#include<stdio.h>
#include<conio.h>
main()
{
int n,s,r,t;
clrscr();
printf("enter n");
scanf("%d",&n);
s=0;t=0;
while(n!=0)
{
r=n%10;
{
if(r%2!=0)
t=t+r;
if(r%2==0)
s=s+r;
}
n=n/10;
}
printf("sum of even position digits%d\n",s);
printf("sum of odd position digits%d\n",t);
getch();
}
To write a program that inputs a number and displays the digits absent in it, you can follow these steps: Convert the input number into a set of its digits. Create a set of all possible digits (0-9). Subtract the set of digits from the complete set to find the missing ones. Display the missing digits. Here’s a simple example in Python: number = input("Enter a number: ") present_digits = set(number) all_digits = set('0123456789') missing_digits = all_digits - present_digits print("Missing digits:", ''.join(missing_digits))
In QBASIC, you can write a simple program to input the number 64751315 and sum its digits as follows: DIM sum AS INTEGER sum = 0 INPUT "Enter a number: "; number FOR i = 1 TO LEN(number) sum = sum + VAL(MID$(number, i, 1)) NEXT PRINT "The sum of the digits is "; sum This program prompts the user to input a number, iterates through each digit, converts it to an integer, and adds it to the total sum, which is then printed out.
One way to do this is to convert the number to a String, then use the corresponding String method to find out the length of the String.
public static void main(String[] args) { int val = 100; int val1 = 50; System.out.println("Number of digits in " + val + " is: " + new String(val + "").length()); System.out.println("Number of digits in " + val1 + " is: " + new String(val1 + "").length()); }
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.
no thanks
To count the number of times a digit occurs in an integer, start by initializing an array of ten counts of digits, such as int digits[10];Then, in a loop while the number is non zero, increment the element in the digits array that corresponds to the units digit, and then divide the number by ten, such as digits[number%10]++ and number/=10;int digits[10];int i;int number = some number;for (i=0; i
An Armstrong number (or narcissistic number) for a given number of digits is a number that is equal to the sum of its own digits raised to the power of the number of digits. Here’s a simple Visual Basic 10 program that checks for Armstrong numbers: Module ArmstrongNumber Sub Main() Dim num As Integer Dim sum As Integer = 0 Console.Write("Enter a number: ") num = Convert.ToInt32(Console.ReadLine()) Dim temp As Integer = num Dim digits As Integer = num.ToString().Length While temp > 0 Dim digit As Integer = temp Mod 10 sum += Math.Pow(digit, digits) temp \= 10 End While If sum = num Then Console.WriteLine(num & " is an Armstrong number.") Else Console.WriteLine(num & " is not an Armstrong number.") End If End Sub End Module This program takes a number as input, calculates the sum of its digits raised to the power of the number of digits, and checks if the sum equals the original number.
#include <stdio.h> int main(int argc, char **argv) { if (argc<1) { printf("Usage: %s number\n",argv[0]); return -1; } int digits=1, i=atoi(argv[1]); while (i/=10) ++digits; printf("%d\n",digits); }
There are many shell programs that will find the sum of the square of individual digits of a number. A small example is: SD=3n=2, sum=2, and SD=2.
One way to do this is to convert the number to a String, then use the corresponding String method to find out the length of the String.
The number 1,234,567,890,123 has thirteen digits.
A Merchant Number is usually between 6 and 15 digits long. Examples: Natwest Streamline Merchant Number = 8 digits HSBC Merchant Number = 8 digits Lloyds TSB Cardnet Merchant Number = 15 digits Bank of Scotland Merchant Number = 15 digits Barclays Merchant Services Merchant Number = 7 digits American Express Merchant Number = 10 digits Diners Club Merchant Number = 10 digits JCB (Japan Credit Bureau) Merchant Number = 13 digits Elavon Merchant Number = 10 digits AIB (Allied Irish Bank) Merchant Number = 11 digits Ulster Bank (For Ireland Vendors only) Merchant Number = 8 digits Chase Payment Tech Merchant Number = 6 digits
The least whole number with 6 digits is greater because it has more digits than 5 digits.
When multiplying numbers with significant digits, count the total number of significant digits in each number being multiplied. The result should have the same number of significant digits as the number with the fewest significant digits. Round the final answer to that number of significant digits.
The largest ten digit number with no repeated digits is '9876543210'.
no, it matters with the number not digits