Assuming you mean in the numbers 1, 2, 3, ..., 998, 999, 1000 then the digit 0. (The digit 1 appears 301 times, the digits 2-9 all appear 300 times each, but the digit 0 only appears 192 times.)
There are 3168 such numbers.
Zero (192 occurances), then 2 to 9 (all equal at 300), and then 1 (301).
All the numbers between 999 and 3000 are four-digit numbers. You need to subtract 3000 - 999 - 1. The extra -1 is to account for the fact that the numbers 999 and 3000, in this case, are not included.
In every 100 there are 10 numbers that have a units digit of 2 and 10 with 7, making 2 with 2 or 7. Between 0 and 1000 there are 10 hundreds, making 10 × 20 = 200 such numbers. BUT 2 and 7 are less than 10, so there are two less such numbers, making 200 - 2 = 198 such numbers.
Zero
There are 900 three-digit whole numbers between 1 and 1000
By including the number 1000, the digit 1.
Assuming you mean in the numbers 1, 2, 3, ..., 998, 999, 1000 then the digit 0. (The digit 1 appears 301 times, the digits 2-9 all appear 300 times each, but the digit 0 only appears 192 times.)
555
1 to 1000. Pilandromic 108
90 of them.
Specify whether you are including the numbers between 0000 and 1000.
There are 3168 such numbers.
Zero (192 occurances), then 2 to 9 (all equal at 300), and then 1 (301).
All the numbers between 999 and 3000 are four-digit numbers. You need to subtract 3000 - 999 - 1. The extra -1 is to account for the fact that the numbers 999 and 3000, in this case, are not included.
numbers between 100 and 1000 = 900 numbers between 100 and 1000 which do not have digit 6 in any place = 8 x 9 x 9 = 648 Unit digit could take any value of the 9 values (0 to 9, except 6) Tens Digit could take any value of the 9 values (0 to 9, except 6) Hundreds digit could take any value of the 8 values (1 to 9, except 6) numbers between 100 and 1000 which have atleast one digit as 6 = 900 - 648 = 252