Xmas Puzzle 2020 (1)

[L D[AR2P#0.0]]

Contruct three fractions all of the same value, using in every group all the nine digits once only. The fractions may be formed in the ways shown below. At least five different cases can be found, although one of the five may be considered contentious


Answers by PM.

 

[drbitboy]

I request a clarification/disambiguation: is e.g. ghj equal to (g*100)+(h*10)+(j) or equal to (g*h*j)?


The former makes the most sense, but my mind went first to the latter.

 

[Steve Bailey]

Perhaps either way. It is, after all, a puzzle and the text hints at a possible "contentious" answer.

 

[L D[AR2P#0.0]]

I request a clarification/disambiguation: is e.g. ghj equal to (g*100)+(h*10)+(j) or equal to (g*h*j)?


The former makes the most sense, but my mind went first to the latter.

Former.

 

[I_Automation]

You forgot something in your calculation:

 

[drbitboy]

which nine digits?


Brute force finds six solutions (actually four, two for each case, but a/b and c/d can be swapped in the left-hand case)

 

[L D[AR2P#0.0]]

For the purposes of this puzzle, you have found four.

 

[drbitboy]

For the purposes of this puzzle, you have found four.

agreed, and nothing contentious.

 

[L D[AR2P#0.0]]

That's where the fifth one comes in.

 

[Peter Nachtwey]

Each combination actually can be counted as two. Just invert b/a=d/c=


@drbitboy, I like brute force but why did you ignore b/a?
Also, once a/b are selected that narrows down the choice for c/d you can continue of a/b!=c/d. There is no point wasting time on the third fraction.


de is calculated twice. First as a decimal number, then as a hex number. Why multiply by 16.

 

[drbitboy]

Each combination actually can be counted as two. Just invert b/a=d/c=


@drbitboy, I like brute force but why did you ignore b/a?

Actually this was already covered here, although admittedly not in detail, and the statement above is wrong:

The alternate cases swap a/b and c/d (i.e. c/d=a/b=...); inverting those ratios (reciprocal) necessitates inverting the third ratio as well because numerator and denominator have different number of digits.


so inverting the ratios does not yield additional cases.

Also, once a/b are selected that narrows down the choice for c/d you can continue of a/b!=c/d. There is no point wasting time on the third fraction.

Actually I included that optimization: cf. the [(a*d)==(b*c)] test in the [if] statement on line 9.

de is calculated twice. First as a decimal number, then as a hex number. Why multiply by 16.

I was checking to see if the "contentious" case referred to hexadecimal numbers.


You didn't see that? Maybe you should not assume you are looking at code written by the paperboy.

 

[L D[AR2P#0.0]]

Update - 5 solutions submitted by Steve Bailey - his contentious solution is different to mine so we now have at least 6 solutions of which two are contentious.


NB digits/numbers are decimal.

 

[Steve Bailey]

Update:

Initially I stopped looking once I found the fifth solution using my "contentious" methodology. Since Simon's response I've found three more solutions using the same contentious approach.

I think Simon's "contentious" solution is a different approach.

 

[Steve Bailey]

I'm now up to thirteen solutions. There are probably more to be found, but I haven't coded a search script.

Four non-contentious solutions.
Four more with my first "contentious" technique. which I agree could be considered contentious but different from Simon's contentious solution.
Five more using my second "contentious" technique, which I don't think is particularly contentious.

 

[drbitboy]

By the way,


LD...: great puzzle!


LD/SteveB: I look forward to seeing these "contentious" approaches; I'm drawing a blank so I got bupkis on that front.