you can image the 3 digits as a counter 0 0 1 0 0 2 ... ... ... 9 9 8 9 9 9 every digit uses 100 times a 6..... 100*3 digits = 300 sixes :)
Assuming you mean whole numbers and leading zeros do not count, the smallest 3 digit number is 100.
100 is a 3 digit number.
100/5 = 20
Three-digit numbers start at 100. Two-digit numbers start at 10. So out of 1 - 99, the first 9 don't count. That leaves 99 - 9 = 90.
The largest 3-digit number is 999The smallest 3-digit number is 100.There would be 999 of them, but the first 99 don't count. So there are 900 of them.
100. Unless you count 0.01. Depends if you're taking "digit" to imply "integer" or not.
you can image the 3 digits as a counter 0 0 1 0 0 2 ... ... ... 9 9 8 9 9 9 every digit uses 100 times a 6..... 100*3 digits = 300 sixes :)
19 or 20 ( if you count 66 as double)
Assuming you mean whole numbers and leading zeros do not count, the smallest 3 digit number is 100.
The number 7 occurs once. The digit 7 occurs 20 times.
100 is a 3 digit number.
100/5 = 20
There are 90 two-digit whole numbers between 1 and 100.
if u count from 1 to 100 u will pass up 19 sixes
Three-digit numbers start at 100. Two-digit numbers start at 10. So out of 1 - 99, the first 9 don't count. That leaves 99 - 9 = 90.
There aren't going to be any 7's in the hundreds digit, so first we look at how many there will be in the ones digit. Since there will be a 7 for every 10 numbers, and there are 501 numbers between 100 and 600 inclusive), we'll have floor(501 / 10) = 50 sevens in the ones digit. For the tens digit, we'll encounter 10 sevens for every 100, from 70 to 79, and so we have floor(501 / 100) = 5 times we'll have 70 - 79 so that's 5 * 10 = 50 sevens in the tens digit. However, we don't want to know how many sevens there are, we want to exclude overlaps. Thus we subtract every instance of 77 we see (overlap!), and we have floor(501 / 100) = 5 overlaps. 50 + 50 - 5 = 95, so a total of 95 numbers will have the digit 7 at least once.