A two-digit positive number has its units digit equal to the square of its tens digit. When its digits are interchanged, the difference between the original and the new number is 54. What is 40% of the original number?

Introduction / Context:This problem blends digit constraints with a fixed difference after digit reversal. Because the units digit equals the square of the tens digit, only a few two-digit candidates exist, which can be checked quickly against the given difference. Once the original number is identified, computing 40% is straightforward.

Given Data / Assumptions:

• Let the number be 10a + b with digits a (tens) and b (units).
• Units equals square of tens: b = a^2.
• |(10a + b) − (10b + a)| = 54 (difference after interchanging digits).
• Digits are from 0–9, and a ≥ 1 for a two-digit number.

Concept / Approach:Since b = a^2 must be a single digit, possible a values are 1, 2, or 3 (yielding b = 1, 4, 9). Form the candidate numbers 11, 24, and 39 and test the absolute difference after swapping digits. Select the one that satisfies the 54 difference condition, then evaluate 40% of that number.

Step-by-Step Solution:

Verification / Alternative check:Reversal check holds for 39 ↔ 93 with difference 54; the digit constraint b = a^2 is satisfied since 9 = 3^2.

Why Other Options Are Wrong:14.4 (which is 40% of 36), 16, 12, and 18 correspond to incorrect original numbers or arbitrary guesses; they do not align with both the digit-square constraint and the 54 difference requirement.

Common Pitfalls:Including a = 4 (which would require b = 16, not a single digit), or forgetting the absolute value in the “difference” statement, which could flip signs and mislead the check.

Final Answer:15.6