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That is pretty much my way too.

Bob

If n is the number of pages in the book and m and m+1 are the torn pages then

As there are two unknowns but only one equation, there's no simple algebraic trick for solving this. You do know that n and m have to be positive integers, but that's still not much help.

So I looked to see if there was a list of the triangle numbers [n(n+1)/2] on the net but couldn't find one. So I made my own using a spreadsheet.

I made column A the counting numbers, and column B the triangle numbers using the formula. Then I had a look.

The first triangle number after 15000 is n = 173, and that comes out to be 15051. So m = 25 and m+1 = 26 seems to work. But so does n = 174 m = 112, m+1 = 113 [15225 = 112 - 113] But that means the first page is even so I'll reject that.

So I made column C = column B - 15000 and column D = column C/2 and then scanned down for a number of the form odd + 0.5

The first I came to was n = 178, n(n+1)/2 = 15931 which makes m = 465 and m=1 = 466. It fits the arithmetic but you cannot tear out page 465 from a 178 page book.

After 178 this issue continues because the sum - 15000 excess gets bigger faster than the sum of pages goes up. So it looks like the first answer I found is the only one.

Bob

You did have a link in post #1.

I've no idea why it has gone. Perhaps you could try editing it back.

When someone puts a link into a post I check it out just in case it's not 'wholesome'. Quite a few spammers have gotten banned because of that. The library looks ok to me; I've rechecked it today.

They say that if you search for a particular book on a specific shelf and note its contents; you'll get the same if you repeat that another day. So it is not just generating random characters. Must check that out and see if Shakespeare is in there. I discovered today that there is an image library as well. Hhmm. Wonder if my mugshot is there?

Bob

hi Hicies87

You will get better by practising. If you post a problem that is causing you difficulty, I'll try to lead you through it. If you say where you're struggling, perhaps we can sort that out, and I'll set you some more, similar questions if you want to practise your skills.

Bob

Why would a factorial gain zeros at all?

Whenever there is a factor of 5 and 2 in the calculation another zero is added.

eg. It isn't until 5! that we get the first zero .... 120.

factors of 2 are plentiful so I looked at when another factor of 5 is acquired

Bob

There is another way to get the book of course, and that's to read it in the library. It is certainly there!

Bob

OK, I'll outline my complete method. I don't really consider it to be trial and error; rather it enables me to home in on possible solutions.

Let the numbers be A and B, with A>B wnlog

A^2 - B = 158

B^2 - A = 108

Subtracting

A^2 - B^2 - B + A = 50

(A-B)(A+B+1) = 50

Assuming we are seeking +ve integer solutions, with the factors of 50 being {50,25,10,5,2,1}

If A+B+1 = 50 then A = 25, B = 24. This is not a solution.

If A+B+1 = 25 then A = 13, B = 11. This is a solution.

If A+B+1 = 10 then A = 7, B = 2. This is not a solution,

It is unnecessary to check lower values of A+B+1 as A-B would come out bigger and lead to impossible soltions.

Bob

I called the ages A and B and made two equations. You can make, say, B the subject of one and substitute it into the other and solve the quadratic. I didn't, so I don't know about a second answer. My guess would be that the second solution is inadmissible in the context of the question (eg negative or out of range).

What I actually did, because it looked easier and was, was to factorise (A-B) out of the equation which left me with two factors = 50. It's reasonable to assume that there's an integer solution and 50 doesn't have that many factors, so I was able to spot one straight away that works.

Bob

The answer 42 first came from the book so it seems you have, without realising it, encountered THGTTG already. The Guide is certainly VERY useful; it saves Arthur Dent's life several times. Enjoy!

Bob

hi calebwater

Welcome to the forum.

I'm glad you like Tanks 2. I'm an administrator for this site. I didn't construct the MIF site and don't have write access to it. I think it's unlikely that Tanks 3 will get written, sorry.

I googled it and found it is also available on some other sites, so I'm guessing the source code was freely available. I then tried Tanks 3 and found sites with such an animal. I suggest you do the same.

Best wishes, Bob

hi lanxiyu

Welcome to the forum.

Is this an exercise where we should prove this or are you asking for help with it as a problem you have been set.

Either way I had better start trying to construct a proof. Hopefully, more when (if) I succeed.

LATER EDIT. From your top line I can show at least one product is not commutative.

Bob

hi

This guy is going to have difficulty clearing up all false statements.

Consider the statement "This statement is FALSE".

At first sight is seems easy to deal with; after all it's telling you it's false so it's got to go hasn't it?

But let's just think about it. It's a statement, so it is either TRUE or FALSE.

Let's try them both.

Assume it is TRUE. In that case it shouldn't be cleared out as we're only after the FALSE ones. But if TRUE, it is TRUE that it is FALSE, so we have a contradiction.

Better try assuming it is FALSE. If it's FALSE then the statement is lying when it says it is FALSE, so it must be TRUE. Once again we have got a contradiction.

Once you get the hang of it, you can make up loads of statements like this.

The Hitchhikers Guide to the Galaxy started out as a BBC radio series. Then there was the book; then a TV series and finally a full length film.

It acquired a cult following when it first happened on radio and fans loyally claim that the radio version is best. Personally, I think the TV version is better; it has all the action and sound effects plus fantastic graphics illustrating the Guide itself. I should explain, the Guide is an electronic book containing loads of helpful advice for the hitch hiking space traveller, like where to get the best drinks and how to get a ride on a Vogan spaceship. It also has the helpful advice on the cover "Don't Panic".

I've checked and a popular on-line company has copies in all formats.

Douglas Adams followed it up with 4 more books in the Trilogy (no error!). I didn't find them so funny ... the original was delightfully different and full of zany ideas ... but there's only so many times you can tell the same joke. If you like it, Adams also wrote two detective novels starring Dirk Gently, and these are equally wonderful (IMHO).

The question about Life, The Universe and Everything comes from the book and helps explain lots of things that puzzle the average human. You will also get the answer to this important question, but you probably won't like it.

Bob

hi CurlyBracket

Bob

There's an ancient philosophical question: If a tree falls in the forest and there's no one around to hear it fall, does it make a sound?

The doubt and uncertainty comes from 'The Hitch Hiker's Guide to the Galaxy' by Douglas Adams. If you haven't read it I heartily recommend you put that right straight away. It's very funny. There is also a film and two BBC series, TV and Radio. You will also learn the ultimate answer to the questions about life, the universe and well, everything.

Bob

'Facts' like these become provable (or not) only when we know the context of the fact. If the statement includes a declaration as to who is making the statement as in Bob says: "I am Bob" then we have a way of testing the truth or otherwise. The library will have such statements, and a lot more where we don't know the context. The truth of those cannot be determined. Schrodinger's cat and trees that crash silently in empty woods fall into this category.

Fortunately, the world keeps turning nevertheless.

"What we demand," said Grunfondle, "is rigidly defined areas of doubt and uncertainty." (Hitchhiker's Guide, Douglas Adams)

Bob

hi CurlyBracket

My understanding is that the fastest flow is in the centre of the pipe, with the slowest where the liquid is next to the pipe edges.

Assume the pipe is circular. Divide the liquid into a series of concentric circles; the outermost is next to the pipe edge; each circle after that is adjacent to a circle 'outside it' and a circle 'inside it, until we reach the centre.

The pipe edge is stationary so the drag is maximum there. Each concentric circle of liquid gets drag from the next circle outwards, so each circle is able to go a little faster, as you work inwards towards the centre of the pipe.

This would cover adhesion and surface tension. Not sure if layers above and below make a difference. It probably depends on the weight of liquid and the pressure at the end to make a flow happen at all.

I expect you can google for a formula.

Bob

hi pi_cubed

I hadn't come across these problems so many thanks for that. The algebra is ok so your argument seems to depend upon

<the re-arranged expression is irreducible> implies <the original is irreducible>

I think that is ok (maybe it's even a theorem) but it's not something I've encountered before, so I'm having a think about that.

LATER EDIT.

It's trivial to show that 2/3 = 4/6

But <2/3 is irreducible> does not imply <4/6 is irreducible>

So a re-arrangement alone is in-sufficient; something more is needed I think.

Bob

Bob

hi William,

I recommend you join the forum as a member.

If you have specific questions about this topic, post them in the help section.

Bob

I'm still not getting there. I started with MIF in 2010 and joined the forum shortly afterwards. Occasionally I try questions from the MIF pages and I know they use Mathopolis but I've never actually created a username for Mathopolis until just now.

I don't see any post option. Nor a search option.

So I still don't understand where on the internet you are.

Bob

hi pi_cubed

I cannot repeat what you are trying. Please post the url for when you start searching in each case.

Thanks,

Bob

ps. I don't have editing control for the site, so I may not be able to address any bug anyway.

I'll make the following conjectures:

(1) The set of true facts in the library is 'countable'. (infinite aleph-nought)

(2) The set of false facts isn't countable. (infinite aleph-one)

https://en.wikipedia.org/wiki/Aleph_number#Aleph-one

At this stage I haven't a proof for either.

Bob

hi Lakretia

I often got asked this question when teaching maths to reluctant teenagers. There are areas of maths that a person may need later in life but I had an additional reason, and the harder maths is for the individual, the stronger the argument.

An athlete may well do weight training even though their sport doesn't require them to lift weights (eg tennis). This is because it is recognised that such training builds up the right kind of strength/muscle/stamina and so helps them prepare for their sport.

Maths makes you think hard. When I was a teenager I did two sorts of extra maths to give my brain a workout.

One was multiplying out a 4 x 4 determinant (letters not numbers) on paper , and then re-assembling it with rows or columns swopped about. You have to be really careful to get all the letters right at every stage so it is good practice for concentration and neat, careful working.

The second was to try and solve quadratics using the formula in my head. I had to evaluate b^2 - 4ac, all without writing anything down. So far not too bad. I allowed myself to look up the square root to 4 decimals places (using tables). Then , still without writing anything down, I had to complete the two calculations, just writing down the final answers. For me it was hard to do the root - b part and then the division always having to remember all the digits. I found this improved my concentration a lot and I like to think enabled me to do harder thinking.

I still try to do this sort of thing to keep my brain going.

Bob

hi Curlybracket

That's an interesting discovery. As it has infinite storage, perhaps we can all use it for cloud memory.

But I checked and "the library of babel does not exist" is in the library. If only someone could come up with a way of sorting the true statements from the false.

Bob

hi Gornador

Welcome to the forum.

Good intro!

If you succeed in making a game that you want to sell, this forum has rules about advertising. Sorry. Tread carefully.

Bob

hi jadewest,

Set a possible factor equal to zero and find x; now try that value of x in the quadratic . If that evaluates to zero you've got a factor.

2x + 3 = 0 means x = -1.5

so this is not the factor.

If x + 2 = 0 then x = -2

so this one is a factor.

This technique uses something called the factor theorem.

You'll find it half way down this page: https://www.mathsisfun.com/algebra/poly … actor.html

Bob