Two-digit number with digit sum 10 A two-digit number has digits that sum to 10. Reversing the digits decreases the number by 36. What is the product of the two digits?

Introduction / Context: Digit problems translate neatly into linear equations. With a two-digit number 10a + b (tens digit a, units digit b), reversing produces 10b + a. Conditions on sum and difference yield a and b directly.

Given Data / Assumptions:

• a + b = 10.
• (10a + b) − (10b + a) = 36 (number decreases by 36 on reversal).
• Digits a, b are integers 0–9; a ≥ 1 for a two-digit number.

Concept / Approach: Subtract to eliminate place-value weights: (10a + b) − (10b + a) = 9a − 9b = 36 ⇒ a − b = 4. Then solve the 2×2 system with the sum condition to get a and b.

Step-by-Step Solution:From a + b = 10 and a − b = 4, add the equations.2a = 14 ⇒ a = 7.Then b = 10 − 7 = 3.Product = a * b = 7 * 3 = 21.

Verification / Alternative check: Original number 73; reversed 37; difference 73 − 37 = 36; digit sum 7 + 3 = 10 — fully consistent.

Why Other Options Are Wrong:24, 36, 42 correspond to other pairs that do not satisfy both the sum and difference conditions simultaneously.

Common Pitfalls: Reversing the difference sign (using 36 the wrong way around); forgetting that a is the tens digit (must be ≥ 1).

Final Answer: 21