#include <cstdio>
int main()
{
int x = 1;
while(x < 10)
{
printf("%d ", x);
x++;
}
char wait;
scanf("%s", wait);
return 0;
}
0 and 1. Integers do not include decimals, where as 9% is equal to 0.09. So the consecutive integers would be 0 and 1.
There are 10 one digit positive integers (0 - 9) and 9 one digit negative integers (-9 to -1) making 19 in all.
-1
For example, 11, -1, -1.
Which is the list of all integers from negative 1 to 1?Here is the list of integers: -1, -2, -3, -4, -5, -6, -7, -8, -9, 0, 1.
1. The only positive integer that equals 9 is 9.
There are 9 integers between -8 and 2. These integers are -7, -6, -5, -4, -3, -2, -1, 0, and 1. The count includes both endpoints, but since -8 is not included and 2 is not included, we start counting from -7 to 1. Thus, the total is 9 integers.
Oh, dude, to find out how many integers from 1 to 2000 are divisible by 9, you just need to divide 2000 by 9 and see how many times it goes in evenly. So, 2000 divided by 9 is like 222 with a remainder, but we only care about the whole number part, which is 222. So, there are 222 integers from 1 to 2000 that are divisible by 9.
Yes, sum of integers 1 - 9.
It is 1.
1 3 5 9
Since they are all integers, it is 1.