wap to print all the arnstrong no. between 100&1000
Just generate the Fibonacci numbers one by one, and print each number's last digit ie number%10.
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 StringReverseExample { public static void main(String[] args) { int num=1001; int n=num, rev; while(num!=0) { int d=num%10; rev= (rev*10)+d; num=num/10; } System.uot.println(rev); } }
Q.1 Write a program to print first ten odd natural numbers. Q.2 Write a program to input a number. Print their table. Q.3 Write a function to print a factorial value.
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.
Just generate the Fibonacci numbers one by one, and print each number's last digit ie number%10.
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.
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To write a Java program that prints the sum of the prime digits of a number, you can follow these steps: First, convert the number to a string to access each digit individually. Then, check if each digit (0-9) is prime (2, 3, 5, 7) and, if so, add it to a sum variable. Finally, print the sum. Here’s a simple code snippet: public class PrimeDigitSum { public static void main(String[] args) { int number = 123456789; // Example number int sum = 0; for (char digit : String.valueOf(number).toCharArray()) { int d = Character.getNumericValue(digit); if (d == 2 || d == 3 || d == 5 || d == 7) { sum += d; } } System.out.println("Sum of prime digits: " + sum); } }
public class StringReverseExample { public static void main(String[] args) { int num=1001; int n=num, rev; while(num!=0) { int d=num%10; rev= (rev*10)+d; num=num/10; } System.uot.println(rev); } }
Q.1 Write a program to print first ten odd natural numbers. Q.2 Write a program to input a number. Print their table. Q.3 Write a function to print a factorial value.
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.
You should visit your bank and they will provide you with a print out of your account number with the IBAN number.
There is no such thing as 'the greatest number'. Here is a program in unix to prove it: echo 'for (i=1; i>=0; i= i*2) print i,"\n";' | bc if you want to actually see the output use it this way: echo 'for (i=1; i>=0; i= i*2) print i,"\n";' | bc | less -S
In python, type the following into a document. NOTE: Sentences following a # symbol are comments, and are not necessary for the program to run. #!/usr/bin/python #This program takes a input from a user and reverses it. number = input("Type a number: ") #This takes input from a user. a = len(number) #This finds the length of the number reverse = "" #This sets the variable reverse to an empty string. for i in number: a = a-1 #The places in a string start from 0. The last value will be the length minus 1.reverse = reverse + number[a] #Makes the number's last digit the new string's first. print("The reverse of", number, "is", reverse + ".") #prints the resulting string. This program will take any sequence of characters and reverse them. The final value is a string, but if an integer is needed, one needs only to add the line reverse = int(reverse) above the print statement. However, this will stop the program from being able to reverse strings, as it is not possible to convert a string to an integer if it is not a number.
If a five-digit number is input through the keyboard, write a program to print a new number by adding one to each of its digits. For example if the number that is input is 12391 then the output should be displayed as 23402. //Author::Mudasir Yaqoob..... #include<stdio.h> #include<conio.h> int main() { long number,t; int i=0; long temp[5]; printf("Enter the five digit number:\n\n"); scanf("%ld",&number); while(i<=4) { t=number%10+1; temp[i]=t; number=number/10; i++; } printf("Reverse number\n\n\n\n"); for(i=4;i>=0;i--) { printf("%ld",temp[i]); } getch(); }
i want to write a simple without using pointer or array c program which will print greatest number when i give 20 number .........How far have you gotten so far?