The answer is 300.
(9,19,29,39,49,59,69,79,89,90,91,92,93,94,95,96,97,98,99) = 20
20*10 (ten hundreds) = 200
Add another 100 for the nine hundred (from 900 to 999)
200+100 = 300
269
17
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 .
-4
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.
1000
In the decimal place value system, each digit is ten times bigger than the digit on its right
11 times
How many times does the digit 1 occur in ten place in the numbers from 1 to 1000?
There are 280.
300 times.
300
146 times.
As a digit in other numbers it appears 20 times
1000
#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.
73
1000
1000