1 - 9 = 1
10-19 = 10
20-99 = 8
So...0 - 99 = 19
100 - 199 = 100 + 19 = 119
200 - 299 = 19
300 - 399 = 19
400 - 499 = 19
500 = 0
So, in 1 - 500, the number 1 shows up this many times:
100 + (19 * 5) = 100 + 95 = 195 times
The number 1 appears 100 times in numbers 1-500. This is because there are 100 numbers that have 1 in the units place (1, 11, 21, ..., 491) and 100 numbers that have 1 in the tens place (10-19, 110-119, ..., 490-499). Additionally, there is one occurrence of the number 1 in the hundreds place (100-199).
200
100 times because if you do the opposite with the numbers you have, 500/5=100
500/15 = 33.33... times.500/15 = 33.33... times.500/15 = 33.33... times.500/15 = 33.33... times.
25 times 20 = 500
500 times
900
200
200
Zero. Only the digits 5 and 0 appear in 500.
200
200
100 times because if you do the opposite with the numbers you have, 500/5=100
An infinite number of times. If you restrict yourself to whole numbers, 500 has 6 factor pairs.
95
1 digit number: only 1 number 2 digits number: 18 numbers 3 digits number: 76 So there are 95 numbers containing 9.
There are infinitely many such numbers.
The answer to the question is 200 times. Don't forget about 100 which always has aone in it going up from 100-199.