我对加密非常陌生，我非常感谢你对这个谜题的投入，因为我被卡住了。
谜题：
<?php> blinker
defined( '_JEXEC' ) or die( 'Restricted access' );
终端：〜一个梦幻般的实习机会......
-bash：解决这个难题
如果您有兴趣那么21613@23151811.co.za = fun@work.co.za
这将引导您171225.13916.co.za
终端：〜答案是什么$
请注意我不会回答我对如何解决这个问题感兴趣。
答案 0 :(得分：4)
答案是job.mip.co.za 拿字母表并将数字与每个字母相关联。
A B C D E F G H I J K L M N O P Q R S T U V W X Y Z
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26
单词work：w = 23; o = 15; r = 18; k = 11 如果你用乐趣这个词来做这件事就不行了。那是因为乐趣是“与工作相反”。所以：f = 6但你必须从字母表的末尾开始计数到6。所以你会得到f = 21。你= 21，但你必须从字母表的末尾开始计算，你会得到u = 6。 n将是13。
现在你取数字171225，你就像乐趣一样。你会得到一份工作。 至于数字13916你做同样的工作，你会得到单词mip。 现在输入job.mip.co.za作为答案，您将被重定向到http://job.mip.co.za/
就是这样
答案 1 :(得分：1)
if 21613@23151811.co.za = fun@work.co.za
Then it is safe to assume that we working with a number and alphabet
correlation where a number equals an alphabet
Mathematically speaking 1 = A, and 2 = B .... and the are only 26 alphabets meaning we can't go higher than the number 26.
The first set of numbers are 5, and "fun" has three Alphabets where the only viable combination is
[21][6][13] = [U][f][M]
[2][16][13] = [A][P][M]
but none of them equals FUN
The second set of number are 8, and "work" has four Alphabets where the only viable combination is
[23][15][18][11] = [W][O][R][K]
Which give us the hint that the first set of numbers do in fact equals the word 'Fun'
we just have to look at the numbers differently.
how can 'F' be equal to [21]
and 'U' be equal to [6]
and 'N' be equal to [13]
or
how can 'F' be equal to [2]
and 'U' be equal to [16]
and 'N' be equal to [13]
we discovered counting from the right to the left side leaves us at undesired position
now lets try counting from the left side to right and see if it leave us at the desired position
[21] from left side to right side leave me at position [F]
[6] from left side to right side leave me at position [U]
[13] from left side to right side leave me at position [N]
[2] from left side to right side leave me at position [Y]
[6] from left side to right side leave me at position [U]
[13] from left side to right side leave me at position [N]
thus the first combination equals my desired value.
now using that same method it's safe to repeat on the next set of numbers.
171225.13916.co.za
[17] from left side to right side leave me at position [J]
[12] from left side to right side leave me at position [O]
[25] from left side to right side leave me at position [B]
and second set of numbers read from right side to left side
[13][9][16] = [M][I][P]
[13][9][1][6] = [M][I][A][F]
So that leaves me with only two possibilities.
'job.mip.co.za' or 'job.miaf.co.za' only of them is takes me where I need to be... my new home. Hi :)
答案 2 :(得分：0)
job.mip.co.za
171225.13916.co.za
M = 16; I = 9，P = 16
L = 17;○= 12; B = 25 这是吊球
取mip：m = 16; i = 9; p = 16，l = 12从字母表的末尾开始计数到12，得到l = 15。 o-15并从字母表的结尾开始计算得到o = 12和b = 25
答案 3 :(得分：0)
job.mip.co.za 171225.13916.co.za
我用mip来表示，其中M = 13 I = 9 P = 16
因此对于作业J = 10并从字母末尾计数到10，我得到j = 17，对于o = 15，我从最后一个字母末尾计数到15，我得到12，对于b = 2，我计数到2从上一个字母的结尾开始，我得到25。
答案 4 :(得分：0)
所以，这就是我所做的。
21613. 23151811.co.za等于fun @ work
21 16 13= F=21 U=6 N=14从数字Z开始计数
26 23 15 18 11= W=10 O=15 R=18 K=11
从数字1的A开始计数 然后：fun @ work
171225.13916.co.za等于job.mip
17 12 25= J= 17 O=12 B=25从数字Z开始计数
26 13 9 16= M=13 I=9 P=16
从数字1的A开始计数
答案 5 :(得分：0)
17 12 25 = job
‘j’ ‘o’ ‘b’
13 9 16 = mip
‘m’ ‘i’ ‘p’
因此171225.13916.co.za = job.mip.co.za