Two-Digit Number from Digit Constraints and Halving Rule A two-digit number has its units digit equal to the tens digit of half the number. Its tens digit is one less than the units digit of half the number. If the sum of the two digits is 7, determine the number.

Introduction / Context:
This puzzle ties the digits of a two-digit number to the digits of half that number. It assesses digit reasoning, parity (to ensure halving stays integral), and translating statements into equations.

Given Data / Assumptions:

• Let the number be 10a + b (tens digit a, units digit b).
• Half the number is (10a + b)/2, which must be an integer, so b is even.
• Units digit of the number equals the tens digit of half: b equals tens( (10a + b)/2 ).
• Tens digit of the number is one less than the units digit of half: a = units( (10a + b)/2 ) − 1.
• a + b = 7.

Concept / Approach:
Let half be 10x + y. Then b = x and a = y − 1. Also 10x + y = (10a + b)/2. Substitute x = b and y = a + 1 to relate a and b directly.

Step-by-Step Solution:
Set 10b + (a + 1) = (10a + b)/2.Multiply by 2: 20b + 2a + 2 = 10a + b.Rearrange: 19b − 8a + 2 = 0.Use a + b = 7 → b = 7 − a.Substitute: 19(7 − a) − 8a + 2 = 0 → 133 − 19a − 8a + 2 = 0 → 135 − 27a = 0 → a = 5.Then b = 2. Number = 52.

Verification / Alternative check:
Half of 52 is 26; tens(26) = 2 equals b, units(26) = 6 and a = 6 − 1 = 5; digit sum 5 + 2 = 7. All conditions satisfied.

Why Other Options Are Wrong:
16, 34, and 25 do not satisfy both digit constraints simultaneously; “Data in inadequate” is incorrect because the constraints determine a unique solution.

Common Pitfalls:
Forgetting half must be an integer; swapping the roles of tens and units; ignoring the digit-sum condition.

Final Answer:
52