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A client of mine died last Sunday - just as he was born on a Sunday. He was already at rather ripe age, and I had the pleasure to present his last will to his loving and caring family. As far as I know, they haven't seen him in years, but they for sure knew one thing: He'd made a lot of inventions in his years and his patents are worth a lot. A real lot. Believe me. They were all in for quite a shock, because the old chum did not leave anything to them. So sad. All his patents will go to the one person who's able to open his safe. And with some luck - and maybe your help - this will be me, making me great again.

The safe is in a big wall - a really big wall - to the south. Close to Mexico, actually. Quite close. Very big. Anyway, it appears to require a 11 digit entry code, and all that my client left for a clue was one of his latest inventions: A picture frame containing a scrambled image, which looks exactly like this:

The code

(Download of file advised)

He also clued that image with a title: the Day a great Man was born to Ye' all.(8)

Other than that: Nothing. So sad. Well, this is fake news. For the family. Because I have a digital copy of some pages from his late diary. Given to me for safekeeping together with his last will when we last met in April. Very good man, very good. Still, I'm not sure his diary entries are all too useful. And they are of very bad digital quality. Maybe they help you?

Page1 Page2 Page3 Page4 Page5 Page6 Page7

Can you determine the 11 Digit Code for me?

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    $\begingroup$ I've no idea what this is all about, but it's a BmyGuest puzzle, so it gets an upvote. :-P $\endgroup$
    – Rand al'Thor
    Commented Jun 18, 2017 at 22:30
  • $\begingroup$ Once again a safe puzzle... $\endgroup$
    – Mordechai
    Commented Jun 19, 2017 at 0:30
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    $\begingroup$ wish I know what am I doing i.sstatic.net/QZX4x.png $\endgroup$
    – Jan Ivan
    Commented Jun 19, 2017 at 5:49
  • $\begingroup$ I'm wandering if perhaps we need a picture of the 'digital portrait' on its own to test on...? Might be necessary might not... depends if what I'm thinking is right which it probably isn't :P $\endgroup$
    – Beastly Gerbil
    Commented Jun 19, 2017 at 6:41
  • $\begingroup$ @BeastlyGerbil All you need should be up there already. But I don't think it can be solved without good use of a computer. (Bonus points for anybody solving this "by hand" for sure!!) $\endgroup$
    – BmyGuest
    Commented Jun 19, 2017 at 7:46

5 Answers 5

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My first attempt at an answer was "Too Good"... so here's a second one.

The PIN is:

33154331121

Jan Ivan has already done all the hard work and deserves all the credit for the answer. However, he has been tripped up by the clever ruse that:

"All that is mine will remain mine" means that the safe will remain locked forever.

Luckily for us however...

The message, "All that is mine will remain mine", has 11 unique letters. Sorting them and counting them gives the following:
ax3 ex3 hx1 ix5 lx4 mx3 nx3 rx1 sx1 tx2 wx1
Yielding the PIN: 33154331121

It looks like Luke pretty much pointed this out on Penguino's answer...

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  • $\begingroup$ Thanks! Now I can rest in peace with another puzzle "checked off". This was my intended solution, although I personally liked your other one even a bit better when I saw it 8x) $\endgroup$
    – BmyGuest
    Commented Feb 28, 2018 at 19:38
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It was a pain to solve this, but fun.

First, think about shifting columns/rows in the picture.
For each number in key you shift the column pixels and then the row pixels. To solve the picture, you reverse this process.

You can test this with the last diary picture:

enter image description here enter image description here
The key for this picture is 9,8,7,6,5,4,3,2,1 as shown as hint in the picture above. First you shift the first row by 9 pixels (to the right), then the second row by 8 pixels and so on…
You do the same for columns (shift downwards). And you do this for whole picture.
What puzzled me was - You have to repeat this process 9+8+7+6+5+4+3+2+1=45 times. (And not 9 times as I first thought.)
Following the above, you have encoded (messed up) the picture this way.
For decode you, simply reverse process (first columns)

Now for the real puzzle:

To decode the main picture you have to find the key first. The hint states it should be a date - properly the birth date of the man himself. You have plenty of hints in pictures/text, like: 80th birthday, week in 2016 (May 13 Friday is 2016) around 18. June = go to 18 June 1936. You can also find in the text "just as he was born on a Sunday" so it is either 14 June 1936 or 21 June 1936. And if you also recognize the 'Trump reference' and know that he was born 14 June 1946, you guess the correct date is 14 June 1936. And I was stuck here for a long time, since people should use ISO format 19360614 or I suspected US format because of reference 06141936, but the correct one (that was also used in puzzle) was 1,4,0,6,1,9,3,6. I knew, I was close because the number of iterations gave always "First red" in the picture, but I didn't know the correct format.

Applying the decode algorithm with the correct key gives you:

enter image description here
But I'm not sure if there is any 11 digit code because of this friendly text....

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  • $\begingroup$ Well, nice image decode, but how do you open the safe? (Or don't you?) $\endgroup$
    – BmyGuest
    Commented Jun 20, 2017 at 7:08
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    $\begingroup$ @BmyGuest Yea I'm fast with finding solution, but not fast in explaining it :) $\endgroup$
    – Jan Ivan
    Commented Jun 20, 2017 at 7:30
  • $\begingroup$ @BmyGuest yea i know about title "When i was born" but didn't noticed DMY. I have no idea how to change 7 words to 11 numbers. So maybe in few hours/days. $\endgroup$
    – Jan Ivan
    Commented Jun 20, 2017 at 8:00
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Jan Ivan has already done all the hard work and deserves all the credit for the answer. However, he has been tripped up by the clever ruse that:

"All that is mine will remain mine" means that the safe will remain locked forever.

Luckily for us:

The pot of gold is at the end of the rainbow. The background of the picture is a rainbow. 7 colours, Red, Orange, Yellow, Green, Blue, Indigo, and Violet.
Converting the first letters to their integer location in the alphabet, starting with A=1, we get 18 15 25 7 2 9 22, 11 Digits.

Therefore the combination of the safe is:

18152572922

Reasoning:

There are clues through about such as "Tinkerer", "Luck" and "Ye'all" and "Chum" that hint loosely that the man is Irish. Leprechauns are an irish fiction that have pots of gold. Leprechauns are said to hide their pots of gold at the end of rainbows.

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  • $\begingroup$ This answer is so cool that I really whished it would have been true. +1 and if there would be a 2nd safe, I would make sure to use THAT code. But nai, the safe unfortunately does not open. Ah, actually, the old tinkerer had monochromatic vision I think... 8c] $\endgroup$
    – BmyGuest
    Commented Jun 20, 2017 at 12:50
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    $\begingroup$ @BmyGuest That's it. Flagged too broad ;) $\endgroup$
    – LeppyR64
    Commented Jun 20, 2017 at 13:14
  • $\begingroup$ No second idea on this? It really isn't that complicated to cose this one. (The 'tricky' part was the image - and realizing that there is (?) a second level.) $\endgroup$
    – BmyGuest
    Commented Jun 22, 2017 at 6:45
  • $\begingroup$ @BmyGuest I haven't got anything. I haven't had an opportunity to devote a proper amount of time and to building a simulator yet. Not sure that the solution still requires me to do that. I don't fully understand "3-3-lucky 7" yet. $\endgroup$
    – LeppyR64
    Commented Jun 26, 2017 at 2:35
  • $\begingroup$ No need to go too deep. In fact, your solution (which I still like and am sad about it's not my soluton) is maybe "deeper" than mine. And you may not understand some of the hints because you didn't solve the image-puzzle. Not saying anymore now though. $\endgroup$
    – BmyGuest
    Commented Jun 26, 2017 at 6:34
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After reading BmyGuest's comment on Hugh Meyers post "[the code] only requires the decoded image" I noticed that:

the doubled 'll' in All, and the doubled 'll' in will are identical (both ll, and both on an orange background, similarly there are two copies of 'mi' on a indigo background and two copies of 'ne' on a violet background. So there are only 11 unique combinations of letters(or letter pairs)-colour: A-Red, ii-Orange, th-Yellow, at-Green, is-Blue, mi-Indigo, ne-Violet, wi=Red, re-Yellow, ma-Green, and in-Blue. If the letters(letter pairs) are sorted alphabetically, and substituted by their associated colors then you would get RGBBOGIVYYR, and substituting (Red=1, orange=2...) you get 14552467331 for the safe code.

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    $\begingroup$ Similarly, there are only 11 unique letters in the image. 3x A, 4x L, 2x T, 1x H, 5x I, 1x S, 3x M, 3x N, 3x E, 1x W, 1x R $\endgroup$
    – Luke
    Commented Sep 14, 2017 at 7:17
  • $\begingroup$ Very nice idea and right on track... @Luke got a bit closer though :c) $\endgroup$
    – BmyGuest
    Commented Sep 14, 2017 at 20:17
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I really wouldn't feel justified in collecting a bounty from this but it occurred to me that:

The date was the first part of the code: 14061936, as found by @JanIvan. We now need three more digits to complete the eleven digit sequence.

I noted two things:

The algorithm has an odd name: 3-3-lucky seven. I thought the last number might be spelled out to hide the fact that this is a three-digit sequence. The key is eight digits, and the algorithm is three digits. This makes up the eleven digits required and giving the key and algorithm sounds like a reasonable solution: . The second thing, which may or may not be a clue, is the statement at the end of the first paragraph saying that the safe might be opened with a little luck and our help. This might be a reference to the "lucky" algorithm name. Or maybe not since the statement is made by the narrator who does not know the solution.
I would therefore try: 14061936337. If that didn't work, 33714061936 may be worth a shot.

I can't see anything else that fits, so I offer this as a possible solution.

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  • $\begingroup$ +1 for picking this up.It is not the solution though. (And, as should be expected, the final solution requires the decoded image.) Hint (because I want to put this to rest): It only requires the decoded image. $\endgroup$
    – BmyGuest
    Commented Sep 13, 2017 at 10:07

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