Step 1: Understanding the Concept:
This is a coding-decoding problem. We are given the rule for encoding a letter based on its position (m) in the alphabet. We are given a coded word and need to find the original word, which means we must apply the decoding rule.
Step 2: Key Formula or Approach:
First, let's write down the encoding rules based on the original letter's position 'm':
- If 'm' is odd, the coded letter's position is \(m-1\). (e.g., C(3) \(\to\) B(2))
- If 'm' is even, the coded letter's position is \(m+1\). (e.g., B(2) \(\to\) C(3))
Now, let's derive the decoding rules. Let the coded letter's position be 'c'. We want to find the original letter's position 'm'.
- If a letter at an odd position 'm' was coded, the result 'c' is \(m-1\), which is an even position. So, if the coded letter is at an even position 'c', the original letter was at position \(m=c+1\), which is odd.
- If a letter at an even position 'm' was coded, the result 'c' is \(m+1\), which is an odd position. So, if the coded letter is at an odd position 'c', the original letter was at position \(m=c-1\), which is even.
Decoding Rules (based on coded letter's position 'c'):
- If 'c' is odd, original position is \(m=c-1\) (which is even).
- If 'c' is even, original position is \(m=c+1\) (which is odd).
Step 3: Detailed Explanation:
The coded word is FZHO. Let's find the position of each letter:
- F is the 6th letter (even).
- Z is the 26th letter (even).
- H is the 8th letter (even).
- O is the 15th letter (odd).
Now, apply the decoding rules:
- F (6, even): The original position is \(m = 6+1 = 7\). The 7th letter is G.
- Z (26, even): The original position is \(m = 26+1 = 27\). Since the alphabet has 26 letters, this rule might need clarification for Z. However, let's assume standard alphabet. If an even letter \(m\) becomes \(m+1\), what becomes Z(26)? No even letter \(m\) can result in \(m+1=26\). Let's re-read the rule carefully. "m-th letter is coded as...". Let's check the Z again. An odd letter \(m\) becomes \(m-1\). For \(m-1=26\), \(m=27\), which is A. An even letter \(m\) becomes \(m+1\). For \(m+1=26\), \(m=25\) (Y). Wait, Y is odd. So, the original must have been A(1), an odd letter, coded as \(1-1=0\), perhaps Z(26)? This is ambiguous. Let's try the simpler decoding rules again. Decoding Z(26, even) \(\to\) \(26+1=27\), which is A. Let's proceed with this.
- H (8, even): The original position is \(m = 8+1 = 9\). The 9th letter is I.
- O (15, odd): The original position is \(m = 15-1 = 14\). The 14th letter is N.
Combining the decoded letters, we get G-A-I-N.
The original word is GAIN.
Step 4: Final Answer:
The decoded word is GAIN, which corresponds to option (B).