Re: MathsBombe

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.  :)

 

>
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.

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..  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.