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Given 9 digits from 0 to 9, write the shortest code you can in Python or C to find the missing digit.

Example:
Input : 756432098
Output: 1

A pernicious number is a positive number where the sum of its binary expansion is a prime number.

For example, 5 is a pernicious number since 5 = 101₂ and 1 + 1 = 2, which is prime.

Print all the pernicious numbers from 0 to 50 inclusive, each on their own line. There are 35 such numbers.

What does this code do? Explain why this program works.

  #include <stdio.h>
  #include <stdlib.h>

  #define SIZEOF(arr) (sizeof(arr)/sizeof(arr[0]))

  #define PrintInt(expr) printf("%s:%d\n",#expr,(expr))
  int main()
  {
      /* The powers of 10 */
      int pot[] = {
          0001,
          0010,
          0100,
          1000
      };
      int i;

      for(i=0;i<SIZEOF(pot);i++)
          PrintInt(pot[i]);
      return 0;
  }

The Collatz conjecture states that, for any positive integer n, it will eventually reach 1 by repeatedly applying the following procedure:

• If n is even, divide it by 2.
• If n is odd, multiply by 3 and then add 1.
  The number of steps needed for n to reach 1 is called its stopping time. For example, the stopping time of 10 is six:

10 → 5 → 16 → 8 → 4 → 2 → 1

Print the stopping times of all the numbers from 1 to 1,000 inclusive, each on their own line.

What does this obfuscated C code do? Make a guess as to its logic/why it works.

int main(int b,char**i){long long n=B,a=I^n,r=(a/b&a)>>4,y=atoi(*++i),_=(((a^n/b)*(y>>T)|y>>S)&r)|(a^r);printf("%.8s\n",(char*)&_);}

To compile
clang -include stdio.h -include stdlib.h -Wall -Weverything -pedantic -DB=6945503773712347754LL -DI=5859838231191962459LL -DT=0 -DS=7 -o prog prog.c

The compile-time flags of -DB, -DI, -DT, -DS are passed to the program.

Hint: Look at the bitwise operations and split it out by line! Look at the variable _.

More details here: burton IOCC