# What is the rule?

I have a secret rule. Can you find out what it is, with this rule, I take in any number, and output two numbers! The answer to this puzzle is the same as the answer to the question: How do I get the two numbers from the one input?

Here are some examples:

IN  | OUT
1   |1,1
2   |2,2
3   |3,3
4   |3,2
5   |2,1
6   |3,2
7   |4,3
8   |5,4
9   |3,3
10  |2,2
37  |10,9
60  |3,4
400 |2,2
700 |3,3
2789|17,20


Hint 1 (light):

This pattern is lossy, meaning that 2 inputs can share an output, but not more than Round((e)^(Max(Output1, Output2))) for a given input share that given output.

• Feel free to request more numbers in the comments, I'll put them up over time Nov 1 '19 at 19:59
• As said in my answer below, I would like to request the output for 400 and 700, please. Nov 1 '19 at 20:59
• @WhatsUp 400 and 700 have been posted Nov 1 '19 at 21:07
• Thanks. I've updated my answer, but still puzzled by the "meaning" of the outputs... Nov 1 '19 at 22:15
• The presence of 'e' in the maximum count in the hint leads me to think of pi which in turns leads me to think of rotations which in combination with the Roman numerals makes me think of hour/minute hands on a clock. Yes it's a stretch... Nov 2 '19 at 0:57

## 2 Answers

Based on @WhatsUp 's progress, here's the solution

• I'm confused. What are rot13("urnqf naq yrtf")? Nov 2 '19 at 21:59

The first component of the output is (almost)

and the second component of the output is (not quite)

Thus my guess is that it is closely related to

Roman numerals.

Based on this hypothesis, and using this website

I identify the weights as follows:

I -> 1, 1
V -> 2, 1
X -> 2, 2
L -> 1, 2
C -> 1, 1
D -> 1, 1
M -> 2, 3

Based on the new given outputs for 400 and 700, all the weights are now uniquely determined, assuming that both components of the output are linear combinations of the Roman numerals.

And I still don't see what these really means. But if my assumptions are correct, then I can already give the outputs for all integers from 1 to 4999, at least.

I'm also interested in the output for e.g. 123456789. That is, if I'm on the right track...