Standard Practice Problems in Java: Process Every Digit Safely

Remainder and division can remove one decimal digit at a time

For temp = 1221, the expression temp % 10 selects the rightmost digit, while temp / 10 removes that digit through integer division. Repeating both operations processes the number from right to left. The digits appear as 1, 2, 2, and 1, then temp becomes 0 and the loop stops.

int temp = 1221;

while (temp > 0) {
    int digit = temp % 10;
    temp = temp / 10;
}

A working copy can reach 0 without destroying the original input

Store 1221 in original, then copy it into temp. The loop changes temp from 1221 to 122, then 12, then 1, and finally 0. The value in original never changes, so it remains available after every digit has been processed.

int original = 1221;
int temp = original;

while (temp > 0) {
    int digit = temp % 10;
    temp = temp / 10;
}

// original is still 1221, while temp is 0

If you divide original directly inside the loop, original eventually becomes 0. That is useful for stopping the loop, but it leaves no copy of 1221 for a later comparison. A palindrome test that needs the original value must preserve it before the loop starts.

The accumulator update decides which digit problem the loop solves

The same extracted digit can update several accumulators. Add 1 to count for every pass, add digit to sum, and place digit into reversed. The loop does not need a new structure for each task. Only the update applied after extraction changes.

int count = 0;
int sum = 0;
int reversed = 0;
int temp = 1221;

while (temp > 0) {
    int digit = temp % 10;
    count = count + 1;
    sum = sum + digit;
    reversed = reversed * 10 + digit;
    temp = temp / 10;
}

reversed = reversed * 10 + ____;

A palindrome decision is valid only after the complete reverse is built

After the loop finishes, temp is 0 but reversed contains the complete number in reverse order. Compare reversed with original only then. For 1221, both values are 1221, so the number is a palindrome.

int original = 1221;
int temp = original;
int reversed = 0;

while (temp > 0) {
    int digit = temp % 10;
    reversed = reversed * 10 + digit;
    temp = temp / 10;
}

if (reversed == original) {
    System.out.println("1221 is a palindrome");
} else {
    System.out.println("1221 is not a palindrome");
}

Comparing reversed with the consumed value of original is a different operation: after direct division, that value is 0. The comparison then rejects 1221 even though its reverse is 1221. Matching the first and last digits alone is also not enough, because the middle digits still need to be processed before the decision is complete.