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Your Japanese co-worker from this problem hands you another piece of paper one day:

1820143 2383341 3245353 8425463 6354607
6627972 9088779 7839589 ??????? ???????

What two seven-digit numbers come after the ones given?

On the back of the paper is a hint:

3

You get stuck, and ask your co-worker for another hint.

He simply writes a zero in front of the existing hint, leaving you with a note that reads:

03

And another:

Again, he just scribbles a plus sign in front of the zero, leaving you with a note that reads:

+03

And a fourth:

This time, he writes a whole row of numbers on the bottom:
12 34 56 71 23 45 67

And a fifth:

He crosses out the 71 23 45 67 and writes 78 9A BC DE in its place.

And a sixth:

He writes "COUNTER MODE" next to the "+03".

And a seventh, just as clarification:

Each letter is exactly two digits long.

I promise this one doesn't involve any obscure numbers like 264, or any other such esoteric facts, unlike my last one in this regard. Everything here is simple mathematics and logic.

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13
  • 1
    $\begingroup$ man i really suck at these, i thought i'd give it a try, been at it for 30 mins and got nothing... How do people do these things, really hurting my ego :P $\endgroup$
    – Vincent
    Commented Apr 14, 2015 at 7:18
  • 5
    $\begingroup$ Cryptography puzzles tend to be easy to make but hard to solve in general. What's really hard is making one that's so devious and sneaky that you can't solve it while you don't know how, but once you do know how, you curse yourself for not seeing it before. $\endgroup$
    – user88
    Commented Apr 14, 2015 at 7:35
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    $\begingroup$ do you have another hint for us? I would like to get this question answered, or at least get some guesses. I have tried to get some sort of pattern by adding, subtracting the numbers or parts of them. I didn't find any pattern so far. $\endgroup$
    – JBSregath
    Commented Apr 23, 2015 at 13:06
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    $\begingroup$ Looks like I successfully drew attention to this puzzle by answering it and bumping it back to the top of the page. Didn't get any rep for it though :-/ $\endgroup$
    – Rand al'Thor
    Commented Jun 25, 2015 at 9:20
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    $\begingroup$ @rand al'thor, some of us are in more desperate need of rep than others. Thanks for bumping up one that I could solve. $\endgroup$
    – Carmi
    Commented Jun 25, 2015 at 11:21

4 Answers 4

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The two seven digit numbers to be added are:

9697011 6131119

1820143 2383341 3245353 8425463 6354607
6627972 9088779 7839589 ??????? ???????

Based on hint 4, split the input into pairs of digits:

18 20 14 32 38 33 41 32 45 35 38 42 54 63 63 54 60 76 62 79 72 90 88 77 97 83 95 89 ?? ?? ?? ?? ?? ?? ??

Graphing these numbers shows that the numbers trend upward by a factor of three for each character. This is what is referred to by hints 1, 2, 3, and 6. (Thanks to Quark for the graph)

enter image description here

Reducing these numbers by the factor of three gives:

15 14 05 20 23 15 20 08 18 05 05 06 15 21 18 06 09 22 05 19 09 24 19 05 22 05 14 05

Noticing that these are all less than 26 convert them to characters yields:

O N E T W O T H R E E F O U R F I V E S I X S E V E N E

Now using hint 4 and 5, and noticing that the sequence of characters ends with "E" then continuation is E I G H T N I N E ... Reverse calculating the new numbers for the plain text is:

15 14 05 20 23 15 20 08 18 05 05 06 15 21 18 06 09 22 05 19 09 24 19 05 22 05 14 05 09 07 08 20 14 09 14

Adding the slope back in and taking the modulo by 100 (two digits only) yields:

18 20 14 32 38 33 41 32 45 35 38 42 54 63 63 54 60 76 62 79 72 90 88 77 97 83 95 89 96 97 01 16 13 11 19

Grouping the numbers back into sevens:

1820143 2383341 3245353 8425463 6354607
6627972 9088779 7839589 9697011 6131119

Thanks to rand'al thor for bumping the puzzle up, and Quark for saving me from making my own graph.

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  • $\begingroup$ The last two digits are supposed to be 89, not 99. Funny how a single letter (or rather, an off-by-one error in addition) changes the entire meaning of the puzzle... >_>; $\endgroup$
    – user88
    Commented Jun 25, 2015 at 14:08
  • $\begingroup$ @JoeZ. Really? You want to have a hanging digit left at the end? "O N E T W O T H R E E F O U R F I V E S I X S E V E N E I G H T N" only has length 69 when converted to it's encrypted values. Add the I and you'll get an extra "111". We can use that first "1" to make 70 but then what to do with the extra two ones? This makes no sense unless there's something I'm missing. $\endgroup$
    – LeppyR64
    Commented Jun 25, 2015 at 14:19
  • $\begingroup$ It's addition modulo 100, not straight addition. $\endgroup$
    – user88
    Commented Jun 25, 2015 at 14:22
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I think I have it. The answer is:

9697011 6131119

The idea is as follows, based on Quark's work, and on Hint 6 ("Counter +03").

First, split the strings into couplets giving:
18 20 14 32 38 33 41 32 45 35 38 42 54 63 63 54 60 76 62 79 72 90 88 77 97 83 95 89 ?? ?? ?? ?? ?? ?? ??.
After that, subtract 3 times the couplet's ordinal place from it giving the new string:
15 14 05 20 23 15 20 08 18 05 05 06 15 21 18 06 09 22 05 19 09 24 19 05 22 05 14 05
From here, mapping each number to a letter is trivial, as it's just its position in the alphabet, so we get the string:
O N E T W O T H R E E F O U R F I V E S I X S E V E N E

My guess is that we continue the next two numbers in plain text, convert them backwards, and get the answers at the top, which gives us:

I G H T N I N
09 07 08 20 14 09 14
+87 +90 +93 +96 +99 +02 +05
96 97 01 16 13 11 19

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8
  • $\begingroup$ I'd appreciate any help with formatting the answer, especially if you could show your work. $\endgroup$
    – Carmi
    Commented Jun 25, 2015 at 7:35
  • $\begingroup$ The question only asks for the next two seven digit numbers. Good work finishing off the rest of the answer. $\endgroup$
    – LeppyR64
    Commented Jun 25, 2015 at 10:16
  • $\begingroup$ Can you add in what the different hints were supposed to hint at? $\endgroup$
    – BmyGuest
    Commented Jun 25, 2015 at 12:21
  • $\begingroup$ I actually did this except I subtracted 3 from the 2nd one, 6 from the 3rd one, etc. I do wonder how some of those hints are supposed to help though (red herrings argh...). Anyway, one possibility (although kinda weird) is to represent the first two digits of each triplet using hex (100 = A0, 112 = B2, etc.) since one of the hints implies that. You can try adding that to your answer if you want. $\endgroup$
    – Quark
    Commented Jun 25, 2015 at 13:08
  • $\begingroup$ To be honest, I only used Hint 6, which is an amalgam of hints 1,2 and 3, and Quark's work which is credited in the answer. I honestly have no idea what hints 4 and 5 are about. Looking at them now, Quark's suggestion of writing the numbers in Hex would explain hint 5. Basically, the linear progression and the word "counter" tripped the solution circuit. $\endgroup$
    – Carmi
    Commented Jun 25, 2015 at 13:30
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I've thought about this for far too long and am fairly confident a leap of logic is required at this point in time without further hints. Here's a partial answer though:

From this graph, there's definitely a linear progression if the numbers are broken up into every two. The second column has the differences in case anyone can see a pattern in those.

enter image description here

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5
  • $\begingroup$ You're on the right track. What's the slope of your trend line? $\endgroup$
    – user88
    Commented Jun 25, 2015 at 2:57
  • $\begingroup$ @JoeZ. ofc that's the first thing I noticed, with the hints as added help. Need more than that though :\ $\endgroup$
    – Quark
    Commented Jun 25, 2015 at 2:57
  • $\begingroup$ Alright, hint number six is up. $\endgroup$
    – user88
    Commented Jun 25, 2015 at 2:59
  • $\begingroup$ @JoeZ. Ok, near 30 minutes of looking up CTR encryption/experimenting and not a bit closer. I hate giving up on puzzles I've spent time on but oh well. Looking forward to the solution, especially if this puzzle is really solvable as originally posted (as it wouldn't make sense for a posted puzzle to start out as initially unsolvable). $\endgroup$
    – Quark
    Commented Jun 25, 2015 at 3:31
  • $\begingroup$ Do you notice how the distribution of letters slopes up? That's because a cipher was applied to it. Turn it back into a flat one. $\endgroup$
    – user88
    Commented Jun 25, 2015 at 3:42
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I'll summarise my progress in an answer. Maybe it'll help others get further.

The fourth and fifth hints suggest we should

turn the sequence of 7-tuples of letters into a sequence of pairs by changing the spacing.

Then the given ciphertext:

1820143 2383341 3245353 8425463 6354607
6627972 9088779 7839599 ??????? ???????

becomes:

18 20 14 32 38 33 41 32 45 35 38 42 54 63 63 54 60 76 62 79 72 90 88 77 97 83 95 99 ?? ?? ?? ?? ?? ?? ??

The first three hints combine to give

+03. Maybe we're meant to add 03 to all or some of the 2-digit numbers we now have?

But neither of

21 23 17 35 41 36 44 35 48 38 41 45 57 66 66 57 63 79 65 82 75 93 91 80 100 86 98 102 and 15 17 11 29 35 30 38 29 42 32 35 39 51 60 60 51 57 73 59 76 69 87 85 74 94 80 92 96

has any obvious pattern that I can see...

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3
  • $\begingroup$ Try this grouping method instead: 18 20 14 32 38 33 41. $\endgroup$
    – user88
    Commented Jun 24, 2015 at 18:54
  • $\begingroup$ @JoeZ. OK, I've updated my answer. Still not really sure where to go next... $\endgroup$
    – Rand al'Thor
    Commented Jun 24, 2015 at 22:40
  • $\begingroup$ Guess you need another hint. Try graphing the numbers in a line plot. Notice anything? $\endgroup$
    – user88
    Commented Jun 24, 2015 at 23:02

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