Use an enum if you are using a c style language. Or a map data structure. Assign each integer an English value and then match it to what the user inputs.
Function sum_odd_digits(ByVal number As Integer) As Integer Dim digit As Integer sum_odd_digits = 0 While number <> 0 digit = number Mod 10 If digit And 1 Then sum_odd_digits = sum_odd_digits + digit number = number / 10 End While End Function
In a three digit integer number the tenthsdigit is always 0 as integer numbers are whole numbers and have no decimal part and tenths are decimal parts:tenths_digit_of_integer_number = 0I suspect you mean "How to find the tens digit of an integer number?"; this is the second from the right, so:tens_digit = (INT(number ÷ 10)) MOD 10For example, in C this would become: tens_digit = (number / 10) % 10;
Just generate the Fibonacci numbers one by one, and print each number's last digit ie number%10.
Assuming you've entered a multi-digit number whole number (an integer), then take the modus (%) of the number and 10. E.g., if the number input was 1234, then 1234 % 10 is 4. Thus the final digit is 4. Note that modus 10 is the same as dividing the number by 10 and taking the remainder.
The tricky part is getting the individual digits. There are basically two ways to do this: 1) Convert the number to a string, and use string manipulation to get the individual digits. 2) Repeatedly divide the number by 10. The digit is the remainder (use the "%" operator). To actually get the highest digit, initially assume that the highest digit is zero (store this to a variable, called "maxDigit" or something similar). If you find a higher digit, replace maxDigit by that.
public class PrintOctal { public static void main(String[] args) { int n = Integer.parseInt(args[0]); System.out.printf("%o\n", n); } }
Function sum_odd_digits(ByVal number As Integer) As Integer Dim digit As Integer sum_odd_digits = 0 While number <> 0 digit = number Mod 10 If digit And 1 Then sum_odd_digits = sum_odd_digits + digit number = number / 10 End While End Function
there could be a part in it like this: int num, digit; int count [10]; do { digit = num%10; num != 10; ++count[digit]; } while (num);
1
dvve evd
The smallest three digit integer is 100
The smallest 5 digit integer is -99999. The largest 5 digit integer is 99999. The sum is therefore 0.
A decimal number can be an integer. All that decimal means is that the place value of each digit is ten times that of the digit to its right.
Input the number as a string. If the string has a length of 4 and contains only digits, convert the string to an integer. If the integer is less than 1000, input another number. Otherwise, copy the integer and divide by 100 to get rid of the two least-significant digits. Divide again by 2 and take the remainder. If the remainder is 1 then the second left digit of the 4-digit integer is odd and the 4-digit integer can be added to the array, otherwise do not add it. Repeat for all n numbers.
It is a 29-digit integer. So what?
'Number' or 'digit'.
A six-digit integer.