0 to the 0 power?
This answer is adapted from an entry in the sci.math Frequently Asked Questions file, which is Copyright (c) 1994 Hans de Vreught (hdev@cp.tn.tudelft.nl).
According to some Calculus textbooks, 0^0 is an "indeterminate form." What mathematicians mean by "indeterminate form" is that in some cases we think about it as having one value, and in other cases we think about it as having another.

When evaluating a limit of the form 0^0, you need to know that limits of that form are "indeterminate forms," and that you need to use a special technique such as L'Hopital's rule to evaluate them. For instance, when evaluating the limit Sin[x]^x (which is 1 as x goes to 0), we say it is equal to x^x (since Sin[x] and x go to 0 at the same rate, i.e. limit as x->0 of Sin[x]/x is 1). Then we can see from the graph of x^x that its limit is 1.

Other than the times when we want it to be indeterminate, 0^0 = 1 seems to be the most useful choice for 0^0 . This convention allows us to extend definitions in different areas of mathematics that would otherwise require treating 0 as a special case. Notice that 0^0 is a discontinuity of the function f(x,y) = x^y, because no matter what number you assign to 0^0, you can't make x^y continuous at (0,0), since the limit along the line x=0 is 0, and the limit along the line y=0 is 1.

This means that depending on the context where 0^0 occurs, you might wish to substitute it with 1, indeterminate or undefined/nonexistent.

Some people feel that giving a value to a function with an essential discontinuity at a point, such as x^y at (0,0), is an inelegant patch and should not be done. Others point out correctly that in mathematics, usefulness and consistency are very important, and that under these parameters 0^0 = 1 is the natural choice.

The following is a list of reasons why 0^0 should be 1.


Rotando & Korn show that if f and g are real functions that vanish at the origin and are analytic at 0 (infinitely differentiable is not sufficient), then f(x)^g(x) approaches 1 as x approaches 0 from the right.
From Concrete Mathematics p.162 (R. Graham, D. Knuth, O. Patashnik):


Some textbooks leave the quantity 0^0 undefined, because the functions 0^x and x^0 have different limiting values when x decreases to 0. But this is a mistake. We must define x^0=1 for all x , if the binomial theorem is to be valid when x=0 , y=0 , and/or x=-y . The theorem is too important to be arbitrarily restricted! By contrast, the function 0^x is quite unimportant.
Published by Addison-Wesley, 2nd printing Dec, 1988.


As a rule of thumb, one can say that 0^0 = 1 , but 0.0^(0.0) is undefined, meaning that when approaching from a different direction there is no clearly predetermined value to assign to 0.0^(0.0) ; but Kahan has argued that 0.0^(0.0) should be 1, because if f(x), g(x) --> 0 as x approaches some limit, and f(x) and g(x) are analytic functions, then f(x)^g(x) --> 1 .
The discussion of 0^0 is very old. Euler argues for 0^0 = 1 since a^0 = 1 for a not equal to 0 . The controversy raged throughout the nineteenth century, but was mainly conducted in the pages of the lesser journals: Grunert's Archiv and Schlomilch's Zeitshrift. Consensus has recently been built around setting the value of 0^0 = 1 .


Magicwand think:
it is not undefined. it is "indeterminate form". they are different.
Mathmathic uses convention like 0! which make no sense but it fits other pattern.
0^0 is no different than 0!

[cyclops]

Rotando & Korn show that if f and g are real functions that vanish at the origin and are analytic at 0 (infinitely differentiable is not sufficient), then f(x)^g(x) approaches 1 as x approaches 0 from the right.
From Concrete Mathematics p.162 (R. Graham, D. Knuth, O. Patashnik):

Strange result. f(-x)^g(-x) then approaches 1 as x approaches 0 from the left. The direction of the approach then doesn't matter.

edit: Strange result2: According to this theorem (-x^2)^x approaches 0 in the limit x to 0. But the function nowhere exists near x=0.

Probably you mean: x approaches 0 from the side(s) where f(x) > 0 (if any).

[cyclops]

_ 1_ suppose 0^0 = 1 and inf=infinity then:

_ 2_ log(0^0) = log (1)
_ 3_ 0 * log (0) = 0 ( seems to make some sense: 0 * something = 0 )
_ 4_ 0 * (-inf) = 0
_ 5_ (1/0) * (1/(-inf)) = 1/0
_ 6_ (-1/0) * (1/inf) = inf
_ 7_ -inf * 0 = inf
_ 8_ 0 * (-inf ) = inf
_ 9_ 0 * log(0) = inf
_10_ log(0^0) = log ( inf)
_11_ e^log(0^0) = inf
_12_ 0^0 = inf

edit: put the line numbers in, couldn't edit it more spacy.
edit2: I included some brackets