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 .
The digit 9 appears in the units place 100 times from 1 to 1000 (9, 19, 29,..., 989, 999). It also appears in the tens place 100 times (90, 91, 92,..., 99, 190, 191,..., 199, 290,..., 999). Therefore, the digit 9 appears a total of 200 times from 1 to 1000.
It is: 10000/2500000 times 100 = 0.4%
In the decimal place value system, each digit is ten times bigger than the digit on its right
It is its face value, which is the place value times the value of the digit.
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.
73
9
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.)
#include<iostream> #include<array> #include<sstream> std::array<int,10> get_frequency (int range_min, int range_max) { if (range_max<range_min) std::swap (range_min, range_max); std::array<int,10> digit {}; for (int count {range_min}; count<=range_max; ++count) { std::stringstream ss {}; ss << count; std::string s {}; ss >> s; for (auto c : s) { ++digit[c-'0']; } } return digit; } int main () { std::array<int,10> digit {}; digit = get_frequency(1, 89); std::cout << "In the range 1 to 89...\n"; for (int d {0}; d<10; ++d) { std::cout << "\tthe digit " << d << " appears " << digit[d] << " times.\n"; } } Output: In the range 1 to 89... the digit 0 appears 8 times. the digit 1 appears 19 times. the digit 2 appears 19 times. the digit 3 appears 19 times. the digit 4 appears 19 times. the digit 5 appears 19 times. the digit 6 appears 19 times. the digit 7 appears 19 times. the digit 8 appears 19 times. the digit 9 appears 9 times.
If you count 11 as 2 instances, the digit 1 appears 18 times if you don't count 10, 19 times if you do. 10,11,12,13,14,15,16,17,18,19,21,31,41,51,61,71,81,91
The digit 1 appears 301 times, as opposed to 300 for the digits 2 to 9.
4003
To determine the number of times the digit 1 appears between 1 and 120, we can consider each place value separately. In the units place, the digit 1 appears 12 times (1, 11, 21, ..., 111). In the tens place, the digit 1 appears 11 times (10, 11, 12, ..., 19). Therefore, the total number of times the digit 1 appears between 1 and 120 is 12 (from the units place) + 11 (from the tens place) = 23 times.
11 times
ten thousand times greater
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 .