When for every 10 numbers, there is one zero.
So between 999/10 = 99 and every 100 numbers, 9 more zero
But then there is 101,102,103, etc. then 201,202,203,etc.
So 9*9=81
Now add those two equations....81+9+99=189
If 0 is not counted, and 1 is, then the answer is 1000.
1000
ther are 99 palindromes in between 1 and 1000.
The number 1 is found 300 times between 1 and 1000.
There are infinitely many numbers between them. Also there are infinitely many sequences that wil have two numbers between 1000 and 1.
If 0 is not counted, and 1 is, then the answer is 1000.
It may not.
1000
ther are 99 palindromes in between 1 and 1000.
There are 168 prime numbers between 1 & 1000.
4 in 1000 = 1 in 250 8 in 1000 = 0.8% 3 in 1000 = 0.003
199. There was no year 0.
Between 1-100- 11 zero's- ie., 10,20,30,40,.... 100... so 11*10=110
500. Since they alternate, half the numbers between 1 and 1000 are odd and the other half are even. Adding zero to the list does not change the number of odds.
There are 17 palindromes between 1 and 1000 that are divisible by 11
The number 1 is found 300 times between 1 and 1000.
Not couting 1 and 1000, there would be 998 numbers.