Step 1: Understanding the Concept:
This is a coding problem where we are given the rule to encode a word and asked to apply it. The rule depends on whether the position of the letter in the alphabet is odd or even.
Step 2: Key Formula or Approach:
The encoding rules are based on the original letter's position 'm':
- If 'm' is odd, the new position is \(m-1\).
- If 'm' is even, the new position is \(m+1\).
We need to apply these rules to each letter of the word 'BUST'.
Step 3: Detailed Explanation:
The word to be coded is BUST. Let's find the position of each letter:
- B is the 2nd letter (even).
- U is the 21st letter (odd).
- S is the 19th letter (odd).
- T is the 20th letter (even).
Now, apply the encoding rules to each letter:
- B (2, even): The new position is \(2+1 = 3\). The 3rd letter is C.
- U (21, odd): The new position is \(21-1 = 20\). The 20th letter is T.
- S (19, odd): The new position is \(19-1 = 18\). The 18th letter is R.
- T (20, even): The new position is \(20+1 = 21\). The 21st letter is U.
Combining the coded letters, we get C-T-R-U.
Note on Discrepancy:
My calculated code is CTRU, which is option (B). The provided key indicates (C) CTRS. Let me re-verify.
B(2, even) \(\to\) 2+1 = 3 (C). Correct.
U(21, odd) \(\to\) 21-1 = 20 (T). Correct.
S(19, odd) \(\to\) 19-1 = 18 (R). Correct.
T(20, even) \(\to\) 20+1 = 21 (U). Correct.
The result is unambiguously CTRU. The option key marking CTRS is incorrect. 'S' is the 19th letter. To get 'S', the new position must be 19.
This would require either an original odd 'm' with \(m-1=19 \implies m=20\)(T, which is even), or an original even 'm' with \(m+1=19 \implies m=18\)(R, which is even). If the original letter was R(18, even), it would be coded as 18+1=19(S). So, if the last letter was R, the code would be CTRS. But the word is BUST, with the last letter T.
The question or the answer key is flawed. The correct encoding for BUST is CTRU.
Step 4: Final Answer:
Following the given rules, B(2, even) \(\to\) C(3), U(21, odd) \(\to\) T(20), S(19, odd) \(\to\) R(18), T(20, even) \(\to\) U(21). The code word is CTRU. This is option (B). The provided key (C) is incorrect.