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Showing largest divisible number instead of all numbers

I've started to learn basics of Java and currently I have to find and print largest divisible number.
For example number 30 can be divided by 2, 3, 6 and 10 so on the console I should show only 10. Another example is number 12 which can be divided by 2, 3 and 6 - should show 6.
I'm stuck at the point where I show the largest number. What I have so far is this one
double isItDivisible = Double.parseDouble(scanner.nextLine());
int num = 2;
if ( isItDivisible % 2 == 0 ) {
System.out.println("The number is divisible by " + num);
} else if ( isItDivisible % 3 == 0 ) {
num = 3;
System.out.println("The number is divisible by " + num);
} else if ( isItDivisible % 6 == 0 ) {
num = 6;
System.out.println("The number is divisible by " + num);
} else if ( isItDivisible % 10 == 0 ) {
num = 10;
System.out.println("The number is divisible by " + num);
} else if ( isItDivisible % 7 == 0 ) {
System.out.println("The number is divisible by " + 7);
}
The snipped display the first occurrence instead of the largest.
You can tell how new I'm based on the snipped so please bear with me.

When you want to find the highest divisible number that is not the input you could do the following:
int highest = 1; //1 should always apply - at least for integers
//iterate from highest to lowest number
for( int d = input-1; d > 1; d-- ) {
//stop at the first number that divides the input without a rest
if( input % d == 0 ) {
highest = d;
break;
}
}
If you want to check specific numbers only, you could modify that to use a sorted collection of numbers.
//TreeSet is sorted and the reserved comparator makes it sort from high to low
SortedSet<Integer> divisors = new TreeSet<>(Comparator.reversed());
//if the collection itself is sorted the order of additions doesn't matter
//if you're using a list and don't sort it after adding elements then order of insertion is relevant though
divisors.add(2);
divisors.add(10);
divisors.add(3);
divisors.add(7);
divisors.add(6);
Integer highest = null;
//iterates in sorted order
for( Integer d : divisors ) {
if( input % d == 0 ) {
highest = d;
break;
}
}

Try this :
public static void main(String[] args)
{
// Scanner Class
Scanner scanner = new Scanner(System.in);
double isItDivisible = Double.parseDouble(scanner.nextLine());
int high = 0;
for (int i = 0; i < isItDivisible; i++) {
if (isItDivisible % i == 0) {
high = i;
}
}
System.out.println(high);
}
Edit : With divide by pecific numbers 2,3,6,7,10:
public static void main(String[] args)
{
// Scanner Class
Scanner scanner = new Scanner(System.in);
double isItDivisible = Double.parseDouble(scanner.nextLine());
int high = 0;
int[] divide = {2,3,6,7,10};
Arrays.sort(divide);
for (int i : divide) {
if (isItDivisible % i == 0) {
high = i;
}
}
System.out.println(high);
}

Trying to compare two ints

This is my first time asking a question and I have tried to search the existing threads first. My program is meant to ask the user to enter a 5-digit number and it will check to see if it is a palindrome by reversing the number and then comparing the original number to the reverse. I also put in some verification steps there to reject the number if it is longer or shorter than 5 digits. Everything seems to work until it gets to the part comparing the original number and the reversed number. Here is my code:
import java.util.Scanner;
public class Palindromes {
public static void main(String args[]) {
int n, reverse = 0;
System.out.println("Enter a 5-digit integer to see if it is a palindrome.");
Scanner in = new Scanner(System.in);
n = in.nextInt();
int length = String.valueOf(n).length();
while (length > 5 || length < 5) {
System.out.println("Error: integer must be 5 digits in length.");
System.out.println("Enter a 5-digit integer.");
n = in.nextInt();
length = String.valueOf(n).length();
}
while (length == 5 && n != 0) {
reverse = reverse * 10;
reverse = reverse + n % 10;
n = n / 10;
}
System.out.println("Reversed number is: " + reverse);
if (n == reverse) {
System.out.println("Congratulations! Your number is a palindrome!");
} else {
System.out.println("Sorry. Your number isn't a palindrome.");
}
}
}

Look at what you are doing here!
while (length == 5 && n != 0) {
reverse = reverse * 10;
reverse = reverse + n % 10;
n = n / 10; // <----- You are changing "n"!
}
Which means that after the loop, n will no longer be the same n that the user entered.
To fix this, copy n to another variable and modify that instead.
int temp = n;
while (length == 5 && temp != 0) {
reverse = reverse * 10;
reverse = reverse + temp % 10;
temp = temp / 10;
}

it's more easy to convert the number to an array and reverse it instead of calculating it:
let reverseNumber = parseInt(12345.toString().split("").reverse().join());

Numbers to stop at the user's input

How do I make my program to stop at the user's input?
Here is my code:
public class H {
public static void main(String[] args) {
Scanner input = new Scanner(System.in);
System.out.println("Input x: ");
int x = input.nextInt();
for (int i = 0; i < x; i++) {
if (i < x)
System.out.print(printFib(i) + " ");
else if (i > x)
break;
}
}
public static int printFib(int number) {
if (number == 0 || number == 1)
return number;
else
return printFib(number - 1) + printFib(number - 2);
}
}
So, if I enter 10 my program should stop before the number. Example:
Input: 10
Output: 0 1 1 2 3 5 8
But instead I get 0 1 1 2 3 5 8 13 21 34
How can I fix it?
int x = input.nextInt();
int fib = 0;
while (fib < x){
System.out.print(printFib(fib)+ " ");
fib++;
}
}

Don't use a for loop which right now you're using to print out Fibonacci numbers until the number of items printed is less than the entered number. Instead use a while loop that stops when the Fibonacci number itself is greater than the entered number.
Since this is likely homework, I'm just going to give this suggestion and not a code solution, but please give a solution a try, and if still stuck, come back with questions.
Pseudocode
Get value of x
create fibonacci variable and assign it 0
while fibonacci is less than x
display current fibonacci number
calculate next fibonacci number and place in variable
end while loop

public class H {
public static void main(String[] args) {
Scanner input = new Scanner(System.in);
System.out.println("Input x: ");
int x = input.nextInt();
int i = 0;
while ( printFib(i) <= x ) {
System.out.print(printFib(i) + " ");
i ++ ;
}
}
public static int printFib(int number) {
if (number == 0 || number == 1)
return number;
else
return printFib(number - 1) + printFib(number - 2);
}
}
While the number return from the printFib() method is less than and equal to the user input, it then runs the method. I've tried the code and it works.

Why isn't my "While Loop" computing and printing out a simple 3N+1 equation?

So I had to write a simple code that would compute the 3N+1 equation; where N is an integer user types in and if it is a positive integer than N = N / 2 and if negative integer than N = N * 3 + 1.
However from what I can understand, my code doesn't work after the first while loop and hence prints nothing. What am I doing wrong? New to programming and still learning so I appreciate your help :)
Code:
import java.util.Scanner;
public class ThreeNplusOneProgram {
public static void main(String[] args) {
int N; Scanner input = new Scanner(System.in); int counter;
System.out.println("Please Enter an integer: ");
N = input.nextInt();
while ( N <= 0 ) {
System.out.println("ERROR: Please Enter an integer greater than zero: ");
N = input.nextInt();
}
//So far we know that N is great than Zero
System.out.println(N);
counter = 1;
while ( N != 1 ) {
if (N == N % 2 )
N = N / 2;
else N = N * 3 + 1;
counter = counter + 1;
}
System.out.println("There were" + counter + "terms in the sequence");
}
}

That is wrong: if (N == N % 2 )
N % 2 returns 1 or 0. You should use if (0 == N % 2 ) to check for being odd / even.

The problem is your if (N == N % 2), you might just want to check if (N >= 0) since you do state that you are looking to check if it is a positive integer.

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;
}
}