I was flipping through some of my old puzzling notebooks, and I found an old puzzle of mine I don't quite remember.

I tried to find when I made it (I put the date I made all of my puzzles on them), but I couldn't.

After looking at the puzzle a bit more, I figured out... that's what I had to find.

enter image description here

I eventually figured out when I made the puzzle... but... can you?

Note: I believe my younger self might have missed something on the geometry part. The shape with all sides equal is a square.

Hint #1:

You need to be a mastermind to solve part one.


5 Answers 5


The answer is

17 May, 2016

Building on the answer from Mohirl, they got parts 1, 3, and 4 correct, but their mistake was in part 2.

Move the first digit to the end. So 534 becomes 345. From the comments, part 1 was presented in roman numerals, so 345 was represented as iiiivv, and splitting it up in the groups the way the x are represented, we get iii i v v, equal to 3 1 5 5, and adding 1 to each number we get 4266 = r. The following math is therefore 4266 = sqrt(2) * (x + 1000), so x = 2016.5175... From part 4, we match the pattern and see that the year is 2016, month is 5, and date is 17.

  • 2
    $\begingroup$ Er... this is correct, proper format and all. Why the downvotes? $\endgroup$ Commented Nov 21, 2018 at 0:42
  • $\begingroup$ Ha! I'd somehow missed this answer and just spent 20 minutes completing my own one! And I still got it wrong, because I rounded instead of truncating. Well done! $\endgroup$
    – Mohirl
    Commented Nov 21, 2018 at 12:29

Updated to complete answer

Final answer:

18th May, 2016

Part 1

I believe the answer to part 1 is


I suspect the hint is talking about

the Mastermind board game

which means that the numbers on the right for each line

indicate how many of the number on the left of the row are 1) correct but in the wrong position or 2) correct and in the correct position. Analysing these should give the correct values for the last row

So, starting with row 5:

We have a 2 on the right hand side, so one number is in the correct position, and the other two are wrong. Either the first digit of the hidden code is 1, the second is 9, or the third is 4. Lets assume 4 is the correct one (since that's the one we have most other information about). That means neither 1 or 9 are in the code.

Then based on row 1:

4 is correct, but in the wrong position, and neither 2 or 7 are in the code.

From row 3:

we already know 4 is in the wrong position, but so is either 3 or 6. 6 is in the last position in both rows 2 and 3, but it can't be right in one but wrong in the other. So 6 is out, and 3 must be the other correct but misplaced digit.

That means that row 4:

is telling us that 3 is in the second position. Which we already knew, since it's not the third (that's 4) and not the first (it was in the wrong place on row 3). So 2 and 8 are out.

And since we now know what the last two digits are, row 2

must be telling us that 5 is correct in the first position. So we have 534 as the answer for the missing row 6.

Part 2

Take the answer from part 1, and

This took a while, but think I finally have it based on @ExcitedRaichu's hints in the comments. The answer to part 1 isn't actually 534,but V III IV. Taking the first character and putting it at the end gives III IV V, but then formatting that as XXX X X X gives III I V V, or 3 1 5 5. Add 1 to each digit to get 4266.

Part 3

If that's a square, then

The diagonal also equals the radius. So sqrt(2)*(x+1000) = r. So take r from Part 2, divide it by sqrt2, and subtract 1000.

Part 4

The result from part 3 should

give us 4 digits before the decimal point as the year, the first after as the month, and the next two the day. 4266/sqrt(2) gives 3016.517529... Subtract 1000, and we get 2016.518 when rounded to 3 decimal places (or 2016.517 truncated). Which based gives us a date of 18th May, 2016.

  • $\begingroup$ Correct so far. +1 $\endgroup$ Commented Nov 20, 2018 at 18:18
  • $\begingroup$ For Part 2, consider how part 1 was presented $\endgroup$ Commented Nov 20, 2018 at 18:33
  • $\begingroup$ try converting back into roman numerals, individually, then combining them together $\endgroup$ Commented Nov 20, 2018 at 18:47

Partial answer:

1. I have no idea what 1 is, it could possibly be with roman numbers.
2. the last entry of 1. (where the first is the last equals somehow xxx x x x = r)
3. r = root( 2 * (x + 1000))
4. probably: write x with 7 digits, 4 in front of the dot and 3 behind it. This should mean x is around 1000 to 9999 big



Step 1:

Converting from Roman numerals to decimal, the diagram becomes:
2 : 4 : 7 | 1 : :
5 : 1 : 6 | 2 : :
3 : 4 : 6 | 1 : 1 :
8 : 3 : 2 | 2 : :
1 : 9 : 4 | 2 : :
###### | 2 : 2 : 2
I can not determine any noticeable pattern here...

Step 2:

Without step 1, I've idea what to do here...

Step 3:

The value of $x+1000$ is given by using Pythagoras. The diagonal of the square is the radius of the circle, $r$, and using the Pythagorean Theorem, we can determine that: $2 \cdot (x+1000)^2 = r^2$.
And therefore, $x = \frac{r}{\sqrt{2}} - 1000$.

Step 4:

I assume that step 3 will return a decimal number for x, and that the date of the puzzle creation can be determined by taking the first 4 digits as the year, the 1st decimal place as the month, and the 2nd and 3rd decimal places as the day of the month.


Very Partial Answer
Here's my initial thinking:

If step 1 is a matrix, it is unsolvable on its own because of the unknown third variable on the right side. If you attempt to solve the first five lines, there is only the trivial solution. Assuming, of course, that the symbols are literal roman numerals. I suspect that this is not the case.
Step 2 seems to suggest that shifting the first character of the answer to step 1 to the end of the string becomes XXXXXX when one is added, or XXXXXX is a set of replacement characters for the string (suggesting it is 6 characters) that becomes the radius of the circle when 1 is added.
Steps 3 and 4 appear to be self-explanatory


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