NationStates Jolt Archive

So what're your favourite little thinking games? Throw them down here if you've got any you'd like to share. ^^

This was a little warm-down exercise from a set of Discrete Maths questions I've been doing for the last 12 hours. I figured some of you might enjoy thinking it over. It's a tricky one though, so I'll give you a few hints if you ask for it.


The Master of Regents' College and his wife invite n Fellows and their spouses to a party. After the party, the Master asks everyone (including his wife) how many people they shook hands with, and receives 2n + 1 different answers. No spouse shook hands with their other half.

How do we know that the Master's wife was not the person who shook the most hands?

How many hands did the Master shake?

Figuring out D&D formulas is taxing enough...

I'm not a big math guy...I actually...hate numbers. I seriously hope that Pythagorus is being forced to do his theorem for all eternity...

I once googled this, and i swear i found it, but i might have been incorrect.

A mathematical formula where 1=take that home and chew it

I once googled this, and i swear i found it, but i might have been incorrect.

A mathematical formula where 1=take that home and chew it
Hehe... It all depends on your choice of Axioms. You could pick values for all of those letters such that they all evaluate out to a 1 that you've given a numerical definition to.

zomg!!1! could be a numerical value had you decided to define your numbers that way.

So what're your favourite little thinking games? Throw them down here if you've got any you'd like to share. ^^

This was a little warm-down exercise from a set of Discrete Maths questions I've been doing for the last 12 hours. I figured some of you might enjoy thinking it over. It's a tricky one though, so I'll give you a few hints if you ask for it.


The Master of Regents' College and his wife invite n Fellows and their spouses to a party. After the party, the Master asks everyone (including his wife) how many people they shook hands with, and receives 2n + 1 different answers. No spouse shook hands with their other half.

How do we know that the Master's wife was not the person who shook the most hands?

How many hands did the Master shake?


We know that the Master's wife shook the least hands because if she was a decent hostess she would be slaving away in the kitchen making drinks and food.

We know that the Master shook 2N hands because if he was a decent host he would greet everybody at the door, shake their hand and take their coat.

Here's one that was on NS a while ago.

I want an algorithm which, if you put in a two digit number, will output the two digit number with it's digits in the reverse order. Eg, F(23) = 32.

This took me about ten minutes but I didn't make an algebraic proof. I only wen't through all the numbers from 10 to 99.

Without looking it up on the internet, supply a proof that all integers can be factorised into a product of prime powers. Eg, 20 = 2^2 * 2^2 * 5^1.

Hint: It's a proof by contradiction.

http://i14.photobucket.com/albums/a334/Yatt1/Puzzle.jpg

Draw one continuous line that crosses through each of the sixteen segments in the picture above. Each segment can only be crossed once. You may begin anywere on the diagram.

So what're your favourite little thinking games? Throw them down here if you've got any you'd like to share. ^^

This was a little warm-down exercise from a set of Discrete Maths questions I've been doing for the last 12 hours. I figured some of you might enjoy thinking it over. It's a tricky one though, so I'll give you a few hints if you ask for it.


The Master of Regents' College and his wife invite n Fellows and their spouses to a party. After the party, the Master asks everyone (including his wife) how many people they shook hands with, and receives 2n + 1 different answers. No spouse shook hands with their other half.

How do we know that the Master's wife was not the person who shook the most hands?

How many hands did the Master shake?
2n people attended the party plus the Master and his wife. Whithout shaking his wife's hand, the Master shook only 2n hands.

How do we know the wife didn't shake the most hands? Because aside from the Master, the rest shook 2n + 1 hands. At worst she could have equalled the rest.

http://i14.photobucket.com/albums/a334/Yatt1/Puzzle.jpg

Draw one continuous line that crosses through each of the sixteen segments in the picture above. Each segment can only be crossed once. You may begin anywere on the diagram.

Do you mean a straight line? If you don't then it's a horse shoe turned to the left.

Do you mean a straight line? If you don't then it's a horse shoe turned to the left.
I mean something like this...
http://i14.photobucket.com/albums/a334/Yatt1/Puzzle1.jpg

I mean something like this...
http://i14.photobucket.com/albums/a334/Yatt1/Puzzle1.jpg

Oh, I see. You mean it has to pass across every line in the picture. I was thinking it had to go through the white part between the lines once. Silly me. :headbang:

Oh, I see. You mean it has to pass across every line in the picture. I was thinking it had to go through the white part between the lines once. Silly me. :headbang:
Crossing the white parts would be too easy.

Crossing the white parts would be too easy.

And there's the small fact that there aren't 16 of them. :D

Crossing the white parts would be too easy.

This question is surprisingly difficult.

What happens if we cross the intersection of two line segments. Does that count as crossing both or neither?

What happens if we cross the intersection of two line segments. Does that count as crossing both or neither?
both

This information would have helped at the beginning. :(

Right. I've got the answer. I don't know how to put it here or anything though. :confused:

i like sudoku puzzles, can get very hard in the higher levels tho...

" Fill in the grid so that every row, every column, and every 3x3 box contains the digits 1 through 9. "

http://www.websudoku.com/

and

http://www.sudoku.com/

Hey. Come on people. The thread is called 'Maths Puzzles'. We're supposed to be posting algebra, geometry, calculus and complex numbers, not Sodoku's and brick walls.

Here's one that was on NS a while ago.

I want an algorithm which, if you put in a two digit number, will output the two digit number with it's digits in the reverse order. Eg, F(23) = 32.

This took me about ten minutes but I didn't make an algebraic proof. I only wen't through all the numbers from 10 to 99.#include<iostream>
using namespace std;

int main {
int first, second;
cout << "Input a positive two digit integer.\n";
cin >> first;
if(first < 0 || first > 99)
cout << "It doesn't work with " << first << ".\n";
else {
second = first / 10;
second += 10 * (first%10);
cout << "The transpose is " << second << ".\n";
}
return 0;
}

#include<iostream>
using namespace std;

int main {
int first, second;
cout << "Input a positive two digit integer.\n";
cin >> first;
if(first < 0 || first > 99)
cout << "It doesn't work with " << first << ".\n";
else {
second = first / 10;
second += 10 * (first%10);
cout << "The transpose is " << second << ".\n";
}
return 0;
}

Almost.

Should be:

second = first / 10
second += 9.9 * (first%10)

But well done. How the hell did you do that? :D

both

If crossing the intersection of three lines also counts as crossing three lines, then I have an answer.


Numbering lines from left to right, then the next row beneath them, and so on, such that the top left horizontal line is 1 and the bottom right horizontal line is 16.

Starting on the inside of the top left block:

1,5,9,13,16,12,15,14,10,3,6,11,(crossing point of 4/7/8),2.

If the first hand of a clock is the hour hand and the second hand of a clock is the minute hand, what is the third hand of a clock?

If the first hand of a clock is the hour hand and the second hand of a clock is the minute hand, what is the third hand of a clock?

:D

Without looking it up on the internet, supply a proof that all integers can be factorised into a product of prime powers. Eg, 20 = 2^2 * 2^2 * 5^1.

Hint: It's a proof by contradiction.

2^2*2^2*5^1 = 4*4*5 = 80. Since there's a contradiction in your theorem, does this count as a proof by contradiction?

http://i14.photobucket.com/albums/a334/Yatt1/Puzzle.jpg

Draw one continuous line that crosses through each of the sixteen segments in the picture above. Each segment can only be crossed once. You may begin anywere on the diagram.

It is impossible because there are more than two intersections with an odd number of line segments intersecting.
If there is an intersection with an odd number of line segments intersecting, it must be either a start or an end. An even number will mean that it is possible to go through it, however, an odd number means you can't go through the intersection completely, so you must start or end there.

Therefore, since there are more than 2 odd-number intersections, it is impossible.

:D

No, smily face is the wrong answer. YOU LOSE!

2^2*2^2*5^1 = 4*4*5 = 80. Since there's a contradiction in your theorem, does this count as a proof by contradiction?

No, he just sucks at factoring a number into its prime factors.

If the first hand of a clock is the hour hand and the second hand of a clock is the minute hand, what is the third hand of a clock?

The third hand is the second hand :D

2^2*2^2*5^1 = 4*4*5 = 80. Since there's a contradiction in your theorem, does this count as a proof by contradiction?

:sniper: :sniper: :sniper: :sniper: :sniper: :sniper:

Don't defile my religion.

Well, this one is more of a physics problem than a math problem, but I'm going to post it anyway. (If anyone has heard this one before, please do not post the answer):

You are in a room with dimensions 3 meters by 3 meters by 3 meters. In the middle of the floor is a round hole 150 millimeters deep and with diameter 25 millimeters. Inside the hole is a ping pong ball of diameter slightly less than 25millimeters (so it fits loosely in the hole).

You are given a 9 volt battery, a horseshoe magnet, a piece of string 450 millimeters long and a copper coin 20 millimeters in diameter. How do you remove the ping pong ball from the hole?

Well, this one is more of a physics problem than a math problem, but I'm going to post it anyway. (If anyone has heard this one before, please do not post the answer):

You are in a room with dimensions 3 meters by 3 meters by 3 meters. In the middle of the floor is a round hole 150 millimeters deep and with diameter 25 millimeters. Inside the hole is a ping pong ball of diameter slightly less than 25millimeters (so it fits loosely in the hole).

You are given a 9 volt battery, a horseshoe magnet, a piece of string 450 millimeters long and a copper coin 20 millimeters in diameter. How do you remove the ping pong ball from the hole?

Threaten to throw money at the ping pong ball unless it comes out.

Threaten to throw money at the ping pong ball unless it comes out.

Some day, you will become a very successful bureaucrat.

Some day, you will become a very successful bureaucrat.
:D

Almost.

Should be:

second = first / 10
second += 9.9 * (first%10)

But well done. How the hell did you do that? :DIf the starting number is 23, first / 10 returns 2.
(first%10) returns 3.
10 * 3 returns 30.
second = 2 + 30.

You are in a room with dimensions 3 meters by 3 meters by 3 meters. In the middle of the floor is a round hole 150 millimeters deep and with diameter 25 millimeters. Inside the hole is a ping pong ball of diameter slightly less than 25millimeters (so it fits loosely in the hole).

You are given a 9 volt battery, a horseshoe magnet, a piece of string 450 millimeters long and a copper coin 20 millimeters in diameter. How do you remove the ping pong ball from the hole?step 1: Loop the string around the ball and then pull it out.

step 2: Lick the battery. Batteries taste like happiness.

If the starting number is 23, first / 10 returns 2.
(first%10) returns 3.
10 * 3 returns 30.
second = 2 + 30.

Pardon me. I forgot that divison in c++ returns an integer if done with integers. Haven't used it in a while. Well done. :fluffle:

How many hands did the Master shake?


Clearly the answer is 42....I think it was first found by Douglas Adams!:)

Did I read that correctly? Did someone just change his mind because of something written in a Nation States post? This is scary.

Clearly the answer is 42....I think it was first found by Douglas Adams!:)

Another hitchhiker. They're like lice aren't they.

2n people attended the party plus the Master and his wife. Whithout shaking his wife's hand, the Master shook only 2n hands.

How do we know the wife didn't shake the most hands? Because aside from the Master, the rest shook 2n + 1 hands. At worst she could have equalled the rest.
The problem with that reasoning is that then there'd only be 2 possible answers to the "How many hands did you shake?" question.

There's actually a pretty simple mathematical analysis you can do. The hint is to consider the largest and smallest numbers of people with whom any given individual could shake hands and to thus consider the people that supplied those answers...

I'll give you my solution later on this evening. ^^;

You are in a room with dimensions 3 meters by 3 meters by 3 meters. In the middle of the floor is a round hole 150 millimeters deep and with diameter 25 millimeters. Inside the hole is a ping pong ball of diameter slightly less than 25millimeters (so it fits loosely in the hole).

You are given a 9 volt battery, a horseshoe magnet, a piece of string 450 millimeters long and a copper coin 20 millimeters in diameter. How do you remove the ping pong ball from the hole?
Define the room to be the inside of the hole.

This one looks interesting though. Could you wrap the coin in string so that it takes up almost exactly 25mm in the whole then use the magnet to pull the coin out, thus creating a pressure gradient that sucks the ball out?

How do you remove the ping pong ball from the hole?

Piss in hole. Lift out the floating ping-pong ball. If the ball is still beyond your grasp, then repeat the urination stage and try again.

he Master of Regents' College and his wife invite n Fellows and their spouses to a party. After the party, the Master asks everyone (including his wife) how many people they shook hands with, and receives 2n + 1 different answers. No spouse shook hands with their other half.

How do we know that the Master's wife was not the person who shook the most hands?

How many hands did the Master shake?here is a solution:

it requires two things,
1. that "Master asks everyone" does exclude the master asking himself and getting an answer from himself.
2. that none of the n invited fellows is also a spouse of one of the other fellows, but that the spouses are extra persons comming to the party.

n=2 six people attend the party, the master (M), his wife (W), guest 1 (G1), guest 1s spouse (S1), guest 2 (G2), guest 2s spouse (S2)

2n+1=5 G1 shook 4 hands, S1 shook 0 hands, G2 shook 3 hands, S2 shook 1 hand, W shook 2 hands, M shook 2 hands. 0;1;2;3;4 are the 5 different answers M gets on asking, This can be easily shown by drawing a hexagon with each corner representing an attendee, every corner can be connected to a maximum of 5 others, but since nobody shakes hands with their spouses (M and W are spouses too) the maximum is reduced to 4.
The construction has to be done in a way, that M and W each shake 2 hands or else M will not receive 5 different answers when asking allthough there are 5 different combinations, because he does not include himself in the survey (see 1st requirement above)

this shows, that n=2 is a solution.

if you take n>2 you will get an octagon at n=3 as representation of the attendees, but you can eliminate a couple, by having one half shake all possible hands (2n) and the other half shake 0 hands, reducing the octagon to a hexagon in which the attendee with the highest handcount would have to shake 2n-1 hands, but he can shake a maximum of 2n-2 hands, because he can neither shake himself nor his spouse, so this becomes unsolvable for n>2


new question: can you explain this?
http://home.arcor.de/genius23/pics/geometry.gif

First off, your puzzle.

new question: can you explain this?
It's an optical illusion. The shape on the top diagram is a quadrilateral; if not for the missing square, so would be the bottom.

The area of the blue triangle is 5 squares. The area of the red is 12. The total area of the two l-shaped blocks is 15 squares.

If the top shape was a triangle, the total area covered would be 1/2(5x13). But it's not; we know that it must be 5+12+15 = 32, as opposed to 65/2 = 32.5.

The bottom shape then represents the other half of the 5x13 rectangle that encloses the first shape; as can be seen if you pair up the triangles along the "Diagonal" and include the other shapes in the 2x8 remaining block. This total area is made up of (5x13 - 32) = 33 squares, of which the shapes only comprise 32.

Hence, the 1 left over.

Now, you took an interesting approach to my puzzle, but you reckoned it couldn't be solved. There is a solution for all values of N. But you're along the right lines.

this shows, that n=2 is a solution.
Correct; when there are 4 guests, the master shakes 2 hands.

if you take n>2 you will get an octagon at n=3 as representation of the attendees, but you can eliminate a couple, by having one half shake all possible hands (2n) and the other half shake 0 hands, reducing the octagon to a hexagon in which the attendee with the highest handcount would have to shake 2n-1 hands, but he can shake a maximum of 2n-2 hands, because he can neither shake himself nor his spouse, so this becomes unsolvable for n>2
One thing you're failing to take into account in this case is the inductive step is for a given N. That is, the couples come in pairs, and reducing the system from an "Octagon" would result in the highest handcount being 2(n-1) as opposed to 2n-1.

you are right about my question, allthough i would not call it an optical illusion, if you have keen eyes, you can actually see, that the the left sides of the "triangles" are not straight lines, but that they have a small dent.

in my train of thought for n=3 i did not mean, that the highest possible handcount in the hexagon (octagon reduced by 2 corners) would be 2n-1, but that the highest count would have to be 2n-1 in order for the survey in the end to yield the correct number of 2n+1 differnet answers. the highest possible count actually is 2n-2, i wrote, which is just the same as 2(n-1), so there is no contradiction between what i wrote and what you wrote.

http://i14.photobucket.com/albums/a334/Yatt1/Puzzle.jpg

Draw one continuous line that crosses through each of the sixteen segments in the picture above. Each segment can only be crossed once. You may begin anywere on the diagram.

Another way of doing this...

Draw a line which follows the external face of the figure until you passed through all the separate lines there, then dive into the figure and cross the three internal intersection points with the line.

Figuring out D&D formulas is taxing enough...

I'm not a big math guy...I actually...hate numbers. I seriously hope that Pythagorus is being forced to do his theorem for all eternity...
Pythagoras (if he existed) didn't use numbers (or algebra) to prove his theorem.

:D Wild adult squirrels have two litters of three pups every summer, and they live for about four years. Given these numbers, a single pair of squirrels could multiply to 63,967 trillion in 33 years if they all survived.

surface area of the earth(ea)=510,000,000,000,000,000m2(thats meters squared)

area covered by each squirrel(sa)=.1m x .2m=0.02m2(once again, thats meters squared.

number of squirrels needed to blanket the earth=ea/sa =25,000,000,000,000,000

it would take about 32.5 years to blanket thte earth in squirrels.

assuming that they float.

but that the highest count would have to be 2n-1 in order for the survey in the end to yield the correct number of 2n+1 differnet answers.
Ah, no it wouldn't. The highest count would have to be 2n; assuming one person shook the hands of everyone but themself and their spouse, with the lowest (their spouse) being 0. This gives 2n+1 different answers. Then, when you reduce the problem to a "hexagon", the highest count is 2n-1, since only one person that any given person shook hands with is being taken away. You know for a fact that the one person who shook hands with everyone must have been the only person who shook hands with the person who only shook one hand, so you therefore have another 0 value. So the maximum number this time is 2n-1 answers.

This continues with the number of members in a polygon decreasing at twice the rate of handshakes, eventually resulting in (2n-n)= n as the highest count in a group of (2n-2n+1)=1 single member. Any n problem reduces to a situation where n=0, and you end up with one person who shook the same number of hands as the master; the master's wife.

So regardless of how large n is, the master's wife, and hence the master, always shakes hands with n of the 2n guests.

:D Wild adult squirrels have two litters of three pups every summer, and they live for about four years. Given these numbers, a single pair of squirrels could multiply to 63,967 trillion in 33 years if they all survived.

surface area of the earth(ea)=510,000,000,000,000,000m2(thats meters squared)

area covered by each squirrel(sa)=.1m x .2m=0.02m2(once again, thats meters squared.

number of squirrels needed to blanket the earth=ea/sa =25,000,000,000,000,000

it would take about 32.5 years to blanket thte earth in squirrels.

assuming that they float.
That's awesome. We should try that. Maybe on some other planet first though.

Without looking it up on the internet, supply a proof that all integers can be factorised into a product of prime powers. Eg, 20 = 2^2 * 2^2 * 5^1.

Hint: It's a proof by contradiction.

2^2 * 2^2 * 5^1 = 40! :rolleyes:

2^2 * 2^2 * 5^1 = 40! :rolleyes:
2^2 * 2^2 * 5^1 = 80!

4 x 4 x 5 = 20 x 4 = 80.

=P

2^2 * 2^2 * 5^1 = 40! :rolleyes:

Man, I love this forum.

Has anyone tried to come up with the proof CC was talking about? I was thinking something along the lines of "Every non-prime number can be factored into other numbers that are either prime or non-prime, and every non-prime number that results can be factored the same way until eventually all the factors are prime." I'm not sure how to make that statement mathematically precise, but it makes sense intuitively.

Well, this one is more of a physics problem than a math problem, but I'm going to post it anyway. (If anyone has heard this one before, please do not post the answer):

You are in a room with dimensions 3 meters by 3 meters by 3 meters. In the middle of the floor is a round hole 150 millimeters deep and with diameter 25 millimeters. Inside the hole is a ping pong ball of diameter slightly less than 25millimeters (so it fits loosely in the hole).

You are given a 9 volt battery, a horseshoe magnet, a piece of string 450 millimeters long and a copper coin 20 millimeters in diameter. How do you remove the ping pong ball from the hole?


I think you should just turn the room upside down. it would be so much easier! than fiddling with batteries and magnets!

2^2 * 2^2 * 5^1 = 80!

4 x 4 x 5 = 20 x 4 = 80.

=P

whoops! i can integrate, differentiate and all that malarky at the drop of a hat, but simple multiplication always stumps me! ha!:p

let me take myabbacus and figger this out:headbang:

I think you should just turn the room upside down. it would be so much easier! than fiddling with batteries and magnets!

Actually, Grampus was right: "pee in the hole" is the simplest solution. All that stuff about batteries and magnets was just to throw you off.

What is the smallest 12 digit number composed only of the digits "1" and "0", that is divisible by 9 and ends in three "0"s??

Without looking it up on the internet, supply a proof that all integers can be factorised into a product of prime powers. Eg, 20 = 2^2 * 2^2 * 5^1.

Hint: It's a proof by contradiction.
*Smirks*

This is a false statement.

1 cannot be factorised into a product of prime powers, as 1 itself is not a prime number.

What is the smallest 12 digit number composed only of the digits "1" and "0", that is divisible by 9 and ends in three "0"s??

111,111,111,000. In all numbers that are divisible by nine, the digits in the number must add up to nine. 111,111,111,000 is the only number made up of ones and zeroes with three zeroes at the end that is divisible by nine.

:D Wild adult squirrels have two litters of three pups every summer, and they live for about four years. Given these numbers, a single pair of squirrels could multiply to 63,967 trillion in 33 years if they all survived.

surface area of the earth(ea)=510,000,000,000,000,000m2(thats meters squared)

area covered by each squirrel(sa)=.1m x .2m=0.02m2(once again, thats meters squared.

number of squirrels needed to blanket the earth=ea/sa =25,000,000,000,000,000

it would take about 32.5 years to blanket thte earth in squirrels.

assuming that they float.
Yeah, but nothing beats blanketing the earth in squirrels:rolleyes:

Actually, Grampus was right: "pee in the hole" is the simplest solution. All that stuff about batteries and magnets was just to throw you off.

Another option is just to suck at the hole.

I guess neither of the two options are particularly dignified, but hey, they would work.

Another option is just to suck at the hole.

I guess neither of the two options are particularly dignified, but hey, they would work.

I don't think suction would work. If there was airflow through the hole (like if the hole went clear through to another room below), and if the diameter of the ball was very close to the diameter of the hole, then suction might work, but those things weren't specified in the question. Since the hole was assumed to be closed off at the bottom, you could suck on it until all the air was gone from the hole and there would still be no reason for the ball to move.

I guess suction might be worth a try though, as long as you sucked on the whole before you peed in it.

It is impossible because there are more than two intersections with an odd number of line segments intersecting.
If there is an intersection with an odd number of line segments intersecting, it must be either a start or an end. An even number will mean that it is possible to go through it, however, an odd number means you can't go through the intersection completely, so you must start or end there.

Therefore, since there are more than 2 odd-number intersections, it is impossible.

http://img165.imageshack.us/img165/9494/solved5qe.jpg

yay!

Without looking it up on the internet, supply a proof that all integers can be factorised into a product of prime powers. Eg, 20 = 2^2 * 2^2 * 5^1.

Hint: It's a proof by contradiction.

*Smirks*

This is a false statement.

1 cannot be factorised into a product of prime powers, as 1 itself is not a prime number.

I agree with Kamsaki. No prime number can be factorised into a product of powers of primes.

I agree with Kamsaki. No prime number can be factorised into a product of powers of primes.

EVERY prime number can be factored into a product of prime numbers. It's simply itself to the first power.

It's been a while, but Im sure you'll correct me if I am incorrect. :)

Prime Factorization. Proof by Induction.

Base Case: Prove for 2.

2 can be factored as 2^1.

Assume true for n >= 2

Prove n+1 holds:

In the case that n+1 is prime -

its prime factorization is (n+1)^1

In the case that n+1 is not prime -

Since n+1 is not prime, there exists a,b in the set of positive integers greater than 1 such that a*b = n+1.

Since 2 <= a <= n and 2<= b <= n, we know both a and b have prime factorizations.

n+1 is the prime factorization of a multiplied by the prime factorization of b.

Therefore, n+1 has a prime factorization.
Therefore, all integers greater than 2 have a prime factorization.

111,111,111,000. In all numbers that are divisible by nine, the digits in the number must add up to nine. 111,111,111,000 is the only number made up of ones and zeroes with three zeroes at the end that is divisible by nine.

Given the exact wording of the question I would suspect that it is possible there is a number which when expressed in binary notation is a great deal less than 111,111,111,000 and yet is divisible by 9 and ends in ...000. Alcohol has however fogged my thinking processes at this time of night, and so discovering if such a figure exists is an exercise left up to the gentle reader.

Frex: 11100001000 (binary) or 1800 in base ten.

Given the exact wording of the question I would suspect that it is possible there is a number which when expressed in binary notation is a great deal less than 111,111,111,000 and yet is divisible by 9 and ends in ...000. Alcohol has however fogged my thinking processes at this time of night, and so discovering if such a figure exists is an exercise left up to the gentle reader.

Frex: 11100001000 (binary) or 1800 in base ten.

I didn't think that the question was dealing with binary, but with base ten numbers that were componed entirely of ones and zeroes. Assuming that that is correct, 11100001000 divided by 9 is 1233333444.444 . . .

In base ten, the only numbers that are divisible by nine are those whose digits add up to nine or a multiple of nine. For instance, 36 is divisible by nine because 3+6 = 9, 113 is not divisible by nine because 1+1+3 = 5, and 4815 is divisible by nine because 4+8+1+5 = 18. Assuming that this is true for all integers (i don't know any proof of it, but in my experience it has always been true), 111111111000 is the only twelve digit number made up of just ones and zeroes that ends in three zeroes and is divisible by nine.

It's been a while, but Im sure you'll correct me if I am incorrect. :)

Prime Factorization. Proof by Induction.

Base Case: Prove for 2.

2 can be factored as 2^1.

Assume true for n >= 2

Prove n+1 holds:

In the case that n+1 is prime -

its prime factorization is (n+1)^1

In the case that n+1 is not prime -

Since n+1 is not prime, there exists a,b in the set of positive integers greater than 1 such that a*b = n+1.

Since 2 <= a <= n and 2<= b <= n, we know both a and b have prime factorizations.

n+1 is the prime factorization of a multiplied by the prime factorization of b.

Therefore, n+1 has a prime factorization.
Therefore, all integers greater than 2 have a prime factorization.
Unfortunately this doesn't work as induction would require that you use n's prime factorization, not some factorization less than n (because then you are assuming the inductive step you're trying to prove). For it to work, then you'd have to show that n+1's prime factorization depends on n, but of course, this doesn't work, for example, 6 = 2 * 3 but 7 is a prime and the two factorizations are not related. (If you do find a way they are related, then you should write a paper because you'll have discovered how to find all primes).

The way to do it, as hinted above, is by contradiction. So, of course, try and think about: what if a number had no prime factorization?

A far more exciting proof though is Euclid's theorem that there are infinite primes. It's exciting because the proof is very elegant. As one lecturer I had said: "It's the perfect proof to show people what mathematics is really about."

There's also endless proofs about the existence or nonexistence of certain primes (i.e. of a certain form), or the existence of a prime between two numbers (usually the greater number depending on the smaller number, e.g. n and (n+1)! ). Even this rather fundamental area of number theory is very rich and not fully explored.

Edit: I wrote 'a way to find any prime' but of course we can do that, what I mean is a formula for every prime, which would be astounding.

Edit 2: Philionius Monk, I should point out you were kind of on the right track. You just need to justify the primeness of the factors of n+1, but the whole induction is unnecessary. Try the method via contradiction.

Summarily, I am confused to hell by all of this.

I much prefer:

An enemy has a Hide skill of 30 ( +5 Dex +15 Skill Points +10 Cloak of Elvenkind)

What is the minimum level a character with 3 Wisdom has to be to have sufficient a sufficient Spot skill modifier (without items or magical means) to be able to take 10 on the Spot check and guarentee success?

What is the smallest 12 digit number composed only of the digits "1" and "0", that is divisible by 9 and ends in three "0"s??

000,000,000,000?
-111,111,111,000?

The way to do it, as hinted above, is by contradiction. So, of course, try and think about: what if a number had no prime factorization?
It would have no prime factors.

And such a number exists.

That number is 1.

It would have no prime factors.

And such a number exists.

That number is 1.

0 works as well.

Let me ask you this people: Does 1 have any factors other than 1 and itself. The answer, quite obviously, is no. 1 is prime.

As for the proof by contradiction, I think one of you had it right:

Let's start by going through some numbers. 1, 2, 3, 4, 5, 6, 7, 8, 9, 10.... All can be factorised into prime powers. Lets assume that we get up to a number n which cannot be factored into prime powers.

If n is prime then it can be factorised into n^1, which is a prime factorisation.

If n is not prime, then it has two integers less that itself which are factors. Since we have factorised all numbers up to n into prime powers, these two numbers can themselves be factored into prime powers.

The product of these two prime factored numbers is n. But a product of prime powers is a prime factorisation.

Contradiction. Therefore the axiom 'not all numbers can be factored into prime powers' is false. We have our proof. Well done to whoever it was that got it.

Let me ask you this people: Does 1 have any factors other than 1 and itself. The answer, quite obviously, is no. 1 is prime.

The problem with including 1 in the set of prime numbers is that so many exceptions have to be written, because 1 has unique properties. For example, positive integers could no longer be represented by a unique product of primes.

It's far nicer to exclude 1 from the set of prime numbers.

This web page has an extremely inefficient formula for generating the nth prime number. It works due to the fact that ((n - 1)! + 1) / n will always return an integer if n is prime and a decimal if n is composite. I don't know why that is though. If anybody here is a mathematician and would care to explain to me why this is, I would greatly appreciate it. :fluffle:

Prime Formula (http://mathworld.wolfram.com/PrimeFormulas.html)

The problem with including 1 in the set of prime numbers is that so many exceptions have to be written, because 1 has unique properties. For example, positive integers could no longer be represented by a unique product of primes.

It's far nicer to exclude 1 from the set of prime numbers.

Good point. Much like 2 being the only even prime, 1 does have unique properties. So, while 1 is a prime number by definition, it make maths much nicer if we just ignore it.

The problem with including 1 in the set of prime numbers is that so many exceptions have to be written, because 1 has unique properties. For example, positive integers could no longer be represented by a unique product of primes.

It's far nicer to exclude 1 from the set of prime numbers.

How does it stop uniqueness in prime factorisation? :confused:

So, while 1 is a prime number by definition [...]

Not if you define a prime as "a positive integer greater than 1 ...", which is the normal definition. :)

How does it stop uniqueness in prime factorisation? :confused:

If 1 is included in the set of primes, we have the following:

6 = 3 x 2
6 = 3 x 2 x 1
6 = 3 x 2 x 1^2

You get the idea.

Mathematics becomes so much messier, which makes mathematicians cry. Not a pretty sight.

0 works as well.

No it doesn't. 0 *1 = 0, 0 * 2 = 0, 0 * 3 = 0. There are, technically, an infinite number of factors of 0. 0 is more unique than 1. :(

Let me ask you this people: Does 1 have any factors other than 1 and itself. The answer, quite obviously, is no. 1 is prime.
Obvious, but false. It does not have 1 AND itself, because the two are the same thing.

Here's what Mathworld has to say on primes (http://mathworld.wolfram.com/PrimeNumber.html):

A good reason not to call 1 a prime number is that if 1 were prime, then the statement of the fundamental theorem of arithmetic would have to be modified since "in exactly one way" would be false because any n==n.1. In other words, unique factorization into a product of primes would fail if the primes included 1.

Check around the web. In particular, the Mersenne prime search denies use of (2^1-1).

Not if you define a prime as "a positive integer greater than 1 ...", which is the normal definition. :)

That is a very good point which we've all neglected. :fluffle:

If 1 is included in the set of primes, we have the following:

6 = 3 x 2
6 = 3 x 2 x 1
6 = 3 x 2 x 1^2

You get the idea.

Mathematics becomes so much messier, which makes mathematicians cry. Not a pretty sight.

If you wan't to be really pedantic mathematicians could say that prime factorisation has to be put in simplest form.

3 x 2 x 1 = 3 x 2
3 x 2 x 1^.5 = 3 x 2
Uniquness still holds.

Obvious, but false. It does not have 1 AND itself, because the two are the same thing.

Here's what Mathworld has to say on primes (http://mathworld.wolfram.com/PrimeNumber.html):



Check around the web. In particular, the Mersenne prime search denies use of (2^1-1).

Well, either define 1 as prime and change the unique factoristion theorem, or accept it as niether prime or composite. The thought occurred to me that this argument is stupid. I can't believe that I let myself into this debate. I feel so idiotic. :headbang:

Prime factorisation has to be put in simplest form.

No it doesn't. The uniqueness theorem states that any positive integer can be written as a product of primes. Furthermore, this product is unique except for the ordering.

There's nothing about being in the simplest form. If 1 is a prime, this theorem is trivially false.

No it doesn't. The uniqueness theorem states that any positive integer can be written as a product of primes. Furthermore, this product is unique except for the ordering.

There's nothing about being in the simplest form. If 1 is a prime, this theorem is trivially false.

I modified my post but it all comes down to how pedantic you want to be.

No it doesn't. 0 *1 = 0, 0 * 2 = 0, 0 * 3 = 0. There are, technically, an infinite number of factors of 0. 0 is more unique than 1. :(

But you cannot write 0 as a product of prime numbers because 0 must always be in the product, and 0 is not a prime number. It's a trivial contradiction of the original statement, because the person who posted the statement didn't qualify it correctly.

Just a though, sorry to continue this debate, but one of the other posts makes the point of 2^(1-1). In this case the uniquness theorem doesn't hold either. So I suppose it has to be simplified to hold.

Just a though, sorry to continue this debate, but one of the other posts makes the point of 2^(1-1). In this case the uniquness theorem doesn't hold either. So I suppose it has to be simplified to hold.

My last point, because this isn't actually terribly interesting, to either of us I suspect. The theorem states "product of primes". Your example isn't a product.


Anyway, here's one of my favourites.

Continue the series: 100 121 144 202 244 400

My last point, because this isn't actually terribly interesting, to either of us I suspect. The theorem states "product of primes". Your example isn't a product.


Anyway, here's one of my favourites.

Continue the series: 100 121 144 202 244 400

I meant something like: 80 = 2^4 x 5^1 x 2^0 x 2^0.....

No. it's not terribly interesting.

My last point, because this isn't actually terribly interesting, to either of us I suspect. The theorem states "product of primes". Your example isn't a product.


Anyway, here's one of my favourites.

Continue the series: 100 121 144 202 244 400

Are the next two: 747 and 1408.

It would have no prime factors.

And such a number exists.

That number is 1.
That's a bit of a nitpick, you know what the real proposition is you're just being pedantic. Yes, I know that mathematics requires rigor, but there's rigor and then there's pedantry. The point was made that the fundamental theorem of arithmetic is not what was stated, but just imagine in your mind that what was asked was "prove the fundamental theorem of arithmetic".

Are the next two: 747 and 1408.

No. If you work it out, you'll know it.

EVERY prime number can be factored into a product of prime numbers. It's simply itself to the first power.

Technically that's not a product.

That's a bit of a nitpick, you know what the real proposition is you're just being pedantic. Yes, I know that mathematics requires rigor, but there's rigor and then there's pedantry. The point was made that the fundamental theorem of arithmetic is not what was stated, but just imagine in your mind that what was asked was "prove the fundamental theorem of arithmetic".
Well ... if I got that question in my final alebgra exams at university I would be marked down for that assumption.

Anyway, here's one of my favourites.

Continue the series: 100 121 144 202 244 400

I was trying palindromes and squares, and then I remembered a quote by Douglas Adams:

Q What's six by nine?
A 42.

I'm going to guess 1210 ?

In Dutch, but I don't get it:(

100, 121, 144, 202, 244, 400

100 in 10-tallig stelsel = 100 in tientallig stelsel
121 in 9-tallig stelsel = 81 + 18 + 1 = 100 in tientallig stelsel
144 in 8-tallig stelsel = 64 + 32 + 4 = 100 in tientallig stelsel
202 in 7-tallig stelsel = 98 + 0 + 2 = 100 in tientallig stelsel
244 in 6-tallig stelsel = 72 + 24 + 4 = 100 in tientallig stelsel
400 in 5-tallig stelsel = 100 in tientallig stelsel

het n-de getal is het getal dat in het (11-n)-tallige stelsel
het zelfde is als 100 in het 10-tallige stelsel

I fantastical little puzzle that I know of is this: Prove (algebraically) that 2 = 1

That's it. It can be done, and the algebra used is flawless. It just falls apart slightly when you use actual numbers. See if you can get it - I expect there's someone out there who knows it. And besides I think I posted it on here a while ago (not this thread, just General).

It can be done. Have a go.

I fantastical little puzzle that I know of is this: Prove (algebraically) that 2 = 1

That's it. It can be done, and the algebra used is flawless. It just falls apart slightly when you use actual numbers. See if you can get it - I expect there's someone out there who knows it. And besides I think I posted it on here a while ago (not this thread, just General).

It can be done. Have a go.
Erm, 2x0 = 0 = 1x0?

In Dutch, but I don't get it:(

100, 121, 144, 202, 244, 400

100 in 10-tallig stelsel = 100 in tientallig stelsel
121 in 9-tallig stelsel = 81 + 18 + 1 = 100 in tientallig stelsel
144 in 8-tallig stelsel = 64 + 32 + 4 = 100 in tientallig stelsel
202 in 7-tallig stelsel = 98 + 0 + 2 = 100 in tientallig stelsel
244 in 6-tallig stelsel = 72 + 24 + 4 = 100 in tientallig stelsel
400 in 5-tallig stelsel = 100 in tientallig stelsel

het n-de getal is het getal dat in het (11-n)-tallige stelsel
het zelfde is als 100 in het 10-tallige stelselIt's the same value but different bases.
base 10: 100 -> 1*10^2+0*10^1+0*10^0 = 100 (in decimal)
base 9 :121 -> 1*9^2+2*9^1+1*9^0 = 100 (in decimal)
etc.

You can represent numbers in different ways, decimal (base 10), octal (base 8), hexadecimal (base 16), binary (base 2), trinary (base 3)
You can even try non-integer bases, but that's just silly.

I fantastical little puzzle that I know of is this: Prove (algebraically) that 2 = 1

That's it. It can be done, and the algebra used is flawless.Actually, it's not flawless at all.

You can do the same trick for any 2 numbers, but it almost always comes back to the same flaw; division by zero. Which just throws away a solution, the correct one in fact.

The proof comes down to 0*2=0*1, so divide by 0 on both side, 2=1. Voila, that's the trick. Except they hide it with variables.
But the correct step is: 0*2=0*1, thus 0=0 or 2=1
Which is true.

Actually, it's not flawless at all.

You can do the same trick for any 2 numbers, but it almost always comes back to the same flaw; division by zero. Which just throws away a solution, the correct one in fact.

The proof comes down to 0*2=0*1, so divide by 0 on both side, 2=1. Voila, that's the trick. Except they hide it with variables.
But the correct step is: 0*2=0*1, thus 0=0 or 2=1
Which is true.

Shhhhhh or they'll notice! :D That's why I said algebraically. And also why I said it falls apart of you use actual numbers :D Even if it doesn't technically work (but what are mere technicalities these days eh?) it's still a funky piece of awesomeness. Since you've spilled the proverbial beans I might as well tell you (by you I mean those who don't know lol, since you, Damor, clearly do know) how to do it. Here goes:

x = y Now multiply both sides by x to get
x^2 = xy Now subtract y^2 to get
x^2 - y^2 = xy - y^2 Now factorise to get
(x + y)(x - y) = y(x - y) Now cancel the appropriate brackets to get
x + y = y But we already know that x = y, so
y + y = y Which is the same as
2y = y Now simply cancel the 'y's to get
2 = 1

And cello! There you go. 2 = 1. Proven. Tell your maths teacher to stick that in their pipe and smoke it* :D But as Damor said it doesn't technically work because the brackets would basically equate to multiplying and dividing by zero, which is clearly meaningless. Still cool though :)


* I accept no responsibilty for any sanctions incurred by pupils who actually follow the advice given here :D

So, children, you like maths do you? Prepare to have scratch your head in wonderment until is is raw and bloody!

[Small numbers mean powers, they just don't work on this.]

When 2x2 is divided by 9¼x3 + 5x2 - 15x + 64, carrying the sum through until the power -4, is the remainder divisible by 587?

And, assuming that a right angled triangle is used and that the remainder is 455 times the hypotenuse and another side is the factor by which is divides, what is the remaining side and all the three angles?

If you really want to confuse your math teachers, there are less obvious tricks.
like http://www-math.mit.edu/~tchow/mathstuff/proof.pdf
Some math teachers won't even understand what it says..

So what're your favourite little thinking games? Throw them down here if you've got any you'd like to share. ^^

This was a little warm-down exercise from a set of Discrete Maths questions I've been doing for the last 12 hours. I figured some of you might enjoy thinking it over. It's a tricky one though, so I'll give you a few hints if you ask for it.


The Master of Regents' College and his wife invite n Fellows and their spouses to a party. After the party, the Master asks everyone (including his wife) how many people they shook hands with, and receives 2n + 1 different answers. No spouse shook hands with their other half.

How do we know that the Master's wife was not the person who shook the most hands?

How many hands did the Master shake?

2n+1 different answers of 2n+1 different people means they all gave a different answer. Since they did not shake hands with themselves nor their spouses, the max number of shakes was 2n. So all numbers from 0 to 2n have been used as an answer.

1) what if she was the one that shook most hands, 2n as shown above? She did not shake hands with: herself, the master and the person that replied 0 . since the master isn't the person that replied 0. So she shook 2n+2-3 hands which isn't according to our hypotheses. She was not the person who shook the most hands (since someone who shook 2n hands has allready been shown to exist).

2) now look at the person that replied 2n, he didn't shake hands with himself (or herself, but i am not going to stress this all the time), his spouse and the person that replied 0. This shows that a: his spouse replied 0. b: he shook hands with the master, setting the min hands the master shook at 1. The person who replied 0 didn't shake hands with the master, setting the max hands the master shook at 2n-1. The person that replied 2n also shook hands with the person that replied 1.

3) The person that replied 2n-1 didn't shake hands with himself, his spouse, the person that replied 0 and the person that replied 1 (who shook hands with the person that replied 2n). This shows that a: his spouse replied 1 (since his spouse can't be the person that replied 0 who is allready married to the person that replied 2n). b: he shook hands with the master, setting the min hands the master shook at 2. The person who replied 1 didn't shake hands with the master, setting the max hands the master shook at 2n-2. The person that replied 2n-1 also shook hands with the person that replied 2, together with the person that replied 2n.

4) see how 3 is a bit of an inductive step. in the end you will get that the master shook hands with n people, his wife shook hands with n people and the person who shook hands with 2n-N people is married to the person that shook hands with N people, N going from 0 to n.

x = 1 + 1 + ..., x termes
multiply with x
=> x^2 = x + x + ...
derive
=> 2*x = 1 + 1 + ... = x
=> 2 = 1

It's the same value but different bases.
base 10: 100 -> 1*10^2+0*10^1+0*10^0 = 100 (in decimal)
base 9 :121 -> 1*9^2+2*9^1+1*9^0 = 100 (in decimal)
etc.

You can represent numbers in different ways, decimal (base 10), octal (base 8), hexadecimal (base 16), binary (base 2), trinary (base 3)
You can even try non-integer bases, but that's just silly.
I know, but I can't count in different bases.:(

I know, but I can't count in different bases.:(It's really easy. For a base b, instead of using a carry when you reach 10, use a carry when you reach b
so base 3
0, 1, 2 (2+1=3, so carry a 1 to the next column),
10, 11, 12 (carry again),
20, 21, 22 (carry for both both columns)
100, 101, 102
etc

And to find the value, wx..yz (n digits) = w * b^(n-1) + x * b^(n-2) + .. + y * b^1 + z

A nice brainbreaker for many is
0.999... = 1
(where the ... means an infinite repeat of the last digit. Let me stress that infinite is not finite. You'd think that's obvious, both most people still argue that it 0.999... has a finite number of trailing nines; despite there being infinitely many per definition)

Of course you can do it for every base b (where b is a positive integer > 1).
Take c=b-1 then 0.ccc... = 1

Here's a nasty one.

Any two positive numbers added together will be a bigger number. So, you could say:

x + y = z

Where x, y and z are all positive numbers. So, (where (r) means reccuring)

4.999 (r) + y = 5

So, it means 4.999 (r) plus what makes five? Logicaly the answer is 0.000 (r) 1, ie. an infinate number of zeros, then a one. It theoretacly proves that infinity is not truly infinate, but rather an incomprehendably titanicaly mind-bendingly huge number. Thus, assuming the number was a million (which is just a stupid example for illustrative purposes), numbers would count upwards thus:

999,998 999,999 1,000,000 -1,000,000 -999,999 -999,998... etc.

Meaning the flat graph sine, cosine and tangent graphs are drawn on are acualy spheres, and so the tangent graph is one continuous line. The only problem is, ten billion trillion raised to the power ten billion trillion doesn't even begin to get anywhere near a quintillionth of a percent of this number. It's very, very big.

If you really want to confuse your math teachers, there are less obvious tricks.
like http://www-math.mit.edu/~tchow/mathstuff/proof.pdf
Some math teachers won't even understand what it says..

In the first line of the proof there's a mistake. The function has an asymptote at y = 0. It tries to find the integral between 0 and 1. When 0 is fed into the definite integral it comes out with:

xe^(-x²/1) - xe^(-x²/0)

The lower limit is undefined, not zero as the proof states.

Here's a nasty one.

Any two positive numbers added together will be a bigger number. So, you could say:

x + y = z

Where x, y and z are all positive numbers. So, (where (r) means reccuring)

4.999 (r) + y = 5

So, it means 4.999 (r) plus what makes five? Logicaly the answer is 0.000 (r) 1, ie. an infinate number of zeros, then a one. It theoretacly proves that infinity is not truly infinate, but rather an incomprehendably titanicaly mind-bendingly huge number. Thus, assuming the number was a million (which is just a stupid example for illustrative purposes), numbers would count upwards thus:

999,998 999,999 1,000,000 -1,000,000 -999,999 -999,998... etc.

Meaning the flat graph sine, cosine and tangent graphs are drawn on are acualy spheres, and so the tangent graph is one continuous line. The only problem is, ten billion trillion raised to the power ten billion trillion doesn't even begin to get anywhere near a quintillionth of a percent of this number. It's very, very big.

1) 0.000 (r) 1 does not exist. 0.000 (r) means that the row of numbers behind the decimal point is (n-th number behind the point): 0 for every n>0, and the limit of the row n -> infinity = 0. No room is left after this since the limit to infinity is 0.

2) Even if we allowed 0.000(r) 1 to exist, 4.999 (r) would still equal 5. since x and y have to be > 0 for z to be > x and > y.
0.000 (r) 1 would have to be the limit of 1/10^n, since every finite approximation is of the form 1/10^n, and the approximation is best if n is bigger.
the limit of this series is 0 since for every epsilon>0 i can find an N for which:
|1/10^n - 0| = 1/10^n < epsilon for n >= N.
so 0.000(r) 1 = 0 and thus isn't >0.

1) 0.000 (r) 1 does not exist. 0.000 (r) means that the row of numbers behind the decimal point is (n-th number behind the point): 0 for every n>0, and the limit of the row n -> infinity = 0. No room is left after this since the limit to infinity is 0.

2) Even if we allowed 0.000(r) 1 to exist, 4.999 (r) would still equal 5. since x and y have to be > 0 for z to be > x and > y.
0.000 (r) 1 would have to be the limit of 1/10^n, since every finite approximation is of the form 1/10^n, and the approximation is best if n is bigger.
the limit of this series is 0 since for every epsilon>0 i can find an N for which:
|1/10^n - 0| = 1/10^n < epsilon for n >= N.
so 0.000(r) 1 = 0 and thus isn't >0.

Yes, that's mathermatically, and technicaly in the strictest sense of the words, true. But, if we cannot adapt and change mathermatical laws it wouldn't advance. Infinity as a number is so big that in every calcualtion done by mankind ever, there hasn't been an error big enough to notice it. It's only this theoretical imaginary maths that proves this, but theoretical and imaginary maths proved that pi existed, when before it was origionaly dismissed as rubbish.

In the first line of the proof there's a mistake. The function has an asymptote at y = 0. It tries to find the integral between 0 and 1. When 0 is fed into the definite integral it comes out with:

xe^(-x²/1) - xe^(-x²/0)

The lower limit is undefined, not zero as the proof states.

limit (xe^(-x^2/y), y = 0) is undefined (since left and right limit aren't equal, see below)

limit (xe^(-x^2/y), y =< 0) = sign(x) infinity (-x^2/y goes to + infinity here)

limit (xe^(-x^2/y), y => 0) = 0, (-x^2/y goes to - infinity here)

now in taking the integral, which isn't a normal Riemann integral, we take the following limit: limit(int(..., y = R ... 1), R=>0), so the lower limit is 0. we take the limit R=>0 since the interval in which we take the integral is [0,1], so we are only allowed to take into account values of y (and R) > 0.

If you don't accept it on this grounds: do the exact same integration for the following function:

f(y) =
x^3/y^2*e^(-x^2/y), y>=0
0, y<0

Now the problematic left limit (which went to infinity first) equals 0 too and you don't have any 'problems' left. See how integration over [0,1] is still exactly the same as in our first example.

edit: things change however when you take x = 0 :p everything above is only accurate for x != 0.

Yes, that's mathermatically, and technicaly in the strictest sense of the words, true. But, if we cannot adapt and change mathermatical laws it wouldn't advance. Infinity as a number is so big that in every calcualtion done by mankind ever, there hasn't been an error big enough to notice it. It's only this theoretical imaginary maths that proves this, but theoretical and imaginary maths proved that pi existed, when before it was origionaly dismissed as rubbish.

Well, if you change mathematical laws you'll find yourself trapped in an alternative math (note: it has been done, non-euclidean geometry for example, and proven to be usefull). Since limits are rather fundamental in a lot of areas, you'll have to start your alternative math from almost scratch. More importantly, i suspect it wont be consistant.

limit (xe^(-x^2/y), y = 0) is undefined (since left and right limit aren't equal, see below)

limit (xe^(-x^2/y), y =< 0) = sign(x) infinity (-x^2/y goes to + infinity here)

limit (xe^(-x^2/y), y => 0) = 0, (-x^2/y goes to - infinity here)

now in taking the integral, which isn't a normal Riemann integral, we take the following limit: limit(int(..., y = R ... 1), R=>0), so the lower limit is 0. we take the limit R=>0 since the interval in which we take the integral is [0,1], so we are only allowed to take into account values of y (and R) > 0.

If you don't accept it on this grounds: do the exact same integration for the following function:

f(y) =
x^3/y^2*e^(-x^2/y), y>=0
0, y<0

Now the problematic left limit (which went to infinity first) equals 0 too and you don't have any 'problems' left. See how integration over [0,1] is still exactly the same as in our first example.

edit: things change however when you take x = 0 :p everything above is only accurate for x != 0.


How old are you, what maths are you currently doing (year 12, first year uni,..), and your notation makes no sense to me, could you explain it in words possibly? And what exactly are you concluding?

How old are you, what maths are you currently doing (year 12, first year uni,..), and your notation makes no sense to me, could you explain it in words possibly?

Sorry, but typing math on a forum is hard, so i usually use maple notation.

In words: the integral is over the interval [0,1], so you may only take into account values of y > 0. this means that you have to take the right limit, not the limit.

to 'make it more clear' (i probably failed :() i added an example that illustrates that you have to take the right limit: take the same function for y>=0, but 0 for y<0. as you can see the integration of this function over [0,1] does not change at all and the left limit now equals the right limit, removing the 'problem'.

edit: i'm currently in my master years at university (engineering physics)

Sorry, but typing math on a forum is hard, so i usually use maple notation.

In words: the integral is over the interval [0,1], so you may only take into account values of y > 0. this means that you have to take the right limit, not the limit.

to 'make it more clear' (i probably failed :() i added an example that illustrates that you have to take the right limit: take the same function for y>=0, but 0 for y<0. as you can see the integration of this function over [0,1] does not change at all and the left limit now equals the right limit, removing the 'problem'.

I see. Thanks. :D

x = 1 + 1 + ..., x termes
multiply with x
=> x^2 = x + x + ...
derive
=> 2*x = 1 + 1 + ... = x
=> 2 = 1
You've missed derived it. Let f( x ) = x + x + ... + x (x times).
Then by the definition of the derivative: f' ( x ) = lim_(h -> 0) [ {f ( x + h ) - f ( x )} / h ]
So f'(x) = lim_(h-> 0 )[{(x + h) + (x + h) + ... + (x + h)} - {x + x + ... + x}]/h
^ x + h times ^ x times

So, grouping the x's and -x 's and cancelling we have
f'(x) = lim_(h ->0)[{h + h + . . . + h} + {(x + h) + (x + h) + ... (x + h)}]/h
^ x times ^ h times
= lim_( h -> 0) [ xh + h(x+h) ] / h
= lim_ (h -> 0 ) [ 2x + h ]
= 2x

So, where your argument fell down as the application of the derivative to a function of variable length. You assumed the lenght remained constant with the derivative, but it doesn't---it doubles.

If you really want to confuse your math teachers, there are less obvious tricks.
like http://www-math.mit.edu/~tchow/mathstuff/proof.pdf
Some math teachers won't even understand what it says..

Is the problem perhaps in the last line? A substitution of x = 0 is performed on the definite integral. I was under the assumption that a substitution can only be performed after the anti-derivative is found and the fundamental theorem of calculus is used.

Is the problem perhaps in the last line? A substitution of x = 0 is performed on the definite integral. I was under the assumption that a substitution can only be performed after the anti-derivative is found and the fundamental theorem of calculus is used.I'm not quite sure where the problem lies. I just googled for different kinds of 1=0 proofs :p
It's a bit suspect they first introduce a derivative, just to remove it once moved into the integral. My guess would be that the step from the second line to the third isn't allowed. (Actual, since the other steps seem alright, I'm quite sure the problem should be there)

I'm not quite sure where the problem lies. I just googled for different kinds of 1=0 proofs :p
It's a bit suspect they first introduce a derivative, just to remove it once moved into the integral. My guess would be that the step from the second line to the third isn't allowed. (Actual, since the other steps seem alright, I'm quite sure the problem should be there)

I'm pretty sure that it's the last line. I went through all the lines(properly I hope) and the last one is the only line that didn't make sense. Didn't they give you the answer on the site you found it on?

I'm pretty sure that it's the last line. I went through all the lines(properly I hope) and the last one is the only line that didn't make sense. Didn't they give you the answer on the site you found it on?Nope. No solution. But I can probably get one in a day if you really want one (I know a few very mathematically gifted people I could ask)

The last step makes sense though, although they skip over some steps in between. (It's basicly just (f*g)' = f'*g+f*g'.)

Nope. No solution. But I can probably get one in a day if you really want one (I know a few very mathematically gifted people I could ask)

The last step makes sense though, although they skip over some steps in between. (It's basicly just (f*g)' = f'*g+f*g'.)

That's what I figured. Your gifted people would help me sleep better at night if they got a solution. If you could ask them I'd appreciate it. :fluffle:

When 2x2 is divided by 9¼x3 + 5x2 - 15x + 64, carrying the sum through until the power -4, is the remainder divisible by 587?

And, assuming that a right angled triangle is used and that the remainder is 455 times the hypotenuse and another side is the factor by which is divides, what is the remaining side and all the three angles?

Does anyone know the answer yet? The works in for tomorrow, so if no-one gets it soon... I'll have to do it... myself! Noooooooo....

Edit: THis is to the problem involving the integral.

The answer is that the step you people suspected is invalid, is indeed invalid. To begin with the integral is improper. The improper integral above is convergent, but the derivative on the inside makes an improper integral that is divergent, and so not there. It's kind of like a divide by zero error in calculus.

I think this has been mentioned. If you want me to I can write it (my reasoning) out in some program (LaTeX I suppose) and post it here.

Edit: THis is to the problem involving the integral.

The answer is that the step you people suspected is invalid, is indeed invalid. To begin with the integral is improper. The improper integral above is convergent, but the derivative on the inside makes an improper integral that is divergent, and so not there. It's kind of like a divide by zero error in calculus.

I think this has been mentioned. If you want me to I can write it (my reasoning) out in some program (LaTeX I suppose) and post it here.

Plain english would help. I've never even heard of LaTex.

A nice brainbreaker for many is
0.999... = 1
(where the ... means an infinite repeat of the last digit. Let me stress that infinite is not finite. You'd think that's obvious, both most people still argue that it 0.999... has a finite number of trailing nines; despite there being infinitely many per definition)
That one's simple. So simple, I'll do it two ways.

First off, algebraically.

Let x=0.999... (etc)
10x=9.999...

take away x from both sides

9x = 9

therefore x=1
therefore 1=0.999...

Next, using sequences and series,

0.999... is the sum of a GP, with a = 0.9 and r = 0.1.
ie,

Tn = 0.9*0.1^(n-1)

Sn = 0.9(1-0.1^n)/1-0.1
= 1-0.1^n

As n approaches infinity, 1-0.1^n approaches 1.

Therefore it can be said that 0.999... = 1.

Okay, the last one doesn't quite cover it, but I wanted to see whether I still understand that part of maths.

Plain english would help. I've never even heard of LaTex.
Oh. LaTeX is just a mathematical typesetting program. It's handy for creating maths documents.

But actually, I was typing it out labouriously in an ASCII file when I realised I was wrong anyway. There's something fishy going on with the 'for all x' assumption. Because if x is zero that derivative on the first line is not what is on the LHS. It basically says that 1 = 0 in the first line... Or more accurately that the derivative of 0 is 1. So I guess the real flaw is that the derivative is not defined for all x as they assume.

Edit: No, wait, I'm wrong there too. There's no reason why it shouldn't be that way... I'm stumped for the moment.

If I actually evaluate the integral and then substitute in x=0 it works fine. I get the exact same answer (edit: the exact same answer was the left-hand side). So it must be that applying that substitution first is where the error is in the proof. I don't have any explanation as to why, yet.

As n approaches infinity, 1-0.1^n approaches 1.

Therefore it can be said that 0.999... = 1.

Okay, the last one doesn't quite cover it, but I wanted to see whether I still understand that part of maths.If you take the limit n to infinity, you're there.
Of course, many people won't be convinced by these proofs. (I've seen a thread about it on a puzzle forum go on for several hundreds of posts)

That one's simple. So simple, I'll do it two ways.
Actually you did it one way. The sum of a GP is basically found using the same method.

Now that I think about it yet again. If x were equal to zero that whole first line is invalid. So I guess there's the error. Phew, I knew I had it there. Seriously. Stick x=0 into the first line. The whole integral changes and the whole rest of it is out the door. Edit: so the exact error is assuming that solution 'for all x'. For x=0 the first line is 0, the derivative of which is zero, and it's all happy... I believe. Hrm, no I'm still not convinced. I'll just wait to see what these smart people say.

Edit: Wow, I'm going around in circles. I had that same answer just half an hour ago and dismissed it. And now I realised I'm wrong again.

Edit 2: Yes, it is in the last line I suppose. If you're fixing x, how the heck can you take the derivative of it? Derivatives are only of variables. That must be it. But an explanation still need more rigour. Back to waiting for the smart folk. :)

I'm quite sure the error is bringing the derivative into the integral.
The first line is easy to proof. Writing out the derivative on the right hand side gives the left hand side. You can't substitute x=0 before applying the derivative anyway (otherwise it would either leave the d/dx, giving an incomplete subtitution; or change it to d/d0 which has no meaning).
The second step is simply substituting the given equivalence at the top. So the problem shouldn't lie there either.
The third step seems iffy to me, but I can't quite put my finger on why yet.
The fourth step is perfectly legal again, it's just writing out the partial derivative. You can bring x out of the integral, and see that for x=0, the whole will be 0. (Regardles of what the rest of the integral is).

I'm trying to find an easier example to show that you can't just bring a differentiation from outside the integral into it. But it always ends up just as complicated :(

I'm quite sure the error is bringing the derivative into the integral.

If you evaluate the integral and then do the substitution it does work though. And I can think of no reason why it should matter, both treat the other as constants. Here:


3x^2 x^3 2x
RHS = INT[ ----.exp{ -(x^2)/y} + ---.--.exp{ -(x^2)/y} dy] from 0 to 1
y^2 y^2 y

3x^2 2x^4
= INT[ ----.exp{ -(x^2)/y} + ----.exp{ -(x^2)/y} dy] from 0 to 1
y^2 y^3

Then we evaluate the integral:

2x^4
= lim [ 3.exp{ -(x^2)/y} - INT {----.exp{ -(x^2)/y} dy} ] from y=t to y=1
t->0+ y^3

2.x^2 x^2
= lim [ 3.exp{ -(x^2)/y} - -----.exp{-(x^2)/y} - 2 INT { ----.exp{ -(x^2)/y} dy} ] from y=t to y=1
t->0+ y y^2

2.x^2
= lim [ 3.exp{ -(x^2)/y} - -----.exp{-(x^2)/y} - 2.exp{ -(x^2)/y} ] from y=t to y=1
t->0+ y

The second term when evaluated using L'Hospitals rule gives:

(x^2).(1/t) -(x^2)(1/t^2) x^2
lim [ ------------ ] = lim --------------------- = lim ------------- = 0
t->0+ exp{(x^2)/t} t->0+ -(1/t^2).exp{(x^2)/t} t->0+ exp{(x^2)/t}

Thus:

RHS = 3.exp{-x^2} - 2(x^2).exp{-(x^2)/y} - 2.exp{-x^2}
= exp{-x^2} - 2(x^2)exp{-(x^2)/y}
= LHS

My friends seem to agree the problem is in the third step. And probably it's because Leibnitz rule is broken.
(http://ca.geocities.com/pezeshki8005@rogers.com/STAT3502/Leibnitz.pdf)

(But they don't all agree yet :p )

Can we have some more fun maths puzzles now?

I walk 5 miles north from a point, then 1 mile south. What is the maximum distance from the point I can have moved?

Something short but 'fun' like that. PLEASE!!!

x = 1 + 1 + ..., x termes
multiply with x
=> x^2 = x + x + ...
derive
=> 2*x = 1 + 1 + ... = x
=> 2 = 1

The first line tells us that x must be an integer (at least, some would argue it must be a natural as well) as all terms must be complete.

The difficulty comes when attempting to take the derivitive of a sequence of integers. Remember that the derivitive is essentially finding the slope of a continuous function. Since x is not continuous, taking the slope is meaningless.

The first line tells us that x must be an integer (at least, some would argue it must be a natural as well) as all terms must be complete.

The difficulty comes when attempting to take the derivitive of a sequence of integers. Remember that the derivitive is essentially finding the slope of a continuous function. Since x is not continuous, taking the slope is meaningless.No that wasn't the problem. In fact, there is no problem, the derivative works perfectly. See my earlier post (http://forums.jolt.co.uk/showpost.php?p=9971720&postcount=123). The problem is the blind application of the derivative rules---it doesn't take into account the variable nature of the length of the expression.

Edit: Although, you are technically right. You can still approximate x with a continuous function and find the derivative satisfactorially.

My friends seem to agree the problem is in the third step. And probably it's because Leibnitz rule is broken.
(http://ca.geocities.com/pezeshki8005@rogers.com/STAT3502/Leibnitz.pdf)

(But they don't all agree yet :p )
I'm not sure that is the case. Notice that in our case the end points are constants, so applying Leibnitz rule the derivative of beta=1 and alpha=0 with respect to t will be zero, so those terms out the front vanish (leaving us with the rule as given). f=1 and f=0 are continuously differentiable functions, so they don't violate that requirement. So Leibnitz rule demonstrates that that step is valid.

I'm 95% sure it's something to do with setting x=0. I did demonstrate above that evaluating that last integral does give the solution on the LHS, so there's something fishy about putting that x=0 in there before evaluating the integral. I'm just not sure what.

If you really want to confuse your math teachers, there are less obvious tricks.
like http://www-math.mit.edu/~tchow/mathstuff/proof.pdf
Some math teachers won't even understand what it says..I did the final integral using the substitution u = x^2 / y and I ended up with exp(-x^2) * (1 - 2x^2). That leaves the substitution of 0 for x in the integral on the very last step as the problem.

I did the final integral using the substitution u = x^2 / y and I ended up with exp(-x^2) * (1 - 2x^2). That leaves the substitution of 0 for x in the integral on the very last step as the problem.
You agree with me! :D

I'm not sure that is the case. Notice that in our case the end points are constants, so applying Leibnitz rule the derivative of beta=1 and alpha=0 with respect to t will be zero, so those terms out the front vanish (leaving us with the rule as given). f=1 and f=0 are continuously differentiable functions, so they don't violate that requirement. So Leibnitz rule demonstrates that that step is valid.

I have to admit it's all a bit over my head, so I can only rely on the people I've learned to trust in these matters.
to quote:
Approach (0+, 0+) by the curve (t, t^2). The partial derivative is unbounded. Approach instead by the curve (0, t), the partial derivative is zero. Therefore it is not continuous at (0, 0), and Leibnitz's rule does not apply.

But suffice it to say, the problem is a lot more confusing then most other 1=0 proofs :D

I did the final integral using the substitution u = x^2 / y and I ended up with exp(-x^2) * (1 - 2x^2). That leaves the substitution of 0 for x in the integral on the very last step as the problem.So then you're saying it just isn't 0 as claimed?

I'm not convinced. Maybe I should try to solve that integral myself :\

Well ok, it seems to work the same way if I do it too :(

However, isn't the substitution u=x^2/y only valid if x is not 0?

Can we have some more fun maths puzzles now?What is 1-1+1-1+1-1.....
Or in other notation sum (-1)^i for i=0 to infinity

And how about 1-2+4-8+16-32... i.e. sum (-2)^i for i=0 to infinity

I have to admit it's all a bit over my head, so I can only rely on the people I've learned to trust in these matters.
to quote:


But suffice it to say, the problem is a lot more confusing then most other 1=0 proofs :D
Yes. Hrm, I'll have to think about it some more. It's been awhile since I've done any analysis, so I'm a bit rusty. Your friends are probably right.

However, isn't the substitution u=x^2/y only valid if x is not 0? [/edit]For now, I'm going with that guess.