Two-digit number from digit constraints and a swap-based condition A two-digit number has a ten's digit that exceeds twice its unit's digit by 2. The number formed by interchanging its digits is 5 more than three times the sum of the digits. Find the number.

Introduction / Context:
This two-digit number problem couples a digit relation with a condition on the swapped number. We form equations in the ten's digit t and unit's digit u to solve cleanly.

Given Data / Assumptions:

• Original number N = 10t + u.
• Digit relation: t = 2u + 2.
• Swapped number N' = 10u + t.
• N' = 3(t + u) + 5.

Concept / Approach:
Use the swap condition to link t and u, then substitute the digit relation to obtain a simple linear equation in u. Compute t and build the number.

Step-by-Step Solution:
N': 10u + t = 3(t + u) + 5 ⇒ 10u + t = 3t + 3u + 5.Rearrange: 7u − 2t = 5.With t = 2u + 2, substitute: 7u − 2(2u + 2) = 5 ⇒ 7u − 4u − 4 = 5 ⇒ 3u = 9 ⇒ u = 3.Then t = 2u + 2 = 8 ⇒ number = 83.

Verification / Alternative check:
Sum of digits = 8 + 3 = 11; 3×sum + 5 = 38; the swapped number is indeed 38, which equals 10u + t, confirming consistency.

Why Other Options Are Wrong:
38 is the swapped number, not the original; 56, 65, 74 do not satisfy both constraints.

Common Pitfalls:
Misplacing “exceeds twice the unit’s digit by 2”; mixing up which number is original vs. swapped.

Final Answer:
83