Numeric Digit Recursion

To process a number digit by digit recursively, two integer operations form the toolkit:

• n % 10 — the last digit of n.
• n / 10 — n with the last digit removed (integer division).

Together they let the method "peel" one digit per call and recurse on the smaller number. Reaching 0 means no digits left — the base case.

On the AP exam, digit recursion appears in digit-sum, digit-count, and digit-reverse FRQs.

The template

public int digitMethod(int n) {
    if (n == 0) return 0;   // or 1 for counts
    int last = n % 10;
    return last_processed + digitMethod(n / 10);
}

Sum digits

public int sumDigits(int n) {
    if (n == 0) return 0;
    return (n % 10) + sumDigits(n / 10);
}

Adds the last digit to the sum of the rest.

Count digits

public int countDigits(int n) {
    if (n == 0) return 0;
    return 1 + countDigits(n / 10);
}

Each call contributes 1 (one digit); base case contributes 0. For n = 1234, returns 4.

Why integer division works here

1234 / 10 = 123 — Java truncates the decimal portion. That's exactly what we want: drop the last digit. The remainder we already captured with % 10.

Edge case: n = 0

For sumDigits(0), the spec usually says "the sum of digits of 0 is 0". The base case handles this. For countDigits(0), opinions differ — sometimes 0, sometimes 1 — so AP problems usually test only n > 0.