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Trying to return user to scanner to re enter a number if value falls out of range (<25)

Im trying to configure my code to tell the user to re enter a number, taking them back to the scanner if it falls outside of my specified range of 25
long number;// declares variables for storing number
long factorial = 1;// declare variable for storing factorial
System.out.println("Enter a number between 1 and 25"); // tells user to enter number
number = scanner.nextLong();
if (number <0)
System.out.println("Positive numbers only");// if number entered is negative
else if (number > 25)
System.out.println("Number to large to print");
else if (number <= 1)// if number entered is 0 or 1
System.out.printf("The factorial of " + number+ " is equal to " + factorial);
else {
// if user enter 10, counter starts at 10 and runs to two
for(long mynumber = number; mynumber >= 1; mynumber--) {
factorial = factorial*mynumber; // mynumber would contain different values and that is multiplied by value present in factorial value and storing again in factorial variable
}
System.out.println("The factorial of " + number +" is equal to " + factorial);
}

To simplify the logic, you could extract it to the separate method. The logic itself is pretty strightforward:
in loop
ask a number
check if the number within the bounds
if not repeate or return if yes
public static void main(String... args) {
Scanner scan = new Scanner(System.in);
int num = getNumberWithin(scan, 1, 25);
}
private static int getNumberWithin(Scanner scan, int lo, int hi) {
while (true) {
System.out.format("Enter a number between %d and %d: ", lo, hi);
int num = scan.nextInt();
if (num >= lo && num <= hi)
return num;
System.err.format("The number should be between %d and %d\n", lo, hi);
System.out.println();
}
}

boolean correctInputn = false;
while(!correctInputn)
{
long number;// declares variables for storing number
long factorial = 1;// declare variable for storing factorial
System.out.println("Enter a number between 1 and 25"); // tells user to enter number
number = scanner.nextLong();
if (number <0) {
System.out.println("Positive numbers only"); // if number entered is negative
correctInputn = false;
continue; // if user enters number less than 0 loops back to code start
} else if (number > 25) {
System.out.println("Number to large to print");
correctInputn = false;
continue; // if user enters number over 25 loops back to code start
} else {
// if user enter 10, counter starts at 10 and runs to two
for(long mynumber = number; mynumber >= 1; mynumber--) {
factorial = factorial * mynumber; // mynumber would contain different values and that is multiplied by value present in factorial value and stored again in factorial variable
}
System.out.println("The factorial of " + number +" is equal to " + factorial);
break;
}
}

the function doesn't take more than 10 letters on scanner

I need to get from the user the length and the value of the array and return the digit that shows the most times. for example: the user gave me the length "4" and the numbers {83,238,8,54} the function will return " the numbers that shows the most times is 8";
but it send an error massage if the user enter more than 10 digits in general.
String sum = "";
// get the length
System.out.println("Enter the size of the array :");
int size = in.nextInt();
// create the array
String[] arr = new String[size+1];
// get value
System.out.println("Enter the elements of the array:(max capity 10 digits) ");
for(int i=0; i<arr.length; i++) {
arr[i] = in.nextLine();
sum+=arr[i];
}
System.out.println("the numbers that you gave me are: "+sum);
int all = Integer.parseInt(sum);
System.out.println("the digit that shows the most time is: "+maxOccurring(all));
}
static int countOccurrences(int x,int d) {
int count = 0;
while (x > 0)
{
if (x % 10 == d)
count++;
x = x / 10;
}
return count;
}
static int maxOccurring( int x)
{
if (x < 0)
x = -x;
int result = 0;
int max_count = 1;
for (int d = 0; d <= 10; d++)
{
int count = countOccurrences(x, d);
if (count >= max_count)
{
max_count = count;
result = d;
}
}
return result;
}
}
```

You get an NumberFormatException at the following line:
int all = Integer.parseInt(sum);
Exception trace is:
Exception in thread "main" java.lang.NumberFormatException: For input string: "25445648942"
at java.base/java.lang.NumberFormatException.forInputString(NumberFormatException.java:68)
at java.base/java.lang.Integer.parseInt(Integer.java:652)
at java.base/java.lang.Integer.parseInt(Integer.java:770)
at scan.ScannerTest.main(ScannerTest.java:27)
This exception is telling you that Integer.parseInt can't interpret the number you're passing through it (obtained by concatenating all the entered digits), in this precise example: "25445648942"
That's because this function is trying to yield an Integer from all those digits, which has a maximum capacity of 2^31 - 1 = 2,147,483,647. The entered number is higher that this, so it logically fails.
You should treat each entered number separately and count the digits in each, then do a sum of all to get your answer.
BTW, you don't ever need to convert to a number type to count the digits. It's simple string processing

Print odd numbers in a descending order

The program needs to take an odd number and output it in a descending order
For example: if the input is 11 the output needs to be 11 , 9 , 7 , 5 , 3, 1.
I tried using a for loop but I can only seem to get it to work with even numbers not odd numbers
public static void main(String[] args) {
Scanner input = new Scanner(System.in);
int number = input.nextInt();
for (int i = number - 1; i >= 0; i--) {
if (i % 2 == 0) {
int descend = i;
System.out.println(descend + " ");
}
}
}
The output is the number in descending order but as even only. If I add a 1 into the descend variable the numbers would seem to descend in an odd manner but its not ideal.

This line returns true if the number is even:
if (i % 2 == 0) {
If you want to know when the number is odd:
if (i % 2 != 0) {
Also, why are you starting your count at 1 less than the input value:
int i = number - 1;
I think you want to do this:
for (int i = number; i > 0; i--) { // tests for numbers starting at the input and stopping when i == 0

Just replace (i%2==0) to (i%2==1)

Asking if the number % 2 is equal to zero is basically asking if the number is even, so what you really have to do is ask if the number % 2 is not equal to zero, or equal to 1
if (i % 2 != 0) {
int descend = i;
System.out.println(descend + " ");
}
Also, there's no need to subtract 1 from the user input so your for loop can be written like this
for (int i = number; i >= 0; i--) {
if (i % 2 == 0) {
int descend = i;
System.out.println(descend + " ");
}
}

public static void main(String[] args) {
Scanner input = new Scanner(System.in);
System.out.print("Enter an an odd number: ");
int number = input.nextInt();
while(number%2==0){
System.out.print("Number must be odd number:" +
"(Ex:1, 3,5)\nTry again: ");
number=input.nextInt();
}
for (int i = number; i >= 0; i--) {
if(number%2!=0){
System.out.println(number);}
number-=1;
}
}

Printing the two highest values from user input

I have an assignment where I have to write a code which lets the user decide an amount of int values to be written in, and then decides what these values should be. There has to be atleast 2 inputs from the user. The program will then compare the values from the input and then print out the two highest values. So far I managed to print out the highest value, but I'm not sure whats wrong with the way I've done it since the output just becomes 0 if I choose to print out 2 numbers and the highest one is entered in first. And I'm also not sure how to keep track of the second highest number either. Would appreciate some help.
import java.util.Scanner;
public class ToStoersteTall{
public static void main(String[] args){
System.out.println("How many numbers? (minimum 2)?:");
Scanner reader = new Scanner(System.in);
if (reader.hasNextInt()) {
int numbers = reader.nextInt();
if (numbers >= 2) {
System.out.println("Enter value #1");
if (reader.hasNextInt()) {
int num1 = reader.nextInt();
System.out.println("Enter value #2");
if (reader.hasNextInt()) {
int num2 = reader.nextInt();
int biggest = 0;
for (int i = 3; i <= tall; i++) {
System.out.println("Enter value #" + i);
int num3 = reader.nextInt();
biggest = num1;
if(biggest < num3){
biggest = num3;
}
}
System.out.println(biggest);
} else {
System.out.println("Please enter an integer");
}
} else {
System.out.println("Please enter an integer");
}
} else {
System.out.println("Please enter an integer equal or higher than 2.");
}
} else {
System.out.print("Vennligst oppgi et heltall større eller lik 2.");
}
}
}

I have an assignment where I have to write a code which lets the user decide an amount of int values to be written in, and then decides what these values should be. There has to be atleast 2 inputs from the user. The program will then compare the values from the input and then print out the two highest values. So far I managed to print out the highest value, but I'm not sure whats wrong with the way I've done it since the output just becomes 0 if I choose to print out 2 numbers and the highest one is entered in first. And I'm also not sure how to keep track of the second highest number either. Would appreciate some help.
A couple things:
good practice to close scanner (and IO-related resources in general)
reduced if-statement blocks bloat for easier readability
you specify 2 guaranteed numbers, so attempt to parse those before looping
can remove system.exit calls or replace system.exit and move bulk of code back into the larger if-else blocks as originally state in OP (but I refer back to the sake of readability)
added check for the first and second numbers input to make sure high1 is highest value, and high2 is second highest value.
keep order while looping and checking values (note: does not use array), if the number is a new high, replace high1 and move high1's value down to high2, or if the number is a second (new) high, replace high2. If values are equal, this logic is excluded and you may want to specify based on your own constraints
import java.io.IOException;
import java.util.Scanner;
public class ToStoersteTall {
public static void main(String[] args) throws IOException {
System.out.println("How many numbers? (minimum 2)?:");
Scanner reader = new Scanner(System.in);
int n = 0;
if (reader.hasNextInt()) {
n = reader.nextInt();
} else {
System.out.println("Vennligst oppgi et heltall større eller lik 2.");
System.exit(-1); // quits execution
}
if (n < 2) {
System.out.println("Please enter an integer equal or higher than 2.");
System.exit(-2);
}
// Since guaranteed 2 numbers, parse and assign now
int high1 = 0, high2 = 0;
System.out.println("Enter value # 1");
if (reader.hasNextInt())
high1 = reader.nextInt();
System.out.println("Enter value # 2");
if (reader.hasNextInt())
high2 = reader.nextInt();
// check to see if a switch to keep correct highest order, swap values if so
if (high1 < high2) {
int t = high2;
high2 = high1;
high1 = t;
}
// loop won't execute if only 2 numbers input, but will if 3 or more specified at start
for (int i = 2; i < n; ++i) {
System.out.println("Enter value #" + (i + 1));
if (reader.hasNextInt()) {
int t = reader.nextInt();
if (t > high1) {
high2 = high1; // throw away high2 value and replace with high1
high1 = t; // replace high1 value with new highest value
} else if (t > high2) {
high2 = t;
}
} else {
System.out.println("Please enter an interger");
}
}
reader.close();
System.out.println("The two highest numbers are: " + high1 + ", " + high2);
}
}

You're already keeping track of the biggest, so why not keep track of the second biggest? Another easy way of solving this problem is to keep all the numbers in a list, sort the list by number size, and grab the two highest entries.

I tried your code and used an array to solve the problem.
import java.util.Scanner;
public class Main {
static int secondHighest(int... nums) {
int high1 = Integer.MIN_VALUE;
int high2 = Integer.MIN_VALUE;
for (int num : nums) {
if (num > high1) {
high2 = high1;
high1 = num;
} else if (num > high2) {
high2 = num;
}
}
return high2;
}
public static void main(String[] args) {
System.out.println("How many numbers? (minimum 2)?:");
Scanner reader = new Scanner(System.in);
if (reader.hasNextInt()) {
int numbers = reader.nextInt();
int[] array = new int[numbers];
if (numbers >= 2) {
System.out.println("Enter value #1");
if (reader.hasNextInt()) {
int num1 = reader.nextInt();
array[0] = num1;
System.out.println("Enter value #2");
if (reader.hasNextInt()) {
int num2 = reader.nextInt();
array[1] = num2;
int biggest = 0;
for (int i = 3; i <= numbers; i++) {
System.out.println("Enter value #" + i);
int num3 = reader.nextInt();
array[i-1] = num3;
}
System.out.println("second largest number is" + secondHighest(array));
int largest = 0;
for(int i =0;i<array.length;i++) {
if(array[i] > largest) {
largest = array[i];
}
}
System.out.println("Largest number in array is : " +largest);
} else {
System.out.println("Please enter an integer");
}
} else {
System.out.println("Please enter an integer");
}
} else {
System.out.println("Please enter an integer equal or higher than 2.");
}
} else {
System.out.print("Vennligst oppgi et heltall større eller lik 2.");
}
}
}
Test
How many numbers? (minimum 2)?:
6
Enter value #1
3
Enter value #2
4
Enter value #3
5
Enter value #4
6
Enter value #5
7
Enter value #6
8
second largest number is7
Largest number in array is : 8

There is a logic error in your program. If numbers is 2, then the for loop never gets executed, and the value of biggest remains zero because it is never updated. Change your declaration of biggest to reflect the current maximum value found so far.
int biggest = num1 > num2 ? num1 : num2;
That way if the for loop never executes then biggest will be the maximum value of the first two numbers.
As for keeping track of the second highest value, you could introduce another variable secondBiggest, initialised in a similar manner to biggest, and then write logic to update this value in your for loop. However, in my opinion, it would be much easier to change your strategy to hold the entered values into an array, then when all inputs have been entered, calculate whichever values you desire from the array. This would lead to a much cleaner solution IMO.
(I have assumed that tall in the for loop is actually meant to be numbers...)
import java.util.Scanner;
public class Foo{
public static void main(String[] args){
System.out.println("How many numbers? (minimum 2)?:");
Scanner reader = new Scanner(System.in);
if(reader.hasNextInt()){
int numbers = reader.nextInt();
if(numbers >= 2){
int[] list = new int[numbers];
for(int i = 0; i < numbers; i++){
System.out.println("Enter value #" + (i + 1));
if(reader.hasNextInt())
list[i] = reader.nextInt();
}//for
int biggest = 0;
int secondBiggest = 0;
// find the values you want
for(int i = 0; i < numbers; i++){
if(list[i] > biggest){
secondBiggest = biggest;
biggest = list[i];
}//if
else if(list[i] > secondBiggest)
secondBiggest = list[i];
}//for
// print your results
System.out.println("The biggest integer is: " + biggest);
System.out.println("The second biggest integer is: " + secondBiggest);
}//if
}//if
}//main
}//class

How does this prime number test in Java work?

The code snippet below checks whether a given number is a prime number. Can someone explain to me why this works? This code was on a study guide given to us for a Java exam.
public static void main(String[] args)
{
int j = 2;
int result = 0;
int number = 0;
Scanner reader = new Scanner(System.in);
System.out.println("Please enter a number: ");
number = reader.nextInt();
while (j <= number / 2)
{
if (number % j == 0)
{
result = 1;
}
j++;
}
if (result == 1)
{
System.out.println("Number: " + number + " is Not Prime.");
}
else
{
System.out.println("Number: " + number + " is Prime. ");
}
}

Overall theory
The condition if (number % j == 0) asks if number is exactly divisible by j
The definition of a prime is
a number divisible by only itself and 1
so if you test all numbers between 2 and number, and none of them are exactly divisible then it is a prime, otherwise it is not.
Of course you don't actually have to go all way to the number, because number cannot be exactly divisible by anything above half number.
Specific sections
While loop
This section runs through values of increasing j, if we pretend that number = 12 then it will run through j = 2,3,4,5,6
int j = 2;
.....
while (j <= number / 2)
{
........
j++;
}
If statement
This section sets result to 1, if at any point number is exactly divisible by j. result is never reset to 0 once it has been set to 1.
......
if (number % j == 0)
{
result = 1;
}
.....
Further improvements
Of course you can improve that even more because you actually need go no higher than sqrt(number) but this snippet has decided not to do that; the reason you need go no higher is because if (for example) 40 is exactly divisible by 4 it is 4*10, you don't need to test for both 4 and 10. And of those pairs one will always be below sqrt(number).
It's also worth noting that they appear to have intended to use result as a boolean, but actually used integers 0 and 1 to represent true and false instead. This is not good practice.

I've tried to comment each line to explain the processes going on, hope it helps!
int j = 2; //variable
int result = 0; //variable
int number = 0; //variable
Scanner reader = new Scanner(System.in); //Scanner object
System.out.println("Please enter a number: "); //Instruction
number = reader.nextInt(); //Get the number entered
while (j <= number / 2) //start loop, during loop j will become each number between 2 and
{ //the entered number divided by 2
if (number % j == 0) //If their is no remainder from your number divided by j...
{
result = 1; //Then result is set to 1 as the number divides equally by another number, hergo
} //it is not a prime number
j++; //Increment j to the next number to test against the number you entered
}
if (result == 1) //check the result from the loop
{
System.out.println("Number: " + number + " is Not Prime."); //If result 1 then a prime
}
else
{
System.out.println("Number: " + number + " is Prime. "); //If result is not 1 it's not a prime
}

It works by iterating over all number between 2 and half of the number entered (since any number greater than the input/2 (but less than the input) would yield a fraction). If the number input divided by j yields a 0 remainder (if (number % j == 0)) then the number input is divisible by a number other than 1 or itself. In this case result is set to 1 and the number is not a prime number.

Java java.math.BigInteger class contains a method isProbablePrime(int certainty) to check the primality of a number.
isProbablePrime(int certainty): A method in BigInteger class to check if a given number is prime.
For certainty = 1, it return true if BigInteger is prime and false if BigInteger is composite.
Miller–Rabin primality algorithm is used to check primality in this method.
import java.math.BigInteger;
public class TestPrime {
public static void main(String[] args) {
int number = 83;
boolean isPrime = testPrime(number);
System.out.println(number + " is prime : " + isPrime);
}
/**
* method to test primality
* #param number
* #return boolean
*/
private static boolean testPrime(int number) {
BigInteger bValue = BigInteger.valueOf(number);
/**
* isProbablePrime method used to check primality.
* */
boolean result = bValue.isProbablePrime(1);
return result;
}
}
Output: 83 is prime : true
For more information, see my blog.

Do try
public class PalindromePrime {
private static int g ,k ,n =0,i,m ;
static String b ="";
private static Scanner scanner = new Scanner( System.in );
public static void main(String [] args) throws IOException {
System.out.print(" Please Inter Data : ");
g = scanner.nextInt();
System.out.print(" Please Inter Data 2 : ");
m = scanner.nextInt();
count(g,m);
}
//
//********************************************************************************
private static int count(int L, int R)
for( i= L ; i<= R ;i++){
int count = 0 ;
for( n = i ; n >=1 ;n -- ){
if(i%n==0){
count = count + 1 ;
}
}
if(count == 2)
{
b = b +i + "" ;
}
}
System.out.print(" Data : ");
System.out.print(" Data : \n " +b );
return R;
}
}