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.)
1-301 times
300
All the numbers between 999 and 3000 are four-digit numbers. You need to subtract 3000 - 999 - 1. The extra -1 is to account for the fact that the numbers 999 and 3000, in this case, are not included.
(Number*Location)Example:Give the value of 6 in 53,426,2456*1000=6000The value of a number is the sum of the values of all of its digits. Starting at the decimal point and going left, add together...the 1st digit's value times 100 (100 = 1)the 2nd digit's value times 101 (101 = 10)the 3rd digit's value times 102 (102 = 100)the 4th digit's value times 103 (103 = 1000)and so on to the most significant digit.If there are digits to the right of the decimal point, starting at the decimal point and going right, add to the total above...the 1st digit's value times 10-1 (10-1 = 1/10)the 2nd digit's value times 10-2 (10-2 = 1/100)the 3rd digit's value times 10-3 (10-3 = 1/1000)and so on to the least significant digit.
By including the number 1000, the digit 1.
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.)
9
To determine how many times the digit 9 is written when writing numbers from 1 to 10000, we can consider the pattern of its occurrence in each place value. In the units place, the digit 9 appears 1000 times (from 9 to 9999). In the tens place, the digit 9 appears 1000 times (from 90 to 99, 190 to 199, and so on). In the hundreds place, the digit 9 appears 1000 times (from 900 to 999, 1900 to 1999, and so on). Therefore, the digit 9 is written 3000 times in total when writing numbers from 1 to 10000.
1-301 times
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.
The digit appears eleven time from 1 to 100.
There are 900 three-digit whole numbers between 1 and 1000
1
How many times does the digit 1 occur in ten place in the numbers from 1 to 1000?
There are 200 positive four digit integers that have 1 as their first digit and 2 or 5 as their last digit. There are 9000 positive four digit numbers, 1000 through 9999. 1000 of them have 1 as the first digit, 1000 through 1999. 200 of them have 2 or 5 as their last digit, 1002, 1005, 1012, 1015, ... 1992, and 1995.
#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.