Re: MathsBombe

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 — 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.