Decoding the ARMY to XOJV Code

The problem requires finding the coding pattern used to transform "ARMY" into "XOJV" and then applying this pattern to encode "CALCULATOR".

We examine the letter positions in the alphabet (A=1, B=2, ..., Z=26):

• A is the 1st letter.
• R is the 18th letter.
• M is the 13th letter.
• Y is the 25th letter.

Similarly, for the coded word XOJV:

• X is the 24th letter.
• O is the 15th letter.
• J is the 10th letter.
• V is the 22nd letter.

By comparing the positions, we find the pattern:

• A (1) becomes X (24). This is equivalent to shifting 3 positions backward ($1 - 3 = -2$, which maps to 24 in modulo 26 arithmetic).
• R (18) becomes O (15). This is a shift of $18 - 3 = 15$.
• M (13) becomes J (10). This is a shift of $13 - 3 = 10$.
• Y (25) becomes V (22). This is a shift of $25 - 3 = 22$.

The consistent pattern is a backward shift of 3 positions ($-3$) for each letter.

Let's confirm the calculation using modular arithmetic:

The formula is: $\text{Position}_{\text{New}} = (\text{Position}_{\text{Old}} - 3) \pmod{26}$.

• For A (1): $(1 - 3) \pmod{26} = -2 \pmod{26} = 24$ (X)
• For R (18): $(18 - 3) \pmod{26} = 15 \pmod{26} = 15$ (O)
• For M (13): $(13 - 3) \pmod{26} = 10 \pmod{26} = 10$ (J)
• For Y (25): $(25 - 3) \pmod{26} = 22 \pmod{26} = 22$ (V)

The pattern is confirmed.

Encoding CALCULATOR

Now, apply the $-3$ shift rule to "CALCULATOR":

• C (3): $(3 - 3) \pmod{26} = 0 \pmod{26} = 26$ (Z)
• A (1): $(1 - 3) \pmod{26} = -2 \pmod{26} = 24$ (X)
• L (12): $(12 - 3) \pmod{26} = 9 \pmod{26} = 9$ (I)
• C (3): $(3 - 3) \pmod{26} = 0 \pmod{26} = 26$ (Z)
• U (21): $(21 - 3) \pmod{26} = 18 \pmod{26} = 18$ (R)
• L (12): $(12 - 3) \pmod{26} = 9 \pmod{26} = 9$ (I)
• A (1): $(1 - 3) \pmod{26} = -2 \pmod{26} = 24$ (X)
• T (20): $(20 - 3) \pmod{26} = 17 \pmod{26} = 17$ (Q)
• O (15): $(15 - 3) \pmod{26} = 12 \pmod{26} = 12$ (L)
• R (18): $(18 - 3) \pmod{26} = 15 \pmod{26} = 15$ (O)

Putting it together, the coded word for CALCULATOR is ZXIZRIXQLO.