Mathematics Puzzles (Please spoiler answers)

[AshDesert]

So, I recently saw one of those equations that have one flaw that's hard to spot that ends up with a logical fallacy, and I was inspired to start a thread here. I figure that this is a decently nerdy forum, so I'm hoping for a good response. Anyways, here's one of the easier ones, see if you can spot the step that breaks the rules.

Also, please spoiler your answers for those of us who want to struggle:smallbiggrin:.

X=Y (multiply both sides by X)
X^2=XY (subtract Y^2 from both sides)
X^2 - Y^2=XY - Y^2 (factor both sides)
(X+Y)(X - Y)=Y(X - Y) [divide both sides by (X - Y)]
Therefore, (X+Y)=Y

If X=1, and Y=1, then
(1+1)=1

[Moff Chumley]

Stumped me. *shrug*

[Cobra_Ikari]

You divided by zero and broke the universe.

[Kyace]

Solution to the first post.
If X=Y then (X-Y) is zero hence you divided by zero on the fourth step, resulting in an undefined result.

Not really a 'puzzle' but here is another common brain tickler:
If 1/3 = .3333... (take ... to mean repeats infinitely times) and
and if 3 * 1/3 = 1
then does 1 = .9999999... ?

The answer below:
Yes, 1 = .99999... are mathematically equal.
A way to help this make sense is if you view
1.00 - .90 (nonrepeating) = .10
1.000 - .990 = .010
1.00000 - .99990 = .00010
1.00000... - .99999... = 0.0000...
We never hit the ...90 at the end for the ...10 at the end of the solution.

[Moff Chumley]

...eh. It's complicated. Not really a clearly defined answer for that one.

[AshDesert]

Solution to the first post.
If X=Y then (X-Y) is zero hence you divided by zero on the fourth step, resulting in an undefined result.

Not really a 'puzzle' but here is another common brain tickler:
If 1/3 = .3333... (take ... to mean repeats infinitely times) and
and if 3 * 1/3 = 1
then does 1 = .9999999... ?

The answer below:
Yes, 1 = .99999... are mathematically equal.
A way to help this make sense is if you view
1.00 - .90 (nonrepeating) = .10
1.000 - .990 = .010
1.00000 - .99990 = .00010
1.00000... - .99999... = 0.0000...
We never hit the ...90 at the end for the ...10 at the end of the solution.

Answer is correct.

Also, on the second one, there is a theoretical end, so, no, 1 =/= 0.99999..... in theoretical terms. But, for practical purposes, you will never get to the end of the decimal, so yes, 1 = 0.99999......

[The Dark Fiddler]

Ah, my Algebra teacher tried to get me with this. Can't trick me again.

If x=y then x-y=0 so you can't divide by x-y.

[SensFan]

Answer is correct.

Also, on the second one, there is a theoretical end, so, no, 1 =/= 0.99999..... in theoretical terms. But, for practical purposes, you will never get to the end of the decimal, so yes, 1 = 0.99999......
Actually, for the second one, the answer is theoretically correct. For the purposes of mathematics, 0.99999999999... is equal to 1.

Proof:
Let n=0.99999999999999999...
Then 10n=9.999999999999999....
10n-n=9n
9.999999999999999...-0.9999999999999...=9
9n=9
n=1

Therefore, 0.99999999999999...=1

[AshDesert]

Actually, for the second one, the answer is theoretically correct. For the purposes of mathematics, 0.99999999999... is equal to 1.

Proof:
Let n=0.99999999999999999...
Then 10n=9.999999999999999....
10n-n=9n
9.999999999999999...-0.9999999999999...=9
9n=9
n=1

Therefore, 0.99999999999999...=1

Ah infinity... the place where even mathematical theory has to round!

[RS14]

(3) Prove that for every positive integer n, there some multiple of n which contains each of the digits {0,1,2,3,4,5,6,7,8,9} at least once.

Edit: Here is another, perhaps easier one.

(4) Suppose I have a d4. I roll it until it lands on 4. What is the average number of rolls this will take?

[The Extinguisher]

Actually, for the second one, the answer is theoretically correct. For the purposes of mathematics, 0.99999999999... is equal to 1.

Proof:
Let n=0.99999999999999999...
Then 10n=9.999999999999999....
10n-n=9n
9.999999999999999...-0.9999999999999...=9
9n=9
n=1

Therefore, 0.99999999999999...=1

There's a much better proof.

0.(9) = 9/10 + 9/100 + 9/1000 + ...
= sum of 9/(10^k) as k gets bigger
= lim(n-->∞) (n)Σ(k=1) 9/(10^k)
= lim(n-->∞) (1 - 1/(10^n))
= 1 - lim(n-->∞) (1/(10^n))
= 1 - 0
= 1

(and to steal wikipedia's picture because it's easier to read)
http://upload.wikimedia.org/math/6/f/a/6fa510b44742046a167b4b8515162825.png

[RS14]

(5) Shamelessly taken from Affine Mess: Given 51 distinct integers between 1 and 100, prove that two of them are relatively prime.

Edit: I believe this property also holds for 27 distinct integers, actually. No, see below.

There is a very nice proof for 51 distinct integers.

[Deathslayer7]

everybody freeze! nobody move!

*points to sig*

we don't want a repeat of what happened last time do we. :smallannoyed:

[Kyace]

*eyes dart side to side*
"Hey look, the Monty Hall problem!"
*runs away*

[thubby]

*eyes dart side to side*
"Hey look, the Monty Hall problem!"
*runs away*

*googles* wait, what, how? :smalleek:

[Tirian]

(5) Shamelessly taken from Affine Mess: Given 51 distinct integers between 1 and 100, prove that two of them are relatively prime.

Edit: I believe this property also holds for 27 distinct integers, actually. But there is a very nice proof for 51 distinct integers.

I agree with your belief. I suspect they meant to say that given 51 integers between 1 and 100, at least one of them is a multiple of another. That would require 51 members.

[RS14]

I agree with your belief. I suspect they meant to say that given 51 integers between 1 and 100, at least one of them is a multiple of another. That would require 51 members.

Actually, on second thought, I was reversing it. :smallredface:

Given 27 distinct integers between 1 and 100, there must be two that are not relatively prime.

But that's not the question.

[Tirian]

Ah, indeed. Then, yes, 51 numbers.

Two more combinatorics problems, just because I lurves them so. Either can be solved by two full pages of formulas or a well thought-out few sentences.

(6) John and Sue are flipping a bunch of coins simultaneously. John flips x, Sue flips x+1. What is the probability that Sue gets more heads than John?

(7) Walter and Laura are running for president of their club, where people vote by putting a piece of paper into a fishbowl. The ballots are tallied by pulling them out of the fishbowl one by one and reading them aloud. Given that Walter will win the election by a score of W-L, what is the probability that he has a strict lead all the way through the tally?

[RS14]

(6) John and Sue are flipping a bunch of coins simultaneously. John flips x, Sue flips x+1. What is the probability that Sue gets more heads than John?



Suppose Sue flips one coin before all else. Two cases:

Suppose it is tails. Then each flip x coins, and we desire the probability that Sue has more heads than John in this set of x, call this P(S).

Suppose it is heads. Then each flip x coins. We desire the probability that John has more heads than Sue in this set of x. By symmetry, this is P(S). Then the probability that in this set of x, Sue has an equal or greater number of heads is 1-P(S). So in this case, Sue has more (total) heads with probability 1-P(S).

Thus overall, Sue has more heads with probability 1/2*(P(S)+[1-P(S)])=1/2

I like it.

[Tirian]

That was the first cool proof I came up with, which made me think that the problem was cool. Then I came up with an even cooler proof.

Sue either has more heads or more tails than John, but cannot have both. Therefore, by symmetry, the probability that she has more heads is 1/2.

[Cyrion]

(8) Suppose that you were to wrap a string around the equator of the earth. Now, lengthen that string by six feet. How far off of the surface of the earth, all the way around, could you stretch the ring of string?

What if you were to repeat the experiment with an orange- how far off of the surface of the orange would the string be?

The answer, though not the proof of why it's true, is:

In both cases, the string would be 1 foot (rounded slightly) off of the surface of the sphere. The size of the sphere does not matter; this would also hold true of a basketball, a marble, etc.

[daggaz]

(8) Suppose that you were to wrap a string around the equator of the earth. Now, lengthen that string by six feet. How far off of the surface of the earth, all the way around, could you stretch the ring of string?

What if you were to repeat the experiment with an orange- how far off of the surface of the orange would the string be?

The answer, though not the proof of why it's true, is:

In both cases, the string would be 1 foot (rounded slightly) off of the surface of the sphere. The size of the sphere does not matter; this would also hold true of a basketball, a marble, etc.

Wrong. I wont even bother to prove it.

[Hannes]

What do you get when you multiply six by nine?

[snoopy13a]

So, I recently saw one of those equations that have one flaw that's hard to spot that ends up with a logical fallacy, and I was inspired to start a thread here. I figure that this is a decently nerdy forum, so I'm hoping for a good response. Anyways, here's one of the easier ones, see if you can spot the step that breaks the rules.

Also, please spoiler your answers for those of us who want to struggle:smallbiggrin:.

X=Y (multiply both sides by X)
X^2=XY (subtract Y^2 from both sides)
X^2 - Y^2=XY - Y^2 (factor both sides)
(X+Y)(X - Y)=Y(X - Y) [divide both sides by (X - Y)]
Therefore, (X+Y)=Y

If X=1, and Y=1, then
(1+1)=1

I figure that if you plug in X=1 and y=1 and divide by (x-y) that you are dividing by zero.

[Douglas]

Wrong. I wont even bother to prove it.
No, he's quite right. More precisely, the string will be 6/(2pi) feet off the surface, or about 11.46 inches.

Feel free to try it yourself with a marble, orange, basketball, cup, or any other small round object you like. Wrap a string around it, measure the length, get a piece of string 6 feet longer, and arrange it in a circle with the object at the center. There will be almost a foot of space between the string and the object's outer edge.

[Cyrion]

No, he's quite right. More precisely, the string will be 6/(2pi) feet off the surface, or about 11.46 inches.

Feel free to try it yourself with a marble, orange, basketball, cup, or any other small round object you like. Wrap a string around it, measure the length, get a piece of string 6 feet longer, and arrange it in a circle with the object at the center. There will be almost a foot of space between the string and the object's outer edge.

Indeed. The scary thing is that I originally saw this as part of a kids' quiz show, and one of the kids knew the answer immediately.

[PirateMonk]

What do you get when you multiply six by nine?

Base 10: 54
Base 11: 4A
B12: 46
B13: 42
B14: 3C
15: 39
16: 36
17: 31
18: 30
19: 2G
20: 2E
21: 2C
22: 2A
23: 28
24: 26
25: 24
26: 22
Note: No one makes jokes in base thirteen.

[littlebottom]

Not really a 'puzzle' but here is another common brain tickler:
If 1/3 = .3333... (take ... to mean repeats infinitely times) and
and if 3 * 1/3 = 1
then does 1 = .9999999... ?

The answer below:
Yes, 1 = .99999... are mathematically equal.
A way to help this make sense is if you view
1.00 - .90 (nonrepeating) = .10
1.000 - .990 = .010
1.00000 - .99990 = .00010
1.00000... - .99999... = 0.0000...
We never hit the ...90 at the end for the ...10 at the end of the solution.

may i just say,by that logic. that if you took the mass within the universe and devided it by the space in the universe (infinity) then you end up with a number infinately small, so small it may as well be assumed to be zero, therefore we dont exist.

[Lamech]

(4)[/B] Suppose I have a d4. I roll it until it lands on 4. What is the average number of rolls this will take? Well... the chance of having at least one is .75^0, the chance of having at least 2 is .75^1 the chance of having at least three is .75^2. Of having at least N is .75^N+1 A now we add up the probablity of having one, having two having three. 1+.75+.75^2... this is a geometrict series that sums too 4.

The answer is 4.

[Moff Chumley]

What do you get when you multiply six by nine?

The meaning of life.

[Cicciograna]

may i just say,by that logic. that if you took the mass within the universe and devided it by the space in the universe (infinity) then you end up with a number infinately small, so small it may as well be assumed to be zero, therefore we dont exist.

There is a flaw in your logic. Universe is not infinite.

[Keris]

may i just say,by that logic. that if you took the mass within the universe and devided it by the space in the universe (infinity) then you end up with a number infinately small, so small it may as well be assumed to be zero, therefore we dont exist.
That comes up in the Hitchhiker's trilogy, during a Guide entry.

Population [of the Universe]: None.

It is known that there are an infinite number of worlds, simply because there is an infinite amount of space for them to be in. However, not every one of them is inhabited. Therefore, there must be a finite number of inhabited worlds. Any finite number divided by infinity is as near to nothing as makes no odds, so the average population of all the planets in the Universe can be said to be zero. From this it follows that the population of the whole Universe is also zero, and that any people you may meet from time to time are merely the products of a deranged imagination.

[littlebottom]

That comes up in the Hitchhiker's trilogy, during a Guide entry.

that kinda was what i was going at, but i couldnt remember it to that much detail :smalleek:

[RS14]

That comes up in the Hitchhiker's trilogy, during a Guide entry.

It is known that there are an infinite number of worlds, simply because there is an infinite amount of space for them to be in. However, not every one of them is inhabited. Therefore, there must be a finite number of inhabited worlds.


And herein lies the a flaw.

By analogy, note that there are an infinite number of integers. But not all of them are even. This does not, however, imply that the even integers are finite in number.

[lesser_minion]

An interesting problem I remember seeing once was this:

Imagine we're playing a game. In this game, we will each pick a sequence of three possible coin flips. I will then flip a coin repeatedly, and when one of the sequences we picked occurs, that person is declared the winner.

Given that I picked TTH, what would you pick, and why?

If it's helpful, here are the possible combinations:

- TTT

- TTH

- THT

- HTT

- THH

- HTH

- HHT

- HHH

[Generic Archer]

Is HHT considered different or the same as HTH?
Because that changes the question quite considerably

Dane

[Tirian]

It is different. Makes it a very interesting puzzle. In fact, it might be that if you were forced to pick a four-flip sequence in response to the first player's three-flip sequence that you still have an advantage, but I'm less certain on that point.

Just to clear out some of the unsolved backlog...


(3) Prove that for every positive integer n, there some multiple of n which contains each of the digits {0,1,2,3,4,5,6,7,8,9} at least once.

This is far larger a number than necessary, but it requires less thought than more subtle arguments. :smallbiggrin:
Let k be the leading digit of n. Make a list of all the multiples of n from 100n through 1000n. Obviously, the digit 0 appears in the first one, and the leading digit will cycle through all of the digits from k through 9 , back to 1, and then up to k again. Therefore, each of the digits appears in our list. But simply concatenating these numbers together will also produce a multiple of n that contains each digit from 0 through 9, just like 3926 is trivially a multiple of 13.

[littlebottom]

An interesting problem I remember seeing once was this:

Imagine we're playing a game. In this game, we will each pick a sequence of three possible coin flips. I will then flip a coin repeatedly, and when one of the sequences we picked occurs, that person is declared the winner.

Given that I picked TTH, what would you pick, and why? Could I have found something that wouldn't have given you an advantage?

If it's helpful, here are the possible combinations:

- TTT

- TTH

- THT

- HTT

- THH

- HTH

- HHT

- HHH


if i remember rightly the answer is HTT because you take the middle one, reverse it, and put it at the front, removing the end result

i believe

[lesser_minion]

if i remember rightly the answer is HTT because you take the middle one, reverse it, and put it at the front, removing the end result

i believe

That's right - unless the first two coin flips are both tails (in which case I win), HTT will always come up before TTH.