Number Set Relation Logic Puzzle

Understanding the Problem and Constraints

The question asks us to find a set of three numbers from the given options that follows the same mathematical relationship as the two example sets: (29, 13, 168) and (17, 28, 180).

Key Constraint: We must perform operations on the numbers as whole entities. We cannot break a number down into its digits (e.g., operating on 29 as '2' and '9' is not allowed).

Analyzing the Given Sets to Find the Pattern

Let's represent a set as (A, B, C).

Set 1: (29, 13, 168)

• A = 29, B = 13, C = 168
• We need to find a relationship between A, B, and C. Let's try simple arithmetic operations.
• Sum: A + B = 29 + 13 = 42
• Difference: A - B = 29 - 13 = 16
• Product: A * B = 29 * 13 = 377
• Consider the sum (A + B) = 42. How can we relate 42 to C = 168?
• We see that $42 \times 4 = 168$. So, the potential pattern is $(A + B) \times 4 = C$.

Set 2: (17, 28, 180)

• A = 17, B = 28, C = 180
• Let's test our potential pattern: $(A + B) \times 4 = C$.
• Sum: A + B = 17 + 28 = 45
• Applying the pattern: $45 \times 4 = 180$.
• This matches C = 180. Therefore, the pattern $(A + B) \times 4 = C$ holds true for both given sets.

Verifying the Options Using the Identified Pattern

Now, we apply the pattern $(A + B) \times 4 = C$ to each option:

Option 1: (81, 45, 144)

• A = 81, B = 45, C = 144
• A + B = 81 + 45 = 126
• $(A + B) \times 4 = 126 \times 4 = 504$. This is NOT equal to 144.

Option 2: (23, 41, 256)

• A = 23, B = 41, C = 256
• A + B = 23 + 41 = 64
• $(A + B) \times 4 = 64 \times 4 = 256$. This IS equal to 256. This option matches the pattern.

Option 3: (14, 25, 78)

• A = 14, B = 25, C = 78
• A + B = 14 + 25 = 39
• $(A + B) \times 4 = 39 \times 4 = 156$. This is NOT equal to 78.

Option 4: (31, 47, 234)

• A = 31, B = 47, C = 234
• A + B = 31 + 47 = 78
• $(A + B) \times 4 = 78 \times 4 = 312$. This is NOT equal to 234.

Conclusion

Only Option 2, (23, 41, 256), satisfies the mathematical relationship $(A + B) \times 4 = C$ derived from the given example sets.