14
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See Part I | See Part III

Nice one, @GarethMcCaughan!

Finally, you've gotten into his computer using the password you managed to decode from his clues... but there's another issue! There is a secondary box asking for a six-character passcode (excluding spaces). Fortunately, you found some strips of paper on his desk that will be helpful... enter image description here

You are starting to get anxious about this mystery... better get cracking!


Hint 1

Use the two outside numbers for each group of three to get the inside number in some way...

Hint 2

The code is six characters... An asterisk counts as a character!

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  • $\begingroup$ That extra 100 seems more than is needed, but I'm not going to complain at getting free Meaningless Internet Points... $\endgroup$ – Gareth McCaughan May 14 at 16:49
  • 1
    $\begingroup$ “Meaningless Internet Points.” :) $\endgroup$ – Voldemort's Wrath May 14 at 16:50
9
+150
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Possibly right answer

To find the number in the middle of a strip,

first take the digit-by-digit differences between the numbers at the ends:
483 672 -> 211
976 132 -> 844
759 324 -> 435
148 381 -> 247
533 215 -> 322
891 467 -> 436
318 649 -> 331
948 682 -> 346
then add up the resulting three digits and divide by 2:
483 672 -> 211 -> 2
976 132 -> 844 -> 8
759 324 -> 435 -> 6
148 381 -> 247 -> 6.5
533 215 -> 322 -> 3.5
891 467 -> 436 -> 6.5
318 649 -> 331 -> 3.5
948 682 -> 346 -> 6.5
and then replace ".5" with "*". The first five lines above then yield the numbers seen in the picture. The last three lines are for the other strips, in the order indicated by the arrows.

This suggests that the required code is

6*3*6* (if I've done my arithmetic correctly).


Certainly wrong answer posted earlier, left here so as not to hide my old mistakes:

I will be astonished if this is the intended answer, but it is consistent with the information provided [EDITED to add: ... no, it turned out not to be, because I'd completely missed one of the provided data points] and it's (I think) a little simpler than RedBaron's answer.

To get from the outer numbers to the inner:

For each number, subtract the sum of the two outer digits from the middle digits. E.g., $(483,672)$ becomes $(1,-1)$. Now subtract the number derived from the larger of the two numbers from the one derived from the smaller, so we get $1-(-1)=2$. Finally, if the result is negative then replace its minus sign with an asterisk and write it upside down. So $(148,381) \rightarrow (-5,4) \rightarrow -9 \rightarrow 6*$.

Thus:

our three mystery paper slips, in the order indicated by the arrows, yield $-5$ (hence 5*), $1$, $13$. Pad everything to two characters (adding a space where the asterisk would go if there were one) and we get the 6-character code 5*1 13.

(An earlier version of this answer had a miscalculation in it, which I have now fixed.)

The most unsatisfactory thing here is of course

that space.

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  • $\begingroup$ Very well done! +65 rep for you! $\endgroup$ – Voldemort's Wrath May 14 at 16:33
  • $\begingroup$ Time to go over to Part VII! $\endgroup$ – Voldemort's Wrath May 14 at 16:35
5
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Is the passcode

268716


A convoluted rule which stretches to fit all

For the three digits number on either side do this for each set of digits:

A = Absolute difference of first digits.
B = Absolute difference of second digits.
C = Absolute difference of third digits.

Compute (A-B), save as D. If D<0, add 1 to D (Only way I could match the lower left number).
D + C gives middle number.

Using this we get for unknown numbers:

891 and 467 = (abs(8-4) - abs(9-6)) + abs(1-7) = 1 + 6 = 7
318 and 649 = (abs(3-6) - abs(1-4)) + abs(8-9) = 0 + 1 = 1
948 and 682 = (abs(9-6) - abs(4-8)) + abs(8-2) = -1 + 6 = 5 + 1 (Add 1 as diff is less than 0) = 6

The three middle numbers are

716

As we need 6 character passcode,

Use the non starred middle numbers to get 6 digits. First three numbers are visible and can be arranged as per visual order 268. The next three digits can be arranged as indicated by arows 716.

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  • $\begingroup$ You’re sort of on the right track, but not close enough for an upvote... $\endgroup$ – Voldemort's Wrath May 3 at 19:15
  • $\begingroup$ @InventPalooza it was a very long shot $\endgroup$ – RedBaron May 4 at 6:04
2
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Another possible approach

1. Put the digits of the larger number in numerical order (e.g, $759 \rightarrow 975$)
2. Subtract the smaller number from the result.
3. The middle number is the first digit of the result, it has an asterisk if the second and third digits have different parity (one even and one odd)

Examples

$(483,672) \rightarrow 762 - 483 = 279 \rightarrow 2$     (as $7$ and $9$ both odd)
$(759, 324) \rightarrow 975-324 = 651 \rightarrow 6$     (as $5$ and $1$ both odd)
$(148, 381) \rightarrow 831 - 148 = 683 \rightarrow 6*$   (as $8$ and $3$ opposite parity)
$(976, 132) \rightarrow 976-132 = 844 \rightarrow 8$      (as $4$ and $4$ both even)
$(533, 215) \rightarrow 533-215 = 318 \rightarrow 3*$   (as $1$ and $8$ opposite parity)

Filling in the unknown boxes

$(891, 467) \rightarrow 981-467 = 514 \rightarrow 5*$
$(318, 649) \rightarrow 964-318 = 646 \rightarrow 6$
$(948, 682) \rightarrow 984-682 = 302 \rightarrow 3$

Which means the code could be

"5*6 3 "
if we take the same reasoning as Gareth McCaughan and consider a space an accepted character.

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  • 1
    $\begingroup$ +1, but no. As I will soon add to my question, spaces don’t count as character. $\endgroup$ – Voldemort's Wrath May 14 at 16:31

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