Check digits are determined (or derived) by a set algorithm using the digits of the account number.
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1.382 OR 0.000 or 2831
<float_literals> -> <digit> { <digit> } [ . ] [ { <digit> } ] <digit> -> "0" | <digit excluding zero> <digit excluding zero> -> "1" | "2" | "3" | "4" | "5" | "6" | "7" | "8" | "9" hi dude i wish you are satisfied with my answer LOL ^^
it depends on capacitance and frequency, both inversely. Check your AC circuits textbook for the exact equation.
#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.
Pi doesn't have a last digit - it goes on for infinity (it also doesn't seem to repeat itself, so there can't even be a philosophical argument for the last digit).