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**wintersolstice**- Replies: 11

here's an interesting puzzle (no idea what difficulty level this would be.)

the idea is to create 10 different expressions each totaling 12, using 5 of the same digit (so 5 0's, 5 1's, 5 2's, 5 6's, 5 5's, 5 7's 5 8's and 5 9's)

rules:

1)you have to use EXACTLY 5 of that number no more no less

2)you can't use any other numbers (e.g.

would add a 2 you can't do that)3) you can use any arithmetic symbols, the √ symbol (since it has no number on it) and factorial (!)

4) you can also use decimal point and recurring sign (e.g. 8 = 8/10 and 8(recurring) = 8/9

5) you can't put two(or more) numbers together (e.g. you can put two 5's together to get 55 and you can't do 55 =55/10

6) you can also use brackets

there may be multiple solutions to some of them:D

heres the first one to start you off

EDIT the following solution is invalid since it works for all digits (except 0)

D/D + ((D+D)/D)

any other such solution is also invalid

vikramhegde wrote:

Ok so the procedure is like this.

First consider a triangular number sequence [ fn = (n+1).n/2] for n = 1,2,3..

The 18th number on this sequence is 171.

And sum of all triangular numbers till 171 (the tetrahedral number) crosses 1000

Now, for each subsequent drop of the first ball (when it doesn't break) we add the previous number on the triangular sequence.

So ball number one is dropped for the first time from Floor number 171, second at (171+153)=324, third at (324+136)=460, fourth at (460+120)=580, fifth at (580+105)=685, sixth at (685+91)=776, seventh at 854, eighth at 920, ninth 975, tenth at 999. Let us call these numbers M#1, M#2, M#3....Now, if it breaks in the first try at 171, we start dropping the ball at floor numbers 18, 18+17, 18+17+16,

Similarly for any break of the first ball on floor M#x, the floor number at which we drop the second ball is given by -

{(M#x-1)+P} and if the second ball doesn't break at this, we continue on the sequence - {(M#x-1)+P+(P-1)}, {(M#x-1)+P+(P-2)}, {(M#x-1)+P+(P-3)}... {(M#x-1)+P+(P-P)} Where P is the position of {M#x - (M#x-1)} on the triangular sequence.Having broken the second ball somewhere, we go back to the last try where we didn't break it and work our way up with tries on each floor till it breaks.

We will find that 19 is the least number of tries required.

I'm sorry I'm not trained in mathematics and hence have to put it in such a round about manner. I'm not so familiar with the notation and the use of sigma functions and had to invent some notation of my own. I hope I've explained it adequately.

Also, I'd be happy if someone could explain it a more simple manner.

just to correct you on that it should be M#x-1 + p + 1 to see why imangine dropping the first ball from 172. You can still do it even if it breaks on that floor other that your solution is spot on:D

NNikolay wrote:

Wintersolstice, I have showed how your proof is wrong and have given a solution. What else can I do?

Second modification has a solution too. 2 questions with 3 possible outcomes give 9 combinations! Of course you loose some information due to Random behaving really randomly and because you don't know the words upfront. But it is enough to separate 6 cases.

Hope you will read the solution before telling it's wrong

I completely misread your last post that's all:D When you said "modification" I thought you meant you had modifiyed your original solution! and when you posted your "second modification" I thought it was describing your solution! (giving clues I mean:D)

Sorry about that!:D

anyway I've had a look at the solution but it's going to take a bit of time to get my head round it (although this has ruined for me! Nevermind there's the other puzzle)

btw have you seen my "variation" that I mentioned early (it's not that difficult really though)

NNikolay wrote:

4. You need a solution in two questions.

but that's impossible, it can't be done in two questions, look at my "proof" above it shows that two question isn't enough, if you need it explaining I'll try and explain it better for you:D

try out EVERY combination of answers you get from the two questions (that give you information) and see if that can tell you who's who, some combinations will tell you but some combinations won't.

btw I haven't looked at your solution yet but with my proof I don't think I need to.

MathsIsFun wrote:

There is an option "Never show smilies as icons for this post" in the Post Reply form (in case you really want to type )

sorry to revive this thread but even what wa said tin the quoted post deson't work!

NNikolay wrote:

Hi wintersolstice,

Your proof is based on the assumption that first of 3 questions can't give you any information except sample word from god's language. This is not correct. I mean, this is correct if you have one question only. But in our case you can get more information as a result of all 3 questions.

what I assumed is that you were using the "one question to determine yes/no followed by two questions to work out who's who (which you've proved wrong:D)

however the reason I believed that is because:

I tried the original solution on this version and realised that when you ask the first question (because you don't know any words) you can't actually find any information from the first question apart from a sample word even after the second tow questions.

here the original solution and you see if you can find information form the first question

this is just one solution but you'll find that using this solution is impossible on your version.

the only thing I can suggest is:

write down your own solution of 3+ question showing what the questions are asked to who and what in cirscumstances is each question asked(without showing anyone of course) and see if it works for all 6 possibilites (of who's who) and all possible answers and true and false questions and what is yes and no etc), this way you know it has a solution:D. That's what I did when I posted my variation of the puzzle:D

btw when I posted my proof I only said I don't "think" it's impossibe I never said I "know" it's impossible:D and it was imcomplete so hopefully I have completed it now:D

anonimnystefy wrote:

Hi wintersoltice

I've no idea what you're saying??? plase explain.

I put a question mark after my answer (-i?) to show I wasn't sure if that was right

bobbym wrote:

Hi;

I thought is was just me. I do not get that either.

I think I do get it actually:D

mathgogocart wrote:

correct i,-1,_

-i?

I don't think this puzzle is theoritcally possible! (though I could be wrong:D)

never heard of it.

like I say I can't find any mention of another country of that letter.

anonimnystefy wrote:

Again,it depends on the source. Google for countries that start with that letter. It should be easier.

just done that and every site I've tried says the same thing that there's only one, Are we talking about the same country here?

anonimnystefy wrote:

It is not a real country. It depends on the source.

OK but you still said there were other countries beginning with that letter, yet according to that list I just saw there aren't any? I don't understand.

anonimnystefy wrote:

The one they wrote isn't considered an independent country,but there are others that begin with that letter.

what do you mean?

I just looked at a list of all the countries in the world and first that country was listed and second there were no other countries of that letter!

I got 10/10 no help from anyone or anything!:D

the 2nd one was a bit tricky but I got there in the end, the last one was a bit tough (useless fact: it's the only country that begins with that letter!!!)

what I looked for was unusually word combinations:D

anna_gg wrote:

Here is my solution: It is based to the fact that each perfect square N^2 is the sum of the first N odd numbers (5^2 = 25 = 1+3+5+7+9).

Thus the difference of any 2 perfect squares should equal to the sum of consecutive odd numbers (and this should equal 60).

Starting from 1, we write down the sums of the odd numbers:

1+3+5+7+9+11+13 = 49 while 1+3+5+7+9+11+13+15 = 64. Thus we cannot make 60 starting from 1.

We do the same with 3: 3+5+7+9+11+13=48 while 3+5+7+9+11+13+15 = 63. Not possible.

Starting from 5:

5+7+9+11+13+15=60 Here we are.

So, one perfect square is 4 and the next is 64, their difference being 60, so the first number we are looking for is 34 (34-30 = 4 and 34+30 = 64, both of them perfect squares).

Similarly, we find that the only other sum of consecutive odd numbers equaling 60 is 29+31.

Therefore one perfect square is 1+3+5+...+27=196 (14^2) and the next is 1+3+5+...+27+29+31=256 (16^2) and the second number we are asking for is 226 (226-30 = 196, 226+30 = 256).

There is no other series of successive odd numbers equaling 60, so these two are the only numbers with this property.

Obviously this is not a proper "proof"; it is more based on a "guess and try" method, but it works!

Have you seen my proof? It's very similar in that it's based on consecutive odd numbers.

anonimnystefy wrote:

And I did it non-experimentally!

I did it diferent again:D

I thought this might be better in a new post

bob bundy wrote:

hi wintersolstice and anonimnystefy

Did you mean:

A says None of us is lying. B says Only two of us are lying. C says Only one of us is lying. D says all of us are lying.

The truth table below shows there is no solution to this.

The middle code must have 84 characters if it is to work. A quick count suggests there are less. But some spaces are double. Your problem is to work out which.

Bob

anonimnystefy wrote:

I also must note that if we add the third person saying that they are all lying then the puzzle turns into a paradox.

I see what you mean

I figured out where the paradox is now

bob bundy wrote:

I think I need to re-do the liars puzzle completely. Sorry about that.

anonimnystefy wrote:

hi wintersoltice

have you solve any other problems?

yes I solved a few of them but instead of posting my answers I checked them against yours (I haven't solved all the ones that you solved yet!:D)

anonimnystefy wrote:

I also must note that if we add the third person saying that they are all lying then the puzzle turns into a paradox.

how come? plus I said 4th person:D

hi wintersolstice

That's exactly what I had and I also was surprised by anonimnystefy's answer. Under the circumstances of the test, being surprised is entirely appropriate.

Bob

bob bundy wrote:

hi Stefy,

Good start.

Bob

bob bundy wrote:

hi

Of course, I agree with amberzak!

But for reasons of logic too.

The deduction "We've reached the end of Thursday, so it must be on Friday" is sound.

but this does not allow you to deduce anything about earlier days.

Because, for example, at the end of Wednesday the logic is "We've reached the end of Wednesday, so it could be on Thursday or Friday". It is always dangerous to keep applying the same logic as the student has done. And MathsIsFun's answer shows why!

Bob

I don't actually agree with this logic because if they reached the end of Wednesday then it must be Thursday because if it was on Friday it wouldn't be a suprise (because if it hasn't happen by the end of Thursday is has to be Friday)

but that's just my opinion:D I have seen another way (apart from my own) of resolving th paradox so I suppose it show there's more than one way to do it:D

the way I resolve this paradox is to say that:

the students use the logic to eliminate every possible day, so whichever day it happens it's a a suprise (they didn't see it coming) because they eliminated every possible day:D