Question

If in a coded language, MASTER is written as 79 7 115 121 31 109 then LAUGH will be written as
(Note: - Do NOT include any spaces in your answer)

Answer 1

Correct Answer: 7371274349

Let's find out how "MASTER" is written as "79 7 115 121 31 109" and decode the logic behind it to apply the same to "LAUGH".
Understanding the Pattern:
By comparing the positions of the letters in "MASTER" with their corresponding encoded values:
1. M (13th letter) → 79
2. A (1st letter) → 7
3. S (19th letter) → 115
4. T (20th letter)→ 121
5. E (5th letter) → 31
6. R (18th letter)→ 109
It appears there is a transformation involved. Let's break it down:
- M (13) -> 79
- A (1) -> 7
- S (19) -> 115
- T (20) -> 121
- E (5) -> 31
- R (18) -> 109
Pattern Identification:
1. M (13) -> 79
  - This might be encoded as \( 13 \times 6 + 1 \)
2. A (1) -> 7
  - This might be encoded as \( 1 \times 6 + 1 \)
3. S (19) -> 115
  - This might be encoded as \( 19 \times 6 + 1 \)
4. T (20) -> 121
  - This might be encoded as \( 20 \times 6 + 1 \)
5. E (5) -> 31
  - This might be encoded as \( 5 \times 6 + 1 \)
6. R (18) -> 109
  - This might be encoded as \( 18 \times 6 + 1 \)
This doesn't work consistently. Instead, let's consider an alternative pattern:
By using the formula \( (Position \times 6) + 1 \):
1. M (13) -> \( (13 \times 6) + 1 = 79 \)
2. A (1) -> \( (1 \times 6) + 1 = 7 \)
3. S (19) -> \( (19 \times 6) + 1 = 115 \)
4. T (20) -> \( (20 \times 6) + 1 = 121 \)
5. E (5) -> \( (5 \times 6) + 1 = 31 \)
6. R (18) -> \( (18 \times 6) + 1 = 109 \)
Now, let's apply this formula to "LAUGH":
1. L (12th letter) -> \( (12 \times 6) + 1 = 73 \)
2. A (1st letter) -> \( (1 \times 6) + 1 = 7 \)
3. U (21st letter) -> \( (21 \times 6) + 1 = 127 \)
4. G (7th letter) -> \( (7 \times 6) + 1 = 43 \)
5. H (8th letter) -> \( (8 \times 6) + 1 = 49 \)
Thus, "LAUGH" would be written as 73 7 127 43 49.