30 times because when you count up in tens this is the answer
10 times for the one's digit, 1-100 10 times for the ten's digit, 60-70 = 20 times
48
Instead of going from 1 to 500, let's just subtract one from the set and come up with "How many 3's appear from 1 to 499?" Total possible combinations in this, including leading zeros, is 5*10*10, i.e. 5 possible in the first set(0,1,2,3,4), and 10 in the second two(0,1,2,,4,5,6,7,8,9). Instead of figuring out how many have a 3, lets figure out how many don't. Well there is 4 possibilities in the first (0,1,2,4), and nine in the second two (0,1,2,4,5,6,7,8,9). So, total = 5 * 10 * 10 = 500 Without 3 = 4 * 9 * 9 = 324 500-324 = 176 There are 176 numbers from 1 to 500 that have the digit 3 in them.
The number 7 occurs once. The digit 7 occurs 20 times.
999-111=888 888 3 digit numbers can be made with numbers between 1 - 9
11 times
19 times.
11 times
300 times.
300
As a digit in other numbers it appears 20 times
Including the one in ' 1 ' and the one in '100', there are 21 1s.Every other digit 2 - 9 appears 20 times between 1 and 100 .
10 times for the one's digit, 1-100 10 times for the ten's digit, 60-70 = 20 times
Infinitely many.
Zero. Only the digits 5 and 0 appear in 500.
There will be unlimited 0's between 1 and 10000 because you also need to count the 0's, such as 1.01, 5934.00000000000004.... etc. Well, there could be an actual number of the digit 0 appear between 1 and 10000, but that number will be so much times larger than a google.
18