A BIT OF MATHS

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A BIT OF MATHS

Postby Bigtone53 » Sun Sep 30, 2007 4:16 pm

As I seem to have clogged this site up for a week, here is something for the mathematicians

EVE / DID = .TALKTALKTALK....

Each letter stands for a different consistent digit, zero included. The fraction is in its lowest form. The decimal has a repeating period of 4 digits. The solution is unique.

No claim for originality by the way:)
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Postby re'born » Sun Sep 30, 2007 11:16 pm

Here is a possible solution. I've tried to eliminate the use of a calculator, but in the end I'm stuck plugging in 6 possibilities and eliminating 5. I'll shrink it for those wanting to avoid spoilers.


First, using a trick from the .999...=1 thread, we multiply by 10000 to get
10000*(.TALKTALK...) = TALK.TALKTALK...
Subtracting .TALKTALK... from both sides we get
9999*(.TALKTALK...) = TALK
and hence EVE/DID = .TALKTALK... = TALK/9999.
This means that we started with the fraction TALK/9999 and reduced it to EVE/DID. Factoring, we see 9999 = 3 * 3 * 11 * 101. The reducing process then must have consisted of dividing by elements of (3,3,11, 101). If we had divided by 101, however, the denominator would have been at most 99, contradicting DID being a 3 digit number. So 101 must be left in the denominator, i.e., DID must be a multiple of 101. Since 11*101 > 1000, 11 cannot also divide DID, so the only possible values for DID are {101, 303, 909}.
Since TALK < 9999, we must have EVE < DID. But this is impossible if DID = 101.

If DID = 909, then TALK = 11*EVE = E E+V E+V E. But then K = E, a contraction.

This means we must be have DID = 303 and TALK = 33 * EVE. Since EVE < DID, we must have E = 1 or E = 2. But if E = 1, then 33 * EVE < 33 * 200 < 1000 < TALK, a contradiction. So E = 2.

Expanding EVE into what it really means, we have 100*E + 10*V + E = 101*E + 10*V. Plugging in E = 2, we get 202 + 10*V. Using the equation above, we have
TALK = 33 * (202 + 10*V) = 6666 + 330*V
As 10 divides 330*V, it's last digit is 0 and hence the last digit of 6666 + 330*V is 6. So K = 6.

Summarizing, EVE/DID = .TALK... is now 2V2/303 = .TAL6...
There are 6 choices left for V, {1,4,5,7,8,9}.
212/303 = .6996..., impossible since the 6 and 9 occur twice
242/303 = .7986..., this is a solution
272/303 = .8976..., this duplicates the 7
282/303 = .9306..., this duplicates the 0 and 3
292/303 = .9636..., this duplicates the 3,6 and 9

So the answer is EVE = 242, DID = 303 and TALK = 7986
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Postby udosuk » Mon Oct 01, 2007 12:52 pm

re'born wrote:But if E = 1, then 33 * EVE < 33 * 200 < 1000 < TALK, a contradiction. So E = 2.

33 * 200 = 6600 < 1000 ?!

I have to scratch my eyes 3 times to think about if I've gone blind or insane.:D

Otherwise an excellent analysis, well done re'born.:)

Here is my full logical analysis, and I didn't use the calculator at all.:D

Triple click below to see what I wrote:EVE/DID = .TALKTALKTALK... = TALK/9999, where EVE/DID is a fully-reduced fraction

=> EVE * M = TALK, DID * M = 9999 = 3*3*11*101 where M is a +ve integer

Also, TALK < 9999 => EVE < DID => E < D => D is at least 1 => DID > 100

If M is a multiple of 101, DID = 9999 / M is at most 99 => contradiction
Therefore M is not a multiple of 101 => DID must be a factor of 9999 which is divisible by 101

Hence DID = 101|303|909, M = 99|33|11

If DID = 101, M = 99 => (E < D) E = 0 = I, contradiction
If DID = 909, M = 11 => EVE * 11 = TALK => E = K, contradiction
Therefore DID = 303, M = 33 => (E < D) E = 1 or 2

If E = 1 => TALK = EVE * 33 = (101 + 10 * V) * 33 => K = 3 = D, contradiction
Therefore E = 2

Now we have TALK = EVE * 33 = (202 + 10 * V) * 33 = 6666 + 330 * V

=> K = 6, TAL = 666 + 33 * V where V, T, A, L must be from {145789}

Since TAL > 666, T must be one of {789} => 33 * V > TAL - 666 > 700 - 666 = 34 => V > 1

Focusing on the last digit, we have 3 * V + 6 = 10 * N + L where N is an integer

Also, V and L must be different digits with V from {45789} and L from {145789}

If V = 7, L = 7 => contradiction
If V = 8, L = 0 => contradiction
If V = 9, L = 3 => contradiction
Therefore V = 4 or 5

If V = 5, TAL = 666 + 33 * 5 = 666 + 165 = 831 => A = 3 = D => contradiction
Therefore V = 4 => TAL = 666 + 33 * 4 = 666 + 132 = 798

"EVE/DID = .TALKTALKTALK..." is thus "242/303 = .798679867986..."
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Postby Bigtone53 » Mon Oct 01, 2007 9:41 pm

Well. lads you have both got straight there. My apologies for posting such a simple question (again!). Let me work on something more challenging for you. Riemann Hypothesis?


PS if the fraction was not specified as being in its lowest form,
212/606 = .349834983498 .. would work.
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Postby re'born » Tue Oct 02, 2007 11:46 am

udosuk wrote:
re'born wrote:But if E = 1, then 33 * EVE < 33 * 200 < 1000 < TALK, a contradiction. So E = 2.

33 * 200 = 6600 < 1000 ?!

I have to scratch my eyes 3 times to think about if I've gone blind or insane.:D

Otherwise an excellent analysis, well done re'born.:)


Uh, yeah. Not sure what happened there. Perhaps I need a refresher course on multiplication. Anybody reading, please take udosuk's proof that E <> 1 and insert it for mine.

Well done finishing off the proof without resorting to the calculator. I am duly impressed:)
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Postby Bigtone53 » Tue Oct 02, 2007 7:13 pm

OKaaay,

On the basis that I do not post something unless I have worked it out for myself (which explains why they are so easy) here is another one, which I hope will last a bit longer than the 4 hours or so it took re'born to do the last one.

I am thinking of a 10-figure number in decimal form. The first digit of the number indicates the total number of zeros in the entire number, the second digit indicates the total number of ones in the entire number ... (you get the picture) ... the tenth digit indicates the total number of nines in the entire number. Zero is a digit.

The number?

As before, it is a unique answer and as before, this is not an original thought of mine.

I appreciate that the nature of these types of question is that you either get the answer or you don't. There is not much chance of clues. Let me see whether I can come up with a more user-friendly problem
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Postby re'born » Tue Oct 02, 2007 11:33 pm

Here you go:

Triple click below to see what I wrote:
6210001000

The key observation for me was to note that the sum of digits must be 10. This immediately implies that for n>=5, then the number in the nth position (I label the positions as the 0th through 9th positions) is either 0 or 1. From here I noted that if 1 occurs in the nth-position , then this will imply a sum of digits of at least 1 + n + n*r, where r is the position in which n lays. If we can deduce that r<>1, then we get a sum of digits at least 4 + n + n*r.
For n >= 5, If a 1 occurs in the nth position, then 10 >= 1 + n + n*r >= 6 + n*r => r = 0. In particular, r<>1. Hence we get 10 >= 4 + n. So, this means that the digits in the 7,8 and 9 positions are 0.

Let n denote the largest position containing a non-zero entry.

For n <= 4, the number of 0's is greater than or equal to 5, meaning that we must have a position greater than or equal to 5 with a 1, contradicting the maximality of n.

Now, for n = 5, the above shows we are in the situation
5 _ _ _ _ 1 0 0 0 0
We are now only allowed one more 0, but then we will get at least one of three of the numbers {1,2,3,4}, an impossibility.

So n = 6, and by the above, we have a 6 in the zero position, so our number looks like
6 _ _ _ _ _ 1 0 0 0
The 1 in the 6th position implies that the number in the 1st position is at least 2, so the 3rd, 4th and 5th positions must be 0, giving
6 _ _ 0 0 0 1 0 0 0, where the number in the 1st position is at least 2.
The only way to finish it is to make the 2nd position a 1 and the 1st position a 2, giving the solution
6 2 1 0 0 0 1 0 0 0

Therefore 6 2 1 0 0 0 1 0 0 0 is the unique solution.
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Postby Bigtone53 » Wed Oct 03, 2007 8:41 am

Very impressive re'born:D
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Postby JPF » Wed Oct 03, 2007 1:09 pm

For information, here are the answers for n digits (2<=n<=10)

Triple click below to see what I wrote:n=2 : no sol.
n=3 : no sol.
n=4 : 1210 ; 2020 (2 sol.)
n=5 : 21200
n=6 : no sol.
n=7 : 3211000
n=8 : 42101000
n=9 : 521001000
n=10: 6210001000
brute force:(

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Postby Bigtone53 » Wed Oct 03, 2007 6:44 pm

Right, a last attempt to try and occupy re'born and others for more than 2 minutes.

This one is original, in that while it appeared as a Brainteaser in the UK's Sunday Times in the early 70s, I was the contributor.

It is well known that certain people suffer from fibobia after drinking beer. If you suffer from this unfortunate affliction, then of any 5 statements you may make, the number of false ones is the same as the number of pints you have had. A,B C D and E all suffer from fibobia and I asked them one night how many pints of beer they had had. These were the answers

A said " I've had 4, E 3, B 5,C 1 and D 2"
B said " I've had 4, C 3, A 5, E 1 and D 2"
C said " I've had 3, E 2, B 5, A 1 and D 4"
D said " I've had 4, E 3, B 5, C 2 and A 1"

E by this time had gone home but the landlord, a teetotaller, told me that nobody had had more than 5 pints and each had drunk a different number. He also said that E had had at least 1 pint but he could not tell me who had had what. But that was fine because now I could work it out for myself.
Who had what?

Top marks for a logical answer rather than brute force. No disrespect to JPF:D

PS how do you guys do the click three times thing?
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Postby re'born » Wed Oct 03, 2007 7:58 pm

Nice puzzle Bigtone. That was the hardest yet for me.

Spoiler Warning: I can't get the code to hide, so I'm just hiding the text. Be warned!

A=5, B=2, C=3, D=4, E=1

Proof:
Note that each of A,B,C,D gave different answer for everybody. Hence none of them could have told the truth exactly 4 times. So E=1. Let's draw a table, where a plus after a number indicates true, and a minus false. Since E=1, anybody who said that is telling the truth and anyone who said either that E<>1, or that someone else is 1 is lying:

Code: Select all
    A  B  C  D  E
A|  4  5  1- 2  3-
B|  5  4  3  2  1+
C|  1- 5  3  4  2-
D|  1- 5  2  4  3-


Since B told the truth, A,C and D must be lying about B.

Code: Select all
    A  B  C  D  E
A|  4  5- 1- 2  3-
B|  5  4  3  2  1+
C|  1- 5- 3  4  2-
D|  1- 5- 2  4  3-


Now all 3 of A,C and D have lied at least 3 times, so B=2. Updating the chart, we get:

Code: Select all
    A  B  C  D  E
A|  4  5- 1- 2- 3-
B|  5  4- 3  2- 1+
C|  1- 5- 3  4  2-
D|  1- 5- 2- 4  3-


Since B tells the truth 3 times, he must be telling the truth when he says A=5 and C=3. This leaves D=4.
Last edited by re'born on Wed Oct 03, 2007 5:49 pm, edited 1 time in total.
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Postby Bigtone53 » Wed Oct 03, 2007 8:14 pm

Great answer re'born, although slightly different reasoning to mine.

AAAAAARRRRRGGGHHHH!:D
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Postby re'born » Wed Oct 03, 2007 9:52 pm

Bigtone53 wrote:Great answer re'born, although slightly different reasoning to mine.

Does your solution start differently? The way I did was the only foothold I could find.
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Postby Bigtone53 » Thu Oct 04, 2007 6:26 pm

This is how I did it. Basically the same



No-one could have had no pints because they would have said so.

No-one who spoke could have had one pint as each have at least two lies (I have had x pints and X has had one). So E was the 1 pinter.

A cannot have had 2 pints because of at least 3 lies (I have had 4, E 3 and C 1)

Ditto C and D

So B had 2 pints and his two lies were 'I have had 4' and 'D 2'

So his other statements were true and it falls into place.
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Postby RW » Thu Oct 04, 2007 7:48 pm

And here's the real logical answer:
Triple click below to see what I wrote:E had went home early. There's two possible reasons for this: either he didn't drink enough to get in the mood, or he drank too much and his friends decided to put him in a cab. However, the landlord said that nobody had had more than 5 pints, which wouldn't be enough to get him in such a bad condition that he would have had to leave => E had 1.

Now there's a little problem. I can see that you are from the UK. In the pubs there you have a strong tradition of buying rounds, which means that all of the remaining people must have had the same amount of pints. The most logical solution is that that the teetotaller must have smoked too many tea leaves to remember exactly what had happened. As you mentioned, this happened in the 70s...

As all of A, B, C and D must have had the same amount of pints and they all mentioned different amounts for all party members, the only possible solution would be that they all drank four (tell the truth once for the person who they claim has had four), but as B also says that E had one, this solution cannot be correct. We may conclude that the landlord was not only a tea smoking junkie, he was a pathological liar. Very odd behaviour from a landlord...

So, to satisfy the equation, either E didn't drink any beer at all or he had more than one. If he didn't drink at all, his friends would certainly have remembered that, so that option is ruled out. If he had more than one, then he would have got into the party mood and wouldn't have left and should still be there somewhere. If he still was there and they were going on the fourth round, then he would have been sitting together with his friends drinking beer.

The only logical solution that also explains the strange behaviour of the landlord is thus: All of them had had more than five pints. E had not left, you were E, but in your drunken condition you couldn't remember who you are and you couldn't understand at all what the landlord (who in fact was a police officer) was saying.

:D

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