A06 Answer

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

/* Puzzle A06 -- print the odd integers from 1 to 99, one per line */
int main(int argc, char *argv[])
{
  int j;
  for (j=1; j<=99; j+=2 )
  {
    printf("%3d\n", j);
  }
  system("PAUSE");	
  return 0;
}

Comment: The above is probably the best solution. The following loop is another solution, but it is more complicated, and does twice as much work for the same effect:

  for (j=1; j<=99; j++ )
  {
    if ( j%2 == 1 )
      printf("%3d\n", j);
  }  

Recall that the modulo operator (%) evaluates to the remainder after integer division. So, 1%2 gives 1, 2%2 gives 0, 3%2 gives 1, 4%2 gives 0, and so on. So even and odd can be detected with %. But you must be careful with negative integers (not a problem in this puzzle).

An overly clever programmer might realize that in C, any integer other than zero counts as true, so that the above might be further simplified as:

  for (j=1; j<=99; j++ )
  {
    if ( j%2 )
      printf("%3d\n", j);
  }  

But this is not really simplification. It does not actually express what you want to say, and is harder to understand, and with a good compiler probably produces the same machine code as the less clever version.