The only number with a 6 between 1 and 9 is 6.
Between 10 and 99, every number that ends in a 6 or is in the sixties will have a 6 in it. The former category has 9 numbers in it, the latter 10, but they overlap at 66 so there are 18 numbers total in the range.
Fromm 100 to 999, the number must have a 6 in either the ones, tens, or hundreds digit. 90 numbers have a 6 in the ones place, 90 again in the tens place, and 100 in the hundreds place. NOT including the number 666, there are 8 overlaps that have 6 in both the ones and tens place, 9 with 6 in the ones and hundreds place, and 9 with 6 in the tens and hundreds place. Finally, the number 666 was counted an extra two time. Subtracting all the double/triple appearances, there are 252 numbers with a 6 in this range.
Finally, add up all these ranges to get that there are 271 numbers numbers with the digit 6 in this range.
There are 2700 digits.
If by "between" you mean "between but not including", then the answer is 9999 - 999 - 1 = 8999.If on the other hand, you mean to include both 9999 and 999, then the answer is 9999 - 999 + 1 = 9001.
0
Two of them which are 1 and 2
1, 3, 9, 27, 37, 111, 333, 999
There are 2700 digits.
It is 4 and 1,000
Only 1 exists, and it is "999"
9000 of them. All of the numbers 1 thru 9999 EXCEPT 1 thru 999. 9999 - 999 = 9000.
Assuming that 001, 080, etc are not allowed (that is a leading zero or two is not permitted), the smallest number with exactly three digits is 100. The largest number with exactly three digits is 999. So there are 999 - 100 + 1 = 900 numbers with exactly three digits.
10001
Well, let's see now . . .-- If you were to count from 1 to 9,999 you would count 9,999 numbers.-- The first 999 of them would have less than 4 digits.-- So (9,999 - 999) = 9,000 of them would have 4 digits.
2893 digits in all, made up as follows: 9 1-digit numbers (1 to 9) = 1*9 = 9 digits 90 2-digit numbers (10 to 99) = 2*90 = 180 digits 900 3-digit numbers (100 to 999) = 3*900 = 270 digit 1 4-digit number (1000) = 4*1 = 4 digits
-999
solution: we know that there are 25 prime numbers are between 1-100 and 168 prime numbers less than 1000. 100 x 100=10000(5 digits) 999 x 999=998001(6 digits) 1000 x 1000=1000000(7 digits) so our answer should be same as the number of prime numbers between 100 to 999. hence, 168-25=143. 143 prime numbers will be there less than 1000 whose square has 5 or 6 digits.
Highest 2-digit number = 99Highest 1-digit number = 999 - 9 = 90 two-digit numbers90.
One possible solution is: 9*8*5*3 - 7*6*2 + 4 - 1 = 1080 - 84 + 3 = 999