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Author Topic: Maths thread.  (Read 307254 times)

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Witherweaver

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Re: Maths thread.
« Reply #350 on: March 18, 2015, 09:37:41 pm »
0

Assuming you mean as x (rather than y) goes from 1 to 2, then yes, shells is a really awkward method to use for that problem.  The height will be a piecewise function in terms of y (and it needs to be in terms of y, since it's shells and it's going around the x-axis), so you'll need to do two separate integrals.

Going from y=0 to y=1/2, the height will just be a constant 1 (2 minus 1).  Then going from y=1/2 to y=1, the height will be 1/y (the x-value) minus 1, because the total distance from the graph to the line x=1 is the same as the distance from the graph to the y-axis (which is the x-value, 1/y), minus the distance from x=1 to the y-axis (which is just 1).  The radius for both parts will just be y.

Therefore, you want to do (2pi)*[(integral from 0 to 0.5 of y)+(integral from 0.5 to 1 of y*((1/y)-1))].

We pretty much got this, but were stupid and thought a few things were wrong but somehow got the right answer (also, doing it piecewise just seems weird).

Also, Witherweaver, we drew pictures like crazy trying to figure things out.

Hmm.. the pictures should make it clear that you have to break it up.  There are two distinct cases: one when the cylindrical shell's "height" is x=1/y-1, and one when it's "height" is x=2-1. 
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Re: Maths thread.
« Reply #351 on: March 18, 2015, 10:03:10 pm »
0

Assuming you mean as x (rather than y) goes from 1 to 2, then yes, shells is a really awkward method to use for that problem.  The height will be a piecewise function in terms of y (and it needs to be in terms of y, since it's shells and it's going around the x-axis), so you'll need to do two separate integrals.

Going from y=0 to y=1/2, the height will just be a constant 1 (2 minus 1).  Then going from y=1/2 to y=1, the height will be 1/y (the x-value) minus 1, because the total distance from the graph to the line x=1 is the same as the distance from the graph to the y-axis (which is the x-value, 1/y), minus the distance from x=1 to the y-axis (which is just 1).  The radius for both parts will just be y.

Therefore, you want to do (2pi)*[(integral from 0 to 0.5 of y)+(integral from 0.5 to 1 of y*((1/y)-1))].

We pretty much got this, but were stupid and thought a few things were wrong but somehow got the right answer (also, doing it piecewise just seems weird).

Also, Witherweaver, we drew pictures like crazy trying to figure things out.

Hmm.. the pictures should make it clear that you have to break it up.  There are two distinct cases: one when the cylindrical shell's "height" is x=1/y-1, and one when it's "height" is x=2-1.

Yeah, I got the breaking it up, it just felt weird to do.  And obsessing over the pictures I got mostly everything, but through some stupid mistakes I thought it was wrong.
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   Quote from: sudgy on June 31, 2011, 11:47:46 pm

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Re: Maths thread.
« Reply #352 on: March 19, 2015, 05:47:06 am »
0

(also, doing it piecewise just seems weird).

What do you understand by the word "function"?  One of the key realisations people go through is that it's any rule that, given any number, gives you another number in return; it's not just any such rule that can be expressed by a nice formula.  So the function that is 1 on [0,1] and 1/x for x in [1,100] is still a function, even if you can't write it without using some sort of case distinction.
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Re: Maths thread.
« Reply #353 on: March 19, 2015, 12:01:31 pm »
0

(also, doing it piecewise just seems weird).

What do you understand by the word "function"?  One of the key realisations people go through is that it's any rule that, given any number, gives you another number in return; it's not just any such rule that can be expressed by a nice formula.  So the function that is 1 on [0,1] and 1/x for x in [1,100] is still a function, even if you can't write it without using some sort of case distinction.

You usually don't need to solve it piecewise for these types of problems, so it felt weird because of that.
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   Quote from: sudgy on June 31, 2011, 11:47:46 pm

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Re: Maths thread.
« Reply #354 on: March 19, 2015, 05:26:04 pm »
0

I checked the logic puzzles thread first this time, and I don't think this one has been done yet (though there was something similar).  Two warm-up questions first.

1. You have three coins, which all look identical.  One is a counterfeit, which weighs more than the others (but you don't know by how much); the other two weigh the same.  You have a balance, so you can put any number of coins on each side and see which group weighs more (or if they weigh the same).  How many weighings do you need to do to find the counterfeit?

2. Same as #1, except now you have nine coins with one counterfeit.

3. Now you have nine coins with one counterfeit, and three balances, one of which is demon-possessed (you don't know which one is demon-possessed).  The demon-possessed balance will always give you a result to try to trick you, so it may give you the correct result, or an incorrect result.  How many weighings are needed in the worst case scenario to find the counterfeit?
No, it is five.  I just didn't think you usually made moat jokes.  Well really, I just didn't think about it until I actually read the spoiler.
I think it is possible to do it with 4 weighings.

Let's introduce a nomenclature to make talking about the results easier. The results the ith measurement indicate that the coins ri could be the counterfeit and not ri do not agree with the result (they could still be the counterfeit due to a lying demonic balance).
After two measurements on different balances B1 and B2 of 3 coins in each bowl, with a clever shift after the first result, we obtain 5 cases.
a) (r1,r2) a coin that agrees with first and second result
b) (not r1,r2) two coins that agree with the second but not with the first result (for this to be true, the B1 must be demonic)
c) (r1,not r2) two coins analogous to b) that agree with the first result (B2 demonic).

As a third measurement, we weigh the coins of b) vs. the ones of c) on the third balance B3. There are three cases:
i) a) is the solution and B3 doesn't lie => trivial
ii) a) is the solution, but B3 pretends that one side of b/c is heavier
iii) either B1 or B2 lied, and B3 correctly identifies which side is heavier

So now we want to do a fourth measurement that differentiates between ii) and iii). Side b) heavier is compatible with two possibilities (not r1,r2|B1 demon) and (r1,r2|B3 demon). Hence, we know that B2 is not demonic if the two b) coins are heavier, so we weigh them on that balance. If they have the same weight (r1,r2) is the counterfeit, otherwise it is the heavier coin. The alternative that side c) was heavier follows analogously.
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Re: Maths thread.
« Reply #355 on: March 19, 2015, 08:12:44 pm »
+1

The friend who gave me that problem never gave me the answer, so it might be possible to do it in 4.  I don't really understand what you're saying though.  Here's how I got 5:

First, weigh ABC against DEF on scales 1 and 2.  If they give a different result, then we know that the third scale is safe, so we just do it again on the third scale, and then whichever group has the counterfeit, test again on the third scale, for 4 total weighings.

The trickier case is if scales 1 and 2 give the same result.  Now any of the scales could be demon-possessed still.  But, we know that the result that the first two scales gave us must be correct.  Therefore, we can choose the group of 3 coins that has the counterfeit in it and test on all three scales to see which one is the counterfeit, for a total of 5 weighings.  (Note that, when we start weighing among the group of three coins which must contain the counterfeit, we still don't know which scale is demon-possessed, so no matter which two scales we test on first, we run into trouble if we get different results, and need to check with the last scale, which is why I think the fifth weighing is required.)


If it is possible to do it in fewer than five weighings, I think you need to try different sets of coins in the first set of 2-3 weighings, but that feels like it would be a lot less efficient (and a lot harder) to me.  That might be what you were trying to describe, but I'm having a lot of trouble following your post.
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Re: Maths thread.
« Reply #356 on: March 19, 2015, 08:57:36 pm »
+3

The first two weighings are ABC vs DEF and ADG vs BEH on two different scales. If the scales were honest, this would tell you exactly which coin is false. With the mapping 1 = left, 2 = right and 3 = equal, we get the following table
11 = A
12 = B
13 = C
21 = D
22 = E
23 = F
31 = G
32 = H
33 = I
Now, we have to take into account that one of the scales may be lying. This leads then to reduction to 5 coins. If you observe 11 for example, the results can be
a) both scales tell the truth => A
b) scale B1 is lying, the true result should be 21 or 31 (D, G)
c) scale B2 is lying, the true result should be 12 or 13 (B, C)
Now we test the BC vs DG on scale 3. The trivial case is that the results are the same. This is only possible, if the counterfeit is A.

If we notice a difference in weight, we know that either B3 is lying (the new result), or one of the first two measurements is wrong. However, D and G are only possible when B1 is lying and vice versa B and C are only possible when B2 is lying. Hence, we know 1 scale that is not demonic, which we use then to test D vs G or B vs C depending on the result of the third measurement. As the scale gives the true result, we can identify the counterfeit. If there is a weight difference it's the heavier coin, otherwise B3 was lying and the original measurement (A) is the counterfeit.
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Re: Maths thread.
« Reply #357 on: March 19, 2015, 09:12:37 pm »
0

Very nice.
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Re: Maths thread.
« Reply #358 on: March 20, 2015, 05:28:03 pm »
+2

Alright, here's an interesting problem:

What is the sum from n = 1 to infinity of s(n)/((n)(n+1))?
Note that s(n) is the sum of the digits of n.

Alright then, my solution:

The first thing we want to do is rewrite s(n)/((n)(n+1)) as s(n)/(n) - s(n)/(n+1).
Now, we are going to evaluate the sum by looking at each digit individually.
If we let t(n) = the last digit of n, we can evaluate the sum of t(n)/(n) - t(n)/(n+1) as follows:

T = 1/1 - 1/2 + 2/2 - 2/3 + 3/3 - 4/3 + ... + 8/8 - 8/9 + 9/9 - 9/10 + 0/10 - 0/11 + 1/11 - 1/12 + ...
   = 1/1 + 1/2 + 1/3 + ... + 1/8 + 1/9 + 1/10 - 1 + 1/11 + 1/12 + ... + 1/19 + 1/20 - 1/2 + 1/21 + ...
   = limit as k approaches infinity of (1 + 1/2 + 1/3 + ... + 1/(10k)) - (1 + 1/2 + 1/3 + ... + 1/k).

It is well known that 1/1 + 1/2 + 1/3 + ... + 1/n - ln(n) converges to a constant for large n (Notice that H_n - ln(n) is monotonically decreasing but always positive).
So, our limit is equal to ln(10k) - ln(k) = ln(10), and T = ln(10).

Now, consider the other digits. If u(n) is the second to last digit of n, then the sum of u(n)/(n) - u(n)/(n+1) as follows:

U = 1/10 - 1/11 + 1/11 - 1/12 + 1/12 - 1/12 + ... + 1/18 - 1/19 + 1/19 - 1/20 + 2/20 - 2/21 + ... + 9/99 - 9/100 + 1/110 + ...
   = 1/10 + 1/20 + 1/30 + ... + 1/90 + 1/100 - 1/10 + 1/110 + ...
   = T/10

If we define similar values for V, W, etc. we find that each term is one tenth of the previous. So, we can evaluate a geometric series to find that the sum we want is ln(10)/(1 - 1/10) = 10ln(10)/9.

I believe this problem was originally posed by O. Shallit.
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Re: Maths thread.
« Reply #359 on: April 16, 2015, 07:12:58 am »
0

There's that Albert, Bernard, Cheryl puzzle going around. Someone shared a harder one that doesn't seem to have a solution. I modified it a bit, and I think it should have a single solution now:

Albert and Bernard, two nerdy logic geniuses, have just become friends with Cheryl, a girl with a very peculiar sense of humour. They ask her when her birthday is, but instead, she tells Albert the sum of the month and the day of her birthday, and she tells Bernard the product, then she scurries off to flirt with a jock while the other two try to outnerd each other.

Albert: I don't know when Cheryl's birthday is, but I know that Bernard does not know too.

Bernard: Not only didn't I know when Cheryl's birthday was before, but I still don't know now, and I don't even know if we will ever be able to know.

Albert: Even after what you just said, I still don't know if we will ever be able to know either.

Bernard: Same here.

Albert: Now I know Cheryl's birthday!

Can you deduce Cheryl's birthday too?
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Re: Maths thread.
« Reply #360 on: April 16, 2015, 08:19:50 am »
+2

Cheryl's birthday is on Smarch 33.
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Re: Maths thread.
« Reply #361 on: April 16, 2015, 10:38:48 am »
+2

There's that Albert, Bernard, Cheryl puzzle going around. Someone shared a harder one that doesn't seem to have a solution. I modified it a bit, and I think it should have a single solution now:

Albert and Bernard, two nerdy logic geniuses, have just become friends with Cheryl, a girl with a very peculiar sense of humour. They ask her when her birthday is, but instead, she tells Albert the sum of the month and the day of her birthday, and she tells Bernard the product, then she scurries off to flirt with a jock while the other two try to outnerd each other.

Albert: I don't know when Cheryl's birthday is, but I know that Bernard does not know too.

Bernard: Not only didn't I know when Cheryl's birthday was before, but I still don't know now, and I don't even know if we will ever be able to know.

Albert: Even after what you just said, I still don't know if we will ever be able to know either.

Bernard: Same here.

Albert: Now I know Cheryl's birthday!

Can you deduce Cheryl's birthday too?

Oooh, this looks interesting. Let's start: Albert doesn't know Cheryl's birthday, so it's not 1/1, 30/12 or 31/12, which have unique sums.

He also knows that Bernard doesn't know it, so the date can't have a unique product. Thus the day cannot be 17, 19, 23, 29 or 31 - because those days would generate unique products - and Albert must be able to tell that from the sum. Thus the sum of the day and month must be less than or equal to 17.
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Re: Maths thread.
« Reply #362 on: April 16, 2015, 02:01:47 pm »
+1

There's that Albert, Bernard, Cheryl puzzle going around. Someone shared a harder one that doesn't seem to have a solution. I modified it a bit, and I think it should have a single solution now:

Albert and Bernard, two nerdy logic geniuses, have just become friends with Cheryl, a girl with a very peculiar sense of humour. They ask her when her birthday is, but instead, she tells Albert the sum of the month and the day of her birthday, and she tells Bernard the product, then she scurries off to flirt with a jock while the other two try to outnerd each other.

Albert: I don't know when Cheryl's birthday is, but I know that Bernard does not know too.

Bernard: Not only didn't I know when Cheryl's birthday was before, but I still don't know now, and I don't even know if we will ever be able to know.

Albert: Even after what you just said, I still don't know if we will ever be able to know either.

Bernard: Same here.

Albert: Now I know Cheryl's birthday!

Can you deduce Cheryl's birthday too?

Oooh, this looks interesting. Let's start: Albert doesn't know Cheryl's birthday, so it's not 1/1, 30/12 or 31/12, which have unique sums.

He also knows that Bernard doesn't know it, so the date can't have a unique product. Thus the day cannot be 17, 19, 23, 29 or 31 - because those days would generate unique products - and Albert must be able to tell that from the sum. Thus the sum of the day and month must be less than or equal to 17.

We can also rule out 14 - 7/7 would make a product of 49, and no other date makes that product.

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Re: Maths thread.
« Reply #363 on: April 16, 2015, 03:38:14 pm »
+1

There's that Albert, Bernard, Cheryl puzzle going around. Someone shared a harder one that doesn't seem to have a solution. I modified it a bit, and I think it should have a single solution now:

Albert and Bernard, two nerdy logic geniuses, have just become friends with Cheryl, a girl with a very peculiar sense of humour. They ask her when her birthday is, but instead, she tells Albert the sum of the month and the day of her birthday, and she tells Bernard the product, then she scurries off to flirt with a jock while the other two try to outnerd each other.

Albert: I don't know when Cheryl's birthday is, but I know that Bernard does not know too.

Bernard: Not only didn't I know when Cheryl's birthday was before, but I still don't know now, and I don't even know if we will ever be able to know.

Albert: Even after what you just said, I still don't know if we will ever be able to know either.

Bernard: Same here.

Albert: Now I know Cheryl's birthday!

Can you deduce Cheryl's birthday too?

Oooh, this looks interesting. Let's start: Albert doesn't know Cheryl's birthday, so it's not 1/1, 30/12 or 31/12, which have unique sums.

He also knows that Bernard doesn't know it, so the date can't have a unique product. Thus the day cannot be 17, 19, 23, 29 or 31 - because those days would generate unique products - and Albert must be able to tell that from the sum. Thus the sum of the day and month must be less than or equal to 17.

We can also rule out 14 - 7/7 would make a product of 49, and no other date makes that product.

On the other hand, if the sum is too small, they wouldn't be able to figure it out at all.  As d/m and m/d are indistinguishable for both Albert and Bernard, and both are <=12, they can never tell which one is the month and which on the day. So the day must either be larger than 12,  Or d=m.  Where we can't have primes, and 1, and 9. So we are left with 4/4,6/6,8/8,12/12 or 18>d>12.  Where not so much is left anyway.

:e shit 2 and 3 should work, too.
« Last Edit: April 16, 2015, 03:45:50 pm by DStu »
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Re: Maths thread.
« Reply #364 on: April 16, 2015, 06:49:39 pm »
0

There's that Albert, Bernard, Cheryl puzzle going around. Someone shared a harder one that doesn't seem to have a solution. I modified it a bit, and I think it should have a single solution now:

Albert and Bernard, two nerdy logic geniuses, have just become friends with Cheryl, a girl with a very peculiar sense of humour. They ask her when her birthday is, but instead, she tells Albert the sum of the month and the day of her birthday, and she tells Bernard the product, then she scurries off to flirt with a jock while the other two try to outnerd each other.

Albert: I don't know when Cheryl's birthday is, but I know that Bernard does not know too.

Bernard: Not only didn't I know when Cheryl's birthday was before, but I still don't know now, and I don't even know if we will ever be able to know.

Albert: Even after what you just said, I still don't know if we will ever be able to know either.

Bernard: Same here.

Albert: Now I know Cheryl's birthday!

Can you deduce Cheryl's birthday too?

I think I am a close, but got a little off somewhere; or maybe I am misunderstanding what I am supposed to be inferring.

I get that if B claims "same here" then A cannot choose between 3/12 and 12/3 and they know they will never know. If B instead says, "I know!" the answer is Jan 14.

One step back I eliminate 1/16, 2/15, and 3/14 because A would know that they could eventually know C's birthday.
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Re: Maths thread.
« Reply #365 on: April 16, 2015, 07:06:55 pm »
0

the answer is Jan 14.
I have that date as well.
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Re: Maths thread.
« Reply #366 on: April 16, 2015, 07:08:11 pm »
0

I don't have that date. Could you go into more detail about how you find it?

EDIT: your answer should be eliminated by Bernard's last sentence.
« Last Edit: April 16, 2015, 07:09:59 pm by pacovf »
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Re: Maths thread.
« Reply #367 on: April 16, 2015, 07:18:28 pm »
+1

Here a bit longer the way to get there
The first hint excludes all sums equal to 2, 14, 16, or >=18. Otherwise there would be always (at least) one option where Bernard could know the birthdat.
The second hint excludes all dates for which only combinations where product and sum are the same exist, e.g. 5/1 and 1/5.

That reduces everything to the following dates:
product 4: 04/01, 02/02, 01/04
product 9: 09/01, 03/03, 01/09
product 14: 07/02, 02/07, 01/14
product 16: 08/02, 04/04, 02/08, 01/16
product: 36: 12/03, 09/04, 06/06, 04/09, 03/12
product 42: 07/06, 06/07, 03/14

Because Albert still doesn't know, we exclude 02/02, 03/03, 04/04, and 06/06 (unique sums). Bernard could now conclude that the puzzle is insolvable if the product were 4, 9, or 36. As he doesn't do so, we can exclude those dates as well.

Of the remaining dates
product 14: 07/02, 02/07, 01/14
product 16: 08/02, 04/04, 02/08, 01/16
product 42: 07/06, 06/07, 03/14
only 01/14 has a unique sum.


Edit:
I noticed that I forgot the product 30: 10/03, 06/05, 05/06, 03/10, 02/15, but that doesn't change the result.
« Last Edit: April 16, 2015, 07:33:20 pm by ghostofmars »
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Re: Maths thread.
« Reply #368 on: April 16, 2015, 07:36:09 pm »
0

Quote
Because Albert still doesn't know, we exclude 02/02, 03/03, 04/04, and 06/06 (unique sums).

While what you are saying is true, what Albert is saying is stronger than just not knowing.

EDIT: damn I just noticed there's a problem here. Let me think.

EDIT 2: Ok, no it's fine, there's still a single solution, only not the one I thought. Keep going!
« Last Edit: April 16, 2015, 07:49:07 pm by pacovf »
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Re: Maths thread.
« Reply #369 on: April 16, 2015, 08:01:14 pm »
+1

Ok, with the current wording of the puzzle, no date leads to that conversation. I hadn't noticed that 2/2, 3/3, 4/4 and 6/6 had unique sums.

Please replace the last two sentences with:

Bernard: Thanks, now I do know Cheryl's birthday.

Albert: So do I!

A shame, the other version was cuter.
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Re: Maths thread.
« Reply #370 on: April 16, 2015, 08:18:11 pm »
+1

I fixed the error in my solution with the new version and get the same date as before.

In addition to 02/02, 03/03, 04/04, and 06/06, Albert excludes all dates that sum up to 5, 9, 10, 11, or 13, because they would lead to an insolvable puzzle. The remaining dates are (sum 15) 01/14, 03/12, and 12/03 and (sum 17) 01/16, 02/15, and 03/14.
Bernard's answer excludes the 03/12 and 12/03 possibility (all other products are unique). Then the only unique sum is 01/14.
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Polk5440

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Re: Maths thread.
« Reply #371 on: April 16, 2015, 08:19:33 pm »
+1

This is what I am doing:

Quote
Albert: I don't know when Cheryl's birthday is

Eliminates dates with unique sums (1/1, 12/30, 12/31).

Quote
Albert: ...but I know that Bernard does not know too.

Eliminates sums where one of the potential dates identifies a unique product. Eliminates dates associated with sums 2, 14, 16, and >=18.

Quote
Bernard: Not only didn't I know when Cheryl's birthday was before, but I still don't know now...

Eliminates dates with unique product after eliminating the dates from Albert's declaration. Eliminates 5/5, 2/13, and 4/13.

Quote
Bernard: ... and I don't even know if we will ever be able to know.

If the birthday is such that there are multiple dates with the same (sum, product) then A and B will never know the birthday. From Bernard's perspective, this eliminates products where there is no chance to identify the birthday.

We are left with date with products 4, 9, 14, 16, 30, 36, 42.


Quote
Albert: Even after what you just said, I still don't know if we will ever be able to know either.

If Albert has a sum that is unique in the remaining set, he would know at this point. This eliminates 2/2, 3/3, 4/4, and 6/6.

If the birthday is such that there are multiple dates with the same (sum, product) then A and B will never know the birthday. From Albert's perspective, this eliminates sums where there is no chance to identify the birthday.

This eliminates sums 5, 9, 10, 11, 13, leaving only sums 15 and 17.
The dates are:
sum 15: 1/14, 3/12, 12/3
sum 17: 1/16, 2/15, 3/14.

But sum 17 each has a unique (sum, product), so A would know that they can know the date if they know (sum, product). This eliminates sum 17, leaving sum 15 dates, 1/14, 3/12, 12/3.

Quote
Bernard: Same here.

Eliminates 1/14, but that leaves 3/12 and 12/3 from which A should conclude they will never know.
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pacovf

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Re: Maths thread.
« Reply #372 on: April 16, 2015, 08:24:12 pm »
0

I fixed the error in my solution with the new version and get the same date as before.

In addition to 02/02, 03/03, 04/04, and 06/06, Albert excludes all dates that sum up to 5, 9, 10, 11, or 13, because they would lead to an insolvable puzzle. The remaining dates are (sum 15) 01/14, 03/12, and 12/03 and (sum 17) 01/16, 02/15, and 03/14.
Bernard's answer excludes the 03/12 and 12/03 possibility (all other products are unique). Then the only unique sum is 01/14.


Correct!


Yes, the puzzle as posted didn't have any solution. Sorry for that.
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SwitchedFromStarcraft

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Re: Maths thread.
« Reply #373 on: April 17, 2015, 08:34:37 am »
0

I fixed the error in my solution with the new version and get the same date as before.

In addition to 02/02, 03/03, 04/04, and 06/06, Albert excludes all dates that sum up to 5, 9, 10, 11, or 13, because they would lead to an insolvable puzzle. The remaining dates are (sum 15) 01/14, 03/12, and 12/03 and (sum 17) 01/16, 02/15, and 03/14.
Bernard's answer excludes the 03/12 and 12/03 possibility (all other products are unique). Then the only unique sum is 01/14.


Correct!


Yes, the puzzle as posted didn't have any solution. Sorry for that.
Is this our first math derphammer?
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pacovf

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Re: Maths thread.
« Reply #374 on: April 17, 2015, 10:17:11 am »
0

Could be! The one I've been sent had the following conversation:

A: I don't know the date, but I know that B doesn't know either.
B: I didn't know before, but I do now!
A: So do I!

But that leaves 3 possible dates, so I tried to change it so that it would have only 1. Turns out I ruled out all of them :P
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