C++ recursive base-13 conversion using a class method: determine the output when the initial value is 130. #include <iostream.h> class Number { int Num; public: Number(int x = 0) { Num = x; } void Display() { cout << Num; } void Modify(); }; void Number::Modify() { int Dec; Dec = Num % 13; Num = Num / 13; if (Num > 0) Modify(); if (Dec == 10) cout << "A"; else if (Dec == 11) cout << "B"; else if (Dec == 12) cout << "C"; else if (Dec == 13) cout << "D"; // unreachable for base-13 remainders else cout << Dec; } int main() { Number objNum(130); objNum.Modify(); return 0; }

Introduction / Context: This program prints a number in base 13 using digits 0–9 and letters A, B, C for 10, 11, 12 respectively. The method Modify transforms Num by repeatedly dividing by 13, recursing for higher places, and then printing the current remainder as a base-13 digit.

Given Data / Assumptions:

• Initial Num = 130.
• Remainder mapping: 10 → A, 11 → B, 12 → C.
• Recursion prints higher-order digits first (after dividing), then the current remainder.

Concept / Approach: Converting to base 13: at each step compute Dec = Num % 13 and Num = Num / 13. Recurse if Num > 0 to print the more significant digits first; then print the symbol for Dec. This is the standard recursive number-to-base printing pattern.

Step-by-Step Solution:

Verification / Alternative check: 130 in base 13 equals 10*13 + 0, so the digits are 10 and 0 → A0. Printing order matches the recursion.

Why Other Options Are Wrong:

• 130: That would be base 10, not base 13 representation.
• B0 or 90: Incorrect remainder mapping or arithmetic.
• None of the above: A0 is exactly produced.

Common Pitfalls: Expecting 13 to produce a remainder of 13 (remainders are 0–12), or forgetting that the routine prints higher-order digits first by recursing before output.

Final Answer: A0