jcpl puzzle 2026
a writeup for those who weren't able to solve it
this post contains spoilers! if you wish to try the puzzle yourself, you may find the puzzle here
stage 0
this stage presents the user with a long string of 1's and 0's
011101000110100001100101001000000111001101101111011
011000111010101110100011010010110111101101110001000
000110100101110011001000000011011000111000001101100
011011100110110001101000011100000110111001101000011
001000110010001101010011001000110111001101110011011
1001101010011011100110000
this is a string of binary. binary is a numbering system, so an idea is to convert the entirety of the string into decimal and try that as a solution
1023933776370555 876710641523592 231886365159532 971295101776707 309077005278107 20264752
which fails. binary, apart from representing numbers, are also used to represent text in a format called ASCII
converting the above string of binary into ASCII
the solution is 6867648742252777570
yields the solution
stage 1
this stage presents the user with a large list of integers
-30
123
-98
126
[...more numbers]
the hint on the page states
i wonder what the sum to all of these numbers are...
as was seen, summing all these numbers together produces the solution
stage 2
this stage presents the user with a list of 0's and 1's, this time explicitly delimited by newlines
00110011
11000111
10110101
01100001
00011001
11110011
11011011
11111111
10101111
00101011
[...more binary]
trying the first solution again and converting each line into ASCII:
3ǵaóÛÿ¯+
[...more gibberish]
produces gibberish. at the top of the screen once more is
i wonder what the sum to all of these numbers are...
this tells us a few things. first it indicates each line is a separate number (not one big number) second, it's a sum. converting the binary into decimal
51
199
181
97
25
243
219
255
175
43
and summing it yields... the wrong answer. so far the conversion from binary has been assumed to be unsigned, where each bit n (0 or 1 value) with place value p (where p=0 is the rightmost place) has the weight 2^p*n (e.g: 1010 = (2^3 * 1) + (2^2 * 0) + (2^1 * 1) + (2^0 * 0)
the most common method of representing signed (positive, negative, and zero) integers is called two's complement. the conversion from a signed 8-bit integer represented by two's complement is identical to the unsigned conversion, except for the most significant bit (the first, reading from left-to-right). the most significant bit has the weight -n*2^p. this is also identical to taking the unsigned version and subtracting 256 if the most significant bit is 1 \
converting the binary into signed 8-bit integers produces
51
-57
-75
97
25
-13
-37
-1
-81
43
[...more numbers]
which when summed yields the solution
stage 3
this stage presents the user with a completely random string of characters
AA<`eI9@YZxLO]nhHhi;2
N246sM8
oYM_\s=UU<xQF:LVI>yCNXFF;e<c:?buTb]yElYo2cLEWMX?N>h\`yfo_:>hMhrrBSakFM4
a hint like so is provided
line 1->92 and line 2->28
looking at the first and second line seems intimidating, and there are a lot of potential red herrings from the random characters but removing all non-numeric characters
92
2468
and putting them side by side to the hint hopefully uncovers a few patterns
92 = 92
2468 = 28
taking the first and last number, combining the two and summing together each line yields the solution