Challenge 2: Clearing a Single Flag Without Disturbing the Others — Possible Solution
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#include <stdio.h>

#define FLAG_READ  (1 << 0)
#define FLAG_WRITE (1 << 1)

int main() {
    unsigned int flags = FLAG_READ | FLAG_WRITE;
    printf("before: %u\n", flags);

    flags &= ~FLAG_WRITE;
    printf("after:  %u\n", flags);

    return 0;
}

Output:

    before: 3
    after:  1

Explanation: ~FLAG_WRITE inverts every bit of FLAG_WRITE (0b10),
producing a mask with every bit set to 1 EXCEPT bit 1 (the WRITE bit),
which becomes 0. ANDing flags with that mask keeps every bit of flags
unchanged except bit 1, which is forced to 0 regardless of its
previous value -- exactly clearing FLAG_WRITE while leaving FLAG_READ
(bit 0) completely untouched. before is 3 (0b11, both flags set);
after is 1 (0b01, only FLAG_READ remains).

WHY THIS WORKS AS AN ANSWER
------------------------------
This shows the numeric before/after values (3 then 1) confirming
exactly one bit changed, and explains the AND-with-NOT'd-mask
mechanism precisely -- inverting isolates the target bit as the only
0 in an otherwise all-1s mask, so ANDing can only ever clear that one
bit and nothing else.