If you like mathematical problems and puzzles...

"MathsBombe is aimed at students up to Year 13 (England and Wales), S6 (Scotland), Year 14 (Northern Ireland). You don't need to be a computer
whizz or a mathematical genius — you just need to keep your wits about you and be good at solving puzzles!"

Starts 16:00 GMT, 22nd January, 2025.

https://www.maths.manchester.ac.uk/mathsbombe/

--
David Entwistle

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On Sun, 19 Jan 2025 19:57:26 -0000 (UTC), David Entwistle wrote:

"MathsBombe is aimed at students up to Year 13 (England and Wales), S6 (Scotland), Year 14 (Northern Ireland). You don't need to be a computer
whizz or a mathematical genius — you just need to keep your wits about
you and be good at solving puzzles!"

Starts 16:00 GMT, 22nd January, 2025.

https://www.maths.manchester.ac.uk/mathsbombe/

Please don't post a direct answer to the question posed, but I'd welcome a
bit of guidance on Mathsbombe question 3.

When I look at the question, my reaction is "that doesn't look possible".
The "any positive integer cost can be paid" part of the question seems problematic. Am I misreading, or misunderstanding the question?

Thanks,
--
David Entwistle

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On 09/02/2025 09:40, David Entwistle wrote:

On Sun, 19 Jan 2025 19:57:26 -0000 (UTC), David Entwistle wrote:

"MathsBombe is aimed at students up to Year 13 (England and Wales), S6
(Scotland), Year 14 (Northern Ireland). You don't need to be a computer
whizz or a mathematical genius — you just need to keep your wits about
you and be good at solving puzzles!"

Starts 16:00 GMT, 22nd January, 2025.

https://www.maths.manchester.ac.uk/mathsbombe/

Please don't post a direct answer to the question posed, but I'd welcome a bit of guidance on Mathsbombe question 3.

When I look at the question, my reaction is "that doesn't look possible".
The "any positive integer cost can be paid" part of the question seems problematic. Am I misreading, or misunderstanding the question?

I agree; it doesn't look possible. I was tempted to cut code, but
I hit two ambiguities. What, precisely, does "no more than 14
coins of every given denomination" mean? It could mean an
up-to-14-coin subset of the available range, or up to 14
totapennies PLUS up to 14 totatuppences PLUS up to 14
totathruppences and so on ad nauseam. And what does "any positive
integer" mean? Does it, for example, include bloodybignumber? If
so, how about bloodybignumber factorial?

I don't care enough, I'm afraid, but if I *did*, then having
resolved those dilemmae, I would probably look at brute forcing a
few thousand candidate x's (3.0000, 3.0001, 3.0002, 3.0003 etc)
and then try to spot a pattern.

I would also look for tricks, eg i.

--
Richard Heathfield
Email: rjh at cpax dot org dot uk
"Usenet is a strange place" - dmr 29 July 1999
Sig line 4 vacant - apply within

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On 09/02/2025 10:13, Richard Heathfield wrote:

On 09/02/2025 09:40, David Entwistle wrote:

On Sun, 19 Jan 2025 19:57:26 -0000 (UTC), David Entwistle wrote:

"MathsBombe is aimed at students up to Year 13 (England and Wales), S6
(Scotland), Year 14 (Northern Ireland). You don't need to be a computer
whizz or a mathematical genius ù you just need to keep your wits about
you and be good at solving puzzles!"

Starts 16:00 GMT, 22nd January, 2025.

https://www.maths.manchester.ac.uk/mathsbombe/

Please don't post a direct answer to the question posed, but I'd welcome a >> bit of guidance on Mathsbombe question 3.

When I look at the question, my reaction is "that doesn't look possible".
The "any positive integer cost can be paid" part of the question seems
problematic. Am I misreading, or misunderstanding the question?

I don't instantly see why it would be impossible. It looks at least plausible to me. The x^k coins
go on without limit, so even for big numbers there will be big coins available for payment.

I agree; it doesn't look possible. I was tempted to cut code, but I hit two ambiguities. What,
precisely, does "no more than 14 coins of every given denomination" mean? It could mean an
up-to-14-coin subset of the available range, or up to 14 totapennies PLUS up to 14 totatuppences
PLUS up to 14 totathruppences and so on ad nauseam.

I agree that could be clearer. I read it as your second interpretation. If your first
interpretation were intended, wouldn't they just say "no more that 14 coins" and leave it at that?
[Plus I strongly suspect the first interpretation would indeed be impossible.]

And what does "any positive integer" mean? Does
it, for example, include bloodybignumber? If so, how about bloodybignumber factorial?

That's surely easy - it means any positive integer, integers being whole number like 1,2,3,4,...
There is no limit to how big integers get! Also there's no limit to how big the coin values x^k get
as k grows.

I don't care enough, I'm afraid, but if I *did*, then having resolved those dilemmae, I would
probably look at brute forcing a few thousand candidate x's (3.0000, 3.0001, 3.0002, 3.0003 etc) and
then try to spot a pattern.

That seems like a dead end - you will just be plagued by issues of rounding errors. You are not
"seeing the problem" in the right way :)

I would also look for tricks, eg i. >

i is not greater than 3.3, and neither is 4i etc.. x > 3.3 entails x being a real number...

I have not yet attempted to solve the problem, but as a BIG starter, if x were transcendental (like
Pi), how could 15 be paid...?

Mike.

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On 09/02/2025 11:57, Mike Terry wrote:

<snip>

And what does "any positive integer" mean? Does it, for
example, include bloodybignumber? If so, how about
bloodybignumber factorial?

That's surely easy - it means any positive integer, integers
being whole number like 1,2,3,4,... There is no limit to how big
integers get!  Also there's no limit to how big the coin values
x^k get as k grows.

But these are actual minted coins, so there must be a finite
number of them, yes? Or does the government mint new coins for
every transaction? Really?

I don't care enough, I'm afraid, but if I *did*, then having
resolved those dilemmae, I would probably look at brute forcing
a few thousand candidate x's (3.0000, 3.0001, 3.0002, 3.0003
etc) and then try to spot a pattern.

That seems like a dead end - you will just be plagued by issues
of rounding errors.  You are not "seeing the problem" in the
right way :)

But the right answer is expressed to 4dp when submitted.

I would also look for tricks, eg i. >

i is not greater than 3.3, and neither is 4i etc..  x > 3.3
entails x being a real number...

3.3i then, or whatever. Besides, it was just an aside.

I have not yet attempted to solve the problem, but as a BIG
starter, if x were transcendental (like Pi), how could 15 be
paid...?

Presumably we're looking at a variation of e^i.pi = -1

But let us say that you can pay 15 with your x, whatever it might
turn out to be, we then have to show that you can WITH THE SAME X
pay 15!, 15!!, 15!!! etc - using no more than 14 coins of any
denomination.

I'm still not finding it plausible.

--
Richard Heathfield
Email: rjh at cpax dot org dot uk
"Usenet is a strange place" - dmr 29 July 1999
Sig line 4 vacant - apply within

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On Sun, 9 Feb 2025 09:40:20 -0000 (UTC), David Entwistle wrote:

When I look at the question, my reaction is "that doesn't look
possible".

I suspect that is a defence mechanism I've evolved to have - "that looks difficult" - assume it is impossible and so avoid doing any work
associated with anything difficult.

The question was posed by University of Manchester mathematicians and
answered to their satisfaction a day after the problem was posed. So, I
should assume it is solvable. There was clearly a bit of to-ing and fro-
ing, but they settled on a solution, which suggests it is not trivial and thereby all the more interesting.
--
David Entwistle

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On 09/02/2025 15:03, Richard Heathfield wrote:

On 09/02/2025 11:57, Mike Terry wrote:

<snip>

And what does "any positive integer" mean? Does it, for example, include bloodybignumber? If so,
how about bloodybignumber factorial?

That's surely easy - it means any positive integer, integers being whole number like 1,2,3,4,...
There is no limit to how big integers get!á Also there's no limit to how big the coin values x^k
get as k grows.

But these are actual minted coins, so there must be a finite number of them, yes? Or does the
government mint new coins for every transaction? Really?

It's a puzzle. If you like, you could assume that the mint will manufacture as many coins as
required, but, dude, IT'S A *MATHS PROBLEM* not a manufacturing problem. :)

I don't care enough, I'm afraid, but if I *did*, then having resolved those dilemmae, I would
probably look at brute forcing a few thousand candidate x's (3.0000, 3.0001, 3.0002, 3.0003 etc)
and then try to spot a pattern.

That seems like a dead end - you will just be plagued by issues of rounding errors.á You are not
"seeing the problem" in the right way :)

But the right answer is expressed to 4dp when submitted.

Yes, that's just to confirm the puzzler has found the correct solution. (The actual solution will
have infinitely many digits, but the puzzle setters cannot ask puzzlers to enter infinitely many
digits. You might say that there is a chance that the puzzler has somehow got the wrong answer, but
it just happened to match to 4dp. That is correct but unlikely.)

I would also look for tricks, eg i. >

i is not greater than 3.3, and neither is 4i etc..á x > 3.3 entails x being a real number...

3.3i then, or whatever. Besides, it was just an aside.

I have not yet attempted to solve the problem, but as a BIG starter, if x were transcendental
(like Pi), how could 15 be paid...?

Presumably we're looking at a variation of e^i.pi = -1

No, x is a real number greater than 3.3.

IF x were transcendental, then no combination of non-unit coins could sum to an integer. (That's
effectively what "transcendental" amounts to.) So the only way to pay 15 would be with 15 unit
coins, which is not allowed by the problem. So x CANNOT be transcendental! (x must be an
"algebraic" number...)

But let us say that you can pay 15 with your x, whatever it might turn out to be, we then have to
show that you can WITH THE SAME X pay 15!, 15!!, 15!!! etc - using no more than 14 coins of any
denomination.

Yes, that's the puzzle!

I'm still not finding it plausible.

If we forget all the rational/irrational stuff and just consider x=10, so we have a decimal coinage
system with coins 1, 10, 100, 1000, ... then clearly every integer amount could be payed with a max
of 9 coins of each denomination, right? But hey, what about 33^(8333!!!!!!!+1) That number is
huge, but then what about [33^(8333!!!!!!!+1)]!!!!!!!!!!!!!!!!!!!!. That's even huger!! but can
obviously be paid with no more than 9 coins of each of our denominations. [Yeah, the mint would
have to make lots of coins to pay it....]

Of course, x=10 is not the solution as x (and x^2, x^3, x^4...) must be irrational.
Mike.

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In article <vo9t64$hlp5$2@dont-email.me>,
David Entwistle <qnivq.ragjvfgyr@ogvagrearg.pbz> wrote:

Please don't post a direct answer to the question posed, but I'd welcome a >bit of guidance on Mathsbombe question 3.

I have guessed the correct answer without understanding the problem.

-- Richard

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On 10/02/2025 00:54, Richard Tobin wrote:

In article <vo9t64$hlp5$2@dont-email.me>,
David Entwistle <qnivq.ragjvfgyr@ogvagrearg.pbz> wrote:

Please don't post a direct answer to the question posed, but I'd welcome a >> bit of guidance on Mathsbombe question 3.

I have guessed the correct answer without understanding the problem.

*APPLAUSE*

--
Richard Heathfield
Email: rjh at cpax dot org dot uk
"Usenet is a strange place" - dmr 29 July 1999
Sig line 4 vacant - apply within

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On 09/02/2025 22:39, Mike Terry wrote:

On 09/02/2025 15:03, Richard Heathfield wrote:

On 09/02/2025 11:57, Mike Terry wrote:

<snip>

And what does "any positive integer" mean? Does it, for
example, include bloodybignumber? If so, how about
bloodybignumber factorial?

That's surely easy - it means any positive integer, integers
being whole number like 1,2,3,4,... There is no limit to how
big integers get!  Also there's no limit to how big the coin
values x^k get as k grows.

But these are actual minted coins, so there must be a finite
number of them, yes? Or does the government mint new coins for
every transaction? Really?

It's a puzzle.  If you like, you could assume that the mint will
manufacture as many coins as required, but, dude, IT'S A *MATHS
PROBLEM* not a manufacturing problem.  :)

No, as stated it defines a manufacturing problem, and an
insoluble one.

A maths problem would not bother to mention coins unless so doing
clarifies the problem statement, which here it clearly doesn't.

Pace! Evidently we differ over this and neither of us is likely
to change the other's mind, so I'll let you grump off into your
corner while I grump off into mine, and we can both simmer
quietly about that guy over there in the other corner. :-)

If Kevin Stone is still watching, I'd like to raise a related
grouse, if I may.

Brainbashers sometimes poses a puzzle of the day along these lines:

David, Marmaduke, Kevin and Sebastian like fried tomatoes in
their cooked breakfast, but Andrew, Harold, Luke and Stuart
prefer baked beans.

What does Richard prefer?

Kevin would have us answer 'fried tomatoes', but it'll be a
frosty day in hell before I eat another fried tomato, no matter
how many consonants my name has. Strange as it may seem, people's
likes and dislikes have no discernible connection to the number
of vowels and consonants in their names.

Yes, "here are example members from two mutually exclusive sets -
into which of the two sets would you place /this/ item?" may be a
lot balder, but it does have the advantage of not lying about the
world.

No, I don't suppose Kevin will change his puzzles for my sake,
and arguably the way they are might well make for a slightly more
entertaining puzzle for more people than it annoys; nevertheless,
I feel better for having dislodged the gripe from my sternum.

--
Richard Heathfield
Email: rjh at cpax dot org dot uk
"Usenet is a strange place" - dmr 29 July 1999
Sig line 4 vacant - apply within

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In article <vo9v4b$hlsf$3@dont-email.me>,
Richard Heathfield <rjh@cpax.org.uk> wrote:

I agree; it doesn't look possible. I was tempted to cut code, but
I hit two ambiguities. What, precisely, does "no more than 14
coins of every given denomination" mean? It could mean an
up-to-14-coin subset of the available range

Let's knock this one on the head. It is not possible to have a set of
coins as described where an arbitrary integer value can be made with at
most 14 coins in total.

Proof:

Let the coins have the denominations 1, x, x^2, ... and add in a coin
with value 0 so we can say exactly 14 coins rather than at most 14.

How many values can we make with the denominations up to x^n? There
are n+2 different such denominations (x^0 .. x^n and 0). So there
are (n+2)^14 ways of choosing our coins in order, and less than that
since the order doesn't matter. (And of course most of them won't
produce an integer anyway.)

How many different values do we need to be able to make with those
coins? All the values less than x^(n+1), since we can't use the x^(n+1) denomination or larger to make a value less than x^(n+1).

The number of values increases exponentially, but the number of coin combinations increases polynomially. So as n increases the number of
values will eventually exceed the number of combinations available to
make them.

-- Richard

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