Great riddle, very close to the Chinese remainder theorem,

Here's my brute force code for original riddle:

int remainder = 1;
int answer = 0;
string result = string.Empty;
for (int number = 1; number <= 10000; number++)
{
for (int divisor = 2; divisor <= 10; divisor++)
{
if (number % divisor == remainder)
{
if (remainder == 9)
{
answer = number;
result += string.Format("{0} ", answer);
break;
}
remainder++;
}
}
remainder = 1;
}

Code for the second riddle:

string result2 = string.Empty;
for (int number = 11; number <= 100000; number++)
{
for (int divisor = 2; divisor <= 10; divisor++)
{
if (number % divisor == 1)
{
if (divisor == 10)
{
++divisor;
if (number % divisor == 0)
{
result2 += string.Format("{0} ", number);
break;
}
}
}
else break;
}
}