Code Breaking

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

Z Y X W V U T S R Q P O N M L K J I H G F E D C B A

The encryption is done using block cipher in 2 steps. First, doublets from the start of the word are taken and reversed. Then, a shift cipher of certain shift is applied for each doublet, if necessary, to match the obtained cipher with the given cipher. Let me demonstrate it via one of the given examples.

First, in "FIND", taking the doublets "FI" and "ND" and reversing them, the new cipher text is "IFDN". Applying a right shift of $12$ to first doublet and right shift of $9$ to the second doublet, the doublets become "UR" and "MW" respectively. So, the final cipher text is "URMW" as given in the problem.

Similarly, in "ME", the one and only doublet is reversed, so the new cipher text is "EM". Applying a right shift of $9$ to the doublet, we get the final cipher text as "NV" as given in the problem.

So, in "FOOL", taking the doublets "FO" and "OL" and reversing them, the new cipher text is "OFLO". Now, applying a right shift of $6$ to the first doublet and no shift on the second one, the doublets become "UL" and "LO" respectively. So, the final cipher text is "ULLO".

P.S. - There is a humorous coincidence in the problem which only Hindi speakers will understand, because the word "ULLO" means owl in Hindi and is also the word which is usually associated with foolish people.

Taking A = 1, B = 2, C = 3, ....., Z = 26 and so on, we get:

F I N D = 6 9 14 4

U R M W =21 18 13 23

If you have noticed, the numbers of the coded letters and the numbers of the original letters sum up to 27.

Similarly, M = 13, E = 5

So, code for M = 27 - 13 = 14 = N

Code for E = 27 - 5 = 22 = V

Using this, we get

F O O L = 6 15 15 12

So, subtracting 27 from each number, we get

21 12 12 15

Which is U L L O.

This is a basic substitution cipher, I think.

You may also have noticed, that the code for each letter is merely the letter 'mirroring' it in the ABCD series.

Just trust your gut on picking the answers.

I used process of elimination in the answer choices since I'm new to ciphers.

Straightforward C code:

//alphabet reverser
#include<stdio.h>
#include<conio.h>
#include<string.h>
#define p printf
#define s scanf

main()
{
        char x[10],y[10];
        int z,n;
    strcpy(x,"FOOL");
    z=strlen(x);
    strupr(x);

    for(n=0; n<z; n++)
    {
        switch(x[n])
        {
            case 'A': y[n]='Z'; break;
            case 'B': y[n]='Y'; break;
            case 'C': y[n]='X'; break;
            case 'D': y[n]='W'; break;
            case 'E': y[n]='V'; break;
            case 'F': y[n]='U'; break;
            case 'G': y[n]='T'; break;
            case 'H': y[n]='S'; break;
            case 'I': y[n]='R'; break;
            case 'J': y[n]='Q'; break;
            case 'K': y[n]='P'; break;
            case 'L': y[n]='O'; break;
            case 'M': y[n]='N'; break;
            case 'N': y[n]='M'; break;
            case 'O': y[n]='L'; break;
            case 'P': y[n]='K'; break;
            case 'Q': y[n]='J'; break;
            case 'R': y[n]='I'; break;
            case 'S': y[n]='H'; break;
            case 'T': y[n]='G'; break;
            case 'U': y[n]='F'; break;
            case 'V': y[n]='E'; break;
            case 'W': y[n]='D'; break;
            case 'X': y[n]='C'; break;
            case 'Y': y[n]='B'; break;
            case 'Z': y[n]='A'; break;
        }
    }

    puts(y);
    getche();
    return 0;
}

We have F=U. If we investigate the serial of the alphabets, we'll see the first letter is in the same place from the first of the sequel of alphabets than the second letter which has the same from the last. As an example, F is the sixth letter from the first and U is the sixth letter from the last. So now we have, O=L and L=O. So FOOL=ULLO

normal A = 1 B=2 C=3 D=4 E=5 F=6 G=7 H=8 I=9 J = 10 K=11 L=12 M=13 N=14 O=15 P=16 Q=17 R=18 S=19 T=20 U=21 V=22 W=23 X=24 Y=25 Z=26

reversed Z=1 Y=2 X=3 W=4 V=5 U=6 T=7 S=8 R=9 Q=10 P=11 O=12 N=13 M=14 L=15 K=16 J=17 I=18 H=19 G=20 F=21 E=22 D=23 C=24 B=25 A=26

FIND = 6 9 14 4 ,from normal arrangement
would become URMW = 6 9 14 4 in reversed arrangement ME = 13 5 , from normal arrangement, would become NV = 13 5 in reversed arrangement

FOOL = 6 15 15 12 ,, would become ULLO ,,, still 6 15 15 12

A=Z B=Y . . . Z=A therefore, FOOL=ULLO

For each char i, let j be its encryption. Let 'A' = 1, 'B' = 2, etc.

Then i + j is always equal to 27.

The second and third letters should be the same. Since F is coded as U, the answer starts with U. The last third letters cannot be U,R,M,W or V since none must refer to L.

Eliminate impossible options and ULLO is the answer. :)

After realizing that the answer must start with U, you can see that none of the other letters are given, but think about it like this:

UFGD: F and G are not the same, yet O and O are, so this choice is wrong. Now, we're between ULLO, UMMO, and UDDW. M decodes into N (FIND --> URMW), so UMMO is not the correct answer. W decodes into D (FIND --> URMW), so UDDW is also not correct. Checking the letters in ULLO, there are no contradictions between FIND/URMW and ME/NV, so ULLO is the answer.

No need to figure out how the words were encrypted!

Elimination used for me. First note that the middle two letters in 'FOOL' are identical so the encrypted form must also have its middle two letters identical, ruling out 'UFGD'.

Now note that M is already the encrypted form of N so M cannot at the same time be the encrypted form of O, ruling out 'UMMO'. Finally, W is already the encrypted form of D so W cannot at the same time be the encrypted form of L, ruling out 'UDDW'.

This leaves us with $\boxed{ULLO}$ as the only possible option among the four.

Since URMW is FInd that mean the code cant have RMW in it and since ME is NV it cant have NV in it either. So if u find one of these letters in one of them their wrong. and it also have to have to of the same latter in the middle

It's unbelievably simple! The code is a=z, z=a

Easy! FIND=URMW So the first is U. Then same letter twice which means ufgd out of the game. W=D is given so uddw is out. M=N is given then ummo is out of the game. ULLO is the winner. I don't even need the ME=NV :)

Write ABCDEF...XYZ and below that line ..........ZYX...FEDCBA; the mapping is clearly seen

Being someone who likes numbers I like to see what letter comes in order. When I saw it used m saying that ME = NV. It was easy for me to see how they created this because M is letter 13 and there are 26 letters in the alphabet. Which means after M the next 13 mirror the last 13.

F = U that's obvious

UFGD is wrong because FG are not same letters, like OO in FOOL

UMMO is wrong because M = N in first example

UDDW is wring because W = D in first example

so ULLO is the answer!

.......

I thought about it like this since it's posted as a logic problem! and I don't know cryptography :D

This is what I call a "reverse" code. As $A$ is coded as $Z$ and $Z$ is coded as $A$ or $A \rightleftharpoons Z$ , $B \rightleftharpoons Y$ , $C \rightleftharpoons X$ , ..., $M\rightleftharpoons N$ .

If $A=1$ , $B=2$ , $C=3$ , ... $Z=26$ then the coding formula $C(n) = 27-n$ . Therefore, $C('F') = C(6) = 21 = 'U'$ , $C('O') = C(15) = 12 = 'L'$ and $C('L') = C(12) = 15 = 'O'$ . $\Longrightarrow C('FOOL') = \boxed{'ULLO'}$