We Keep Coding

How to find the 5th perfect number (which is 33550336)? The problem is taking forever to run

I am trying to write a Java method that checks whether a number is a perfect number or not.
A perfect number is a number that is equal to the sum of all its divisor (excluding itself).
For example, 6 is a perfect number because 1+2+3=6. Then, I have to write a Java program to use the method to display the first 5 perfect numbers.
I have no problem with this EXCEPT that it is taking forever to get the 5th perfect number which is 33550336.
I am aware that this is because of the for loop in my isPerfectNumber() method. However, I am very new to coding and I do not know how to come up with a better code.
public class Labreport2q1 {
public static void main(String[] args) {
//Display the 5 first perfect numbers
int counter = 0,
i = 0;
while (counter != 5) {
i++;
isPerfectNumber(i);
if (isPerfectNumber(i)) {
counter++;
System.out.println(i + " ");
}
}
}
public static boolean isPerfectNumber(int a) {
int divisor = 0;
int sum = 0;
for (int i = 1; i < a; i++) {
if (a % i == 0) {
divisor = i;
sum += divisor;
}
}
return sum == a;
}
}
This is the output that is missing the 5th perfect number

Let's check the properties of a perfect number. This Math Overflow question tells us two very interesting things:
A perfect number is never a perfect square.
A perfect number is of the form (2k-1)×(2k-1).
The 2nd point is very interesting because it reduces our search field to barely nothing. An int in Java is 32 bits. And here we see a direct correlation between powers and bit positions. Thanks to this, instead of making millions and millions of calls to isPerfectNumber, we will be making less than 32 to find the 5th perfect number.
So we can already change the search field, that's your main loop.
int count = 0;
for (int k = 1; count < 5; k++) {
// Compute candidates based on the formula.
int candidate = (1L << (k - 1)) * ((1L << k) - 1);
// Only test candidates, not all the numbers.
if (isPerfectNumber(candidate)) {
count++;
System.out.println(candidate);
}
}
This here is our big win. No other optimization will beat this: why test for 33 million numbers, when you can test less than 100?
But even though we have a tremendous improvement, your application as a whole can still be improved, namely your method isPerfectNumber(int).
Currently, you are still testing way too many numbers. A perfect number is the sum of all proper divisors. So if d divides n, n/d also divides n. And you can add both divisors at once. But the beauty is that you can stop at sqrt(n), because sqrt(n)*sqrt(n) = n, mathematically speaking. So instead of testing n divisors, you will only test sqrt(n) divisors.
Also, this means that you have to start thinking about corner cases. The corner cases are 1 and sqrt(n):
1 is a corner case because you if you divide n by 1, you get n but you don't add n to check if n is a perfect number. You only add 1. So we'll probably start our sum with 1 just to avoid too many ifs.
sqrt(n) is a corner case because we'd have to check whether sqrt(n) is an integer or not and it's tedious. BUT the Math Overflow question I referenced says that no perfect number is a perfect square, so that eases our loop condition.
Then, if at some point sum becomes greater than n, we can stop. The sum of proper divisors being greater than n indicates that n is abundant, and therefore not perfect. It's a small improvement, but a lot of candidates are actually abundant. So you'll probably save a few cycles if you keep it.
Finally, we have to take care of a slight issue: the number 1 as candidate. 1 is the first candidate, and will pass all our tests, so we have to make a special case for it. We'll add that test at the start of the method.
We can now write the method as follow:
static boolean isPerfectNumber(int n) {
// 1 would pass the rest because it has everything of a perfect number
// except that its only divisor is itself, and we need at least 2 divisors.
if (n < 2) return false;
// divisor 1 is such a corner case that it's very easy to handle:
// just start the sum with it already.
int sum = 1;
// We can stop the divisors at sqrt(n), but this is floored.
int sqrt = (int)Math.sqrt(n);
// A perfect number is never a square.
// It's useful to make this test here if we take the function
// without the context of the sparse candidates, because we
// might get some weird results if this method is simply
// copy-pasted and tested on all numbers.
// This condition can be removed in the final program because we
// know that no numbers of the form indicated above is a square.
if (sqrt * sqrt == n) {
return false;
}
// Since sqrt is floored, some values can still be interesting.
// For instance if you take n = 6, floor(sqrt(n)) = 2, and
// 2 is a proper divisor of 6, so we must keep it, we do it by
// using the <= operator.
// Also, sqrt * sqrt != n, so we can safely loop to sqrt
for (int div = 2; div <= sqrt; div++) {
if (n % div == 0) {
// Add both the divisor and n / divisor.
sum += div + n / div;
// Early fail if the number is abundant.
if (sum > n) return false;
}
}
return n == sum;
}
These are such optimizations that you can even find the 7th perfect number under a second, on the condition that you adapt the code for longs instead of ints. And you could still find the 8th within 30 seconds.
So here's that program (test it online). I removed the comments as the explanations are here above.
public class Main {
public static void main(String[] args) {
int count = 0;
for (int k = 1; count < 8; k++) {
long candidate = (1L << (k - 1)) * ((1L << k) - 1);
if (isPerfectNumber(candidate)) {
count++;
System.out.println(candidate);
}
}
}
static boolean isPerfectNumber(long n) {
if (n < 2) return false;
long sum = 1;
long sqrt = (long)Math.sqrt(n);
for (long div = 2; div <= sqrt; div++) {
if (n % div == 0) {
sum += div + n / div;
if (sum > n) return false;
}
}
return n == sum;
}
}
The result of the above program is the list of the first 8 perfect numbers:
6
28
496
8128
33550336
8589869056
137438691328
2305843008139952128
You can find further optimization, notably in the search if you check whether 2k-1 is prime or not as Eran says in their answer, but given that we have less than 100 candidates for longs, I don't find it useful to potentially gain a few milliseconds because computing primes can also be expensive in this program. If you want to check for bigger perfect primes, it makes sense, but here? No: it adds complexity and I tried to keep these optimization rather simple and straight to the point.

There are some heuristics to break early from the loops, but finding the 5th perfect number still took me several minutes (I tried similar heuristics to those suggested in the other answers).
However, you can rely on Euler's proof that all even perfect numbers (and it is still unknown if there are any odd perfect numbers) are of the form:
2i-1(2i-1)
where both i and 2i-1 must be prime.
Therefore, you can write the following loop to find the first 5 perfect numbers very quickly:
int counter = 0,
i = 0;
while (counter != 5) {
i++;
if (isPrime (i)) {
if (isPrime ((int) (Math.pow (2, i) - 1))) {
System.out.println ((int) (Math.pow (2, i -1) * (Math.pow (2, i) - 1)));
counter++;
}
}
}
Output:
6
28
496
8128
33550336
You can read more about it here.
If you switch from int to long, you can use this loop to find the first 7 perfect numbers very quickly:
6
28
496
8128
33550336
8589869056
137438691328
The isPrime method I'm using is:
public static boolean isPrime (int a)
{
if (a == 1)
return false;
else if (a < 3)
return true;
else {
for (int i = 2; i * i <= a; i++) {
if (a % i == 0)
return false;
}
}
return true;
}

Java program taking forever to run with large numbers

I am writing a Java program that calculates the largest prime factor of a large number. But I have an issue with the program's complexity, I don't know what has caused the program to run forever for large numbers, it works fine with small numbers.
I have proceeded as follow :
import java.util.ArrayList;
import java.util.Arrays;
import java.util.Collections;
public class Largest_prime_factor {
public static void main(String[] args)
{
//ArrayList primesArray = new ArrayList();
ArrayList factorArray = new ArrayList();
long largest = 1;
long number = 600851475143L ;
long i, j, k;
//the array list factorArray will have all factors of number
for (i = 2; i < number; i++)
{
if( number % i == 0)
{
factorArray.add(i);
}
}
Here, the Array List will have all the factors of the number.
So I'll need to get only the prime ones, for that, I used a method that checks if a number is prime or not, if it's not a prime number, I remove it from the list using the following method :
java.util.ArrayList.remove()
So the next part of the code is as follow :
for (i = 2; i < number; i++)
{
if (!isPrime(i))
{
factorArray.remove(i);
System.out.println(factorArray);
}
}
System.out.println(Collections.max(factorArray));
}
The last line prints the largest number of factorArray, which is what I am looking for.
public static boolean isPrime(long n)
{
if(n > 2 && (n & 1) == 0)
return false;
for(int i = 3; i * i <= n; i += 2)
if (n % i == 0)
return false;
return true;
}
}
The function above is what I used to determine if the number is a prime or not before removing it from the list.
This program works perfectly for small numbers, but it takes forever to give an output for large numbers, although the last function is pretty fast.
At first, I used to check if a number is prime or not inside of the first loop, but it was even slower.

You are looping over 600851475143 numbers.
long number = 600851475143L ;
for (i = 2; i < number; i++)
Even if we assume that each iteration takes very very small time (as small as 1 microsecond), it'll still take days before the loop finishes.
You need to optimise your prime-finding logic in order for this program to run faster.
One way to reduce the iterations to reasonable number is to loop until square root of number.
for (i = 2; i < Math.sqrt(number); i++)
or
for (i = 2; i*i < number; i++)

The calculation of the prime factors of 600851475143L should take less than a milli-second (with a not totally inefficient algorithm). The main parts your code is currently missing:
The border should be sqrt(number) and not number.
The current value should be checked in a while-loop (to prevent that non-prime-factors are added to the list, reduces range to check).
The max. value should be decreased (as well as the border) to number/factor after finding a factor.
Further improvements are possible, e.g. to iterate only over non-even numbers (or only iterate over numbers that are neither a multiple of 2 and 3) etc.
An example implementation for the same question on codereview (link):
public static long largestPrimeFactor(
final long input) {
////
if (input < 2)
throw new IllegalArgumentException();
long n = input;
long last = 0;
for (; (n & 1) == 0; n >>= 1)
last = 2;
for (; n % 3 == 0; n /= 3)
last = 3;
for (long v = 5, add = 2, border = (long) Math.sqrt(n); v <= border; v += add, add ^= 6)
while (n % v == 0)
border = (long) Math.sqrt(n /= last = v);
return n == 1 ? last : n;
}

for (i = 2; i < number; i++)
{
if( number % i == 0)
{
factorArray.add(i);
}
}
For an large input size, you will be visiting up to the value of the number. Same for the loop of removing factors.
long number = 600851475143L ;
this is a huge number, and you're looping through this twice. Try putting in a count for every 10,000 or 100,000 (if i%10000 print(i)) and you'll get an idea of how fast it's moving.

One of the possible solutions is to only test if the the prime numbers smaller than the large number divide it.
So I checked
for (i=2; i < number; i++)
{
if(isPrime(i))
{
if( number % i == 0)
{
factorArray.add(i);
}
}
}
So here I'll only be dividing by prime numbers instead of dividing by all numbers smaller than 600851475143.
But this is still not fast, a complete modification of the algorithm is necessary to obtain an optimal one.

#Balkrishna Rawool suggestion is the right way to go. For that I would suggest to change the iteration like this: for (i = 3; i < Math.sqrt(number); i+=2) and handle the 2 manually. That will decrease your looping because none of the even numbers except 2 are prime.

Decrease execution time of the Java program

I wrote the following program for the second problem of project Euler, for the question: "Project Euler #3: Largest prime factor".It is supposed to print out all the highest prime factors of the provided inputs.
import java.util.Scanner;
public class euler_2 {
public static boolean isPrime(int n) {
if (n % 2 == 0) return false;
for (int i = 3; i * i <= n; i += 2) {
if (n % i == 0)
return false;
}
return true;
}
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
int a = sc.nextInt();
for (int i = 0; i < a; i++) {
int b = sc.nextInt();
for (int j = b; j >= 1; j--) {
boolean aa = isPrime(j);
if (aa == true && b % j == 0) {
b = j;
break;
}
}
System.out.println(b);
}
}
}
What changes can I make to the program to make it execute faster? What would be a better algorithm for this problem?

The problem with your approach is that for every number N, you try each number smaller or equal to N whether it is a prime and after that whether it is a divisor of N.
Obvious improvement is to check whether it is a divisor first and only then whether it is a prime. But most probably this will not help that much.
What you can do instead is just to start checking each number whether it is a divisor of a number. If it is a divisor, divide it. You continue this till sqrt(N).
I have not done anything with java in a long time, but here is Go implementation, which most probably any Java person will be able to transform to Java.
func biggestPrime(n uint64) uint64 {
p, i := uint64(1), uint64(0)
for i = 2; i < uint64(math.Sqrt(float64(n))) + uint64(1); i++ {
for n % i == 0 {
n /= i
p = i
}
}
if n > 1 {
p = n
}
return p
}
Using my algorithm it will take you O(sqrt(N)) to find the biggest prime of a number. In your case it was O(N * sqrt(N))

Attempt to factor the number into 2 factors. Repeat on the largest factor found so far until you find one that can't be factored -- that is the largest prime factor.
There are many different ways you might try to factor the numbers, but since they are only ints, then Fermat's method or even trial division (going down from sqrt(N)) will probably do. See http://mathworld.wolfram.com/FermatsFactorizationMethod.html

Returning String of prime factors in Java

I know this is a classic problem. I solved it in Java. I have my solution below. However, when I used this solution in codefights.com, It went beyond the execution time limit. I would appreciate if anyone could give me suggestions on improving this code in any way possible. Please feel free to criticize my code so that I can improve on my coding skills. Thanks
You're given number n.
Return n as a product of its prime factors.
Example
For n = 22 the output should be "2*11".
For n = 120 the output should be "2*2*2*3*5".
For n = 17194016 the output should be "2*2*2*2*2*7*59*1301".
[input] integer n
An integer number smaller than 109. [output] string
The Prime factors of n which splitted by * symbol. prime factors
should be in increasing order.
Solution (JAVA):
public String primefactors(int n) {
String factors = "";
for (int i = 2; i <= n / 2; i++) {
if (isPrime(i)) {
while (n % i == 0) {
n /= i;
if (isPrime(n) && n != 1) {
factors = factors + Integer.valueOf(i).toString() + "*"
+ Integer.valueOf(n).toString();
break;
} else if (n == 1)
factors = factors + Integer.valueOf(i).toString();
else
factors = factors + Integer.valueOf(i).toString() + "*";
}
}
}
return factors;
}
public boolean isPrime(int n) {
boolean prime = true;
if (n == 1)
return false;
else if (n % 2 == 0 && n!=2)
return false;
else if (n % 3 == 0 && n!=3)
return false;
else {
for (int j = 2; j < n / 2; j++) {
if (n % j == 0) {
return false;
}
}
}
return prime;
}

Since n is smaller than a fixed number (109), simply use a table containing all prims <= 109, instead of generating them dynamically. Or atleast generate the prims first using the sieve of erathostenes or atkin. The hardcoded table would be preferable, but using a sieve for generating the table dynamically would aswell speed things up alot. The isPrime() function you implemented is a performance killer.

The function isPrime() is called too much times in primefactors. For example, i == 2 and there are many divisors 2 in n. The top call (isPrime(i)) is fine. However, inside the loop while (n % i == 0) you check isPrime(n) after each dividing n /= 2;. So, if initial n is 100 the function isPrime() is called for 50 and on the next loop for 25. That does not make any sense. I think that it is the biggest problem here, since even if isPrime works in linear time it is too much to call it many times in the internal loop.
It is possible to exit from the loop for i in two cases: n is equal 1 after divisions or n is for sure prime if i is larger than sqrt(n).
public String primefactors(int n) {
String factors = "";
int max_divisor = sqrt(n);
for (int i = 2; i <= max_divisor; i++) {
if (isPrime(i)) {
while (n % i == 0) {
n /= i;
if (n == 1)
factors = factors + Integer.valueOf(i).toString();
else
factors = factors + Integer.valueOf(i).toString() + "*";
}
max_divisor = sqrt(n);
}
}
// check for the last prime divisor
if (n != 1)
factors = factors + Integer.valueOf(n).toString();
return factors;
}
Even after that improvement (and sqrt(n) as the max limit in isPrime()) your algorithm will have linear complexity O(n), since there are at most sqrt(n) loops for i and the maximum number of probes for prime in isPrime is also sqrt(n).
Yes, it can be done better by choosing better algorithms for isPrime(). Even if you are not allowed to use hardcoded table of prime numbers it is possible to generate such look up table in runtime (if there are enough memory). So, it is possible to use automatically generated list of primes organized in ascending order to probe given number up to sqrt(n). If i becomes larger than sqrt(n) it means that the next prime is found and it should be appended to the look up table and isPrime() should return true.
Example
Suppose isPrime is called for 113. At that moment the look up table has a list of previous prime numbers: 2,3,5,7,11,13.... So, we try to divide 113 by items from that list up to sqrt(113) (while (i <= 10)). After trying 2,3,5,7 the next item on the list 11 is too large, so 113 is appended to the list of primes and the function returns true.
The other algorithms may give better performance in the worst case. For example the sieve of Eratosthenes or sieve of Atkin can be used to effectively precomputed list of prime numbers up to given n with the best O(n) complexity for the best implementation. Here you need to find all primes up to sqrt(n), so it takes O(sqrt(n)) to generate such list. Once such list is generated you need to try to divide your input by numbers is the list that takes at most sqrt(n) probes. So, the algorithm complexity is O(sqrt(n)). However, suppose that your input is 1024 that is 2 to the power of 10. In that case the first algorithm will be better, since it will not go to primes larger than 2.
Do you really need the function isPrime()?
With flexible thinking If we look closer it appears that you do not have to search for all primes in some range. You only needed to find all prime divisors of one given integer. However if we try to divide n by all integers in range up to sqrt(n) that is also good solution. Even if such integer is not prime it will be skipped due to the condition n % i == 0, since all primes lower than the integer under test are already removed from n, so that simple modular division does here the same as isPrime(). The full solution with O(sqrt(n)) complexity:
public String primefactors(int n) {
String factors = "";
int max_divisor = sqrt(n);
for (int i = 2; i <= max_divisor; i++) {
while (n % i == 0) {
n /= i;
max_divisor = sqrt(n);
if (n == 1)
factors = factors + Integer.valueOf(i).toString();
else
factors = factors + Integer.valueOf(i).toString() + "*";
}
}
// check for the last prime divisor
if (n != 1)
factors = factors + Integer.valueOf(n).toString();
return factors;
}
It is also possible to split the function to avoid if (n == 1) check in the inner loop, however it does not change the algorithm complexity.

Russian Doll Primes

This question was asked in an interview (about prime numbers)
Russian Doll Primes
They are more commonly known as Truncatable Primes.
I found this code on wiki
public static void main(String[] args){
final int MAX = 1000000;
//Sieve of Eratosthenes (using BitSet only for odd numbers)
BitSet primeList = new BitSet(MAX>>1);
primeList.set(0,primeList.size(),true);
int sqroot = (int) Math.sqrt(MAX);
primeList.clear(0);
for(int num = 3; num <= sqroot; num+=2)
{
if( primeList.get(num >> 1) )
{
int inc = num << 1;
for(int factor = num * num; factor < MAX; factor += inc)
{
//if( ((factor) & 1) == 1)
//{
primeList.clear(factor >> 1);
//}
}
}
}
//Find Largest Truncatable Prime. (so we start from 1000000 - 1
int rightTrunc = -1, leftTrunc = -1;
for(int prime = (MAX - 1) | 1; prime >= 3; prime -= 2)
{
if(primeList.get(prime>>1))
{
//Already found Right Truncatable Prime?
if(rightTrunc == -1)
{
int right = prime;
while(right > 0 && primeList.get(right >> 1)) right /= 10;
if(right == 0) rightTrunc = prime;
}
//Already found Left Truncatable Prime?
if(leftTrunc == -1 )
{
//Left Truncation
String left = Integer.toString(prime);
if(!left.contains("0"))
{
while( left.length() > 0 ){
int iLeft = Integer.parseInt(left);
if(!primeList.get( iLeft >> 1)) break;
left = left.substring(1);
}
if(left.length() == 0) leftTrunc = prime;
}
}
if(leftTrunc != -1 && rightTrunc != -1) //Found both? then Stop loop
{
break;
}
}
}
System.out.println("Left Truncatable : " + leftTrunc);
System.out.println("Right Truncatable : " + rightTrunc);
}
This gives the output:
Left Truncatable : 998443
Right Truncatable : 796339
But I am not able to solve this particular Russian doll prime number problem like if you have a prime number and you remove either left or right digit of this prime number then if that resulting number is prime number or not?
I am new to this so please pardon any mistake.

Let's start from the beginning:
According to the link you supplied with your question:
"Russian Doll Primes are
prime numbers whose right digit can be repeatedly removed, and are
still prime."
I will assume that you have a function boolean isPrime(int) to find out if a number is prime.
Googling, we will find from Wikipedia that the number of right-truncatable prime numbers up to 73,939,133 is 83, which makes brute-force a viable option; but a few optimization techniques can be employed here:
Since we right-truncate, we can safely eliminate even numbers (since any even number won't be prime, and so any number generated upon it will never be a russian doll prime).
Since any number that starts with 5 is divisible by 5, then based on the same rule I mentioned in the previous point, we can eliminate 5.
That leaves us with numbers that contain 1, 3, 7, and 9.
Now if we wanted to generate all possible combinations of these 4 numbers that do not exceed the maximum you mentioned (1,000,000), it would take only 4,096 iterations.
The downside of this technique is that we now have 4,096 numbers that contain possible non-prime numbers, or prime numbers that are formed from non-prime numbers and hence are not russian doll primes. We can eliminate these numbers by looping through them and checking; or better yet, we can examine russian doll primes more closely.
Upon examining the rule I quoted from your link above, we find that a russian doll primes are prime numbers whose right digit can be repeatedly removed, and are still prime (and hence are still russian doll prime, given the word repeatedly)!
That means we can work from the smallest single-digit russian doll primes, work our generation magic that we used above, and since any prime number that is formed from russian doll prime numbers is a russian doll prime number, we can eliminate non-primes early on, resulting in a clean list of russian doll prime numbers, while reducing the running time of such a program dramatically.
Take a look at the generation code below:
void russianDollPrimesGeneration(int x) {
x *= 10;
if (x * 10 >= 1000000) return;
int j;
for (int i=1; i<=9; i+=2) {
if (i == 5) continue;
j = x + i;
if (isPrime(j)) {
addToRussianDollPrimesList(j);
russianDollPrimesGeneration(j);
}
}
}
Provided that void addToRussianDollPrimesList(int x) is a function that adds x to a list that we previously preserved to store the russian doll prime numbers.
UPDATED NOTE
Note that you can put the call to void russianDollPrimesGeneration(int x) that we made inside the if condition inside the void addToRussianDollPrimesList(int x) function, because whenever we call the former function, we will always call the latter function with the same arguments. I'm separating them here to emphasize the recursive nature of the generation function.
Also note that you must run this function with the integer 0.
A final note is that there are a number of cases that the generation function void russianDollPrimesGeneration(int x) above won't count, even though they are Russian Doll Primes.
Remember when we omitted 2 and 5, because even numbers and numbers divided by 5 cannot be primes and so cannot be Russian Doll Primes? and consequently cannot form Russian Doll Primes? Well, that case does not apply to 2 and 5, because they are prime, and since they are single digits, therefore they are Russian Doll Primes, and are eligible to form Russian Doll Primes, if placed in the left-side, like 23 and 53.
So how to correct our code to include these special cases?
We can make a wrapper function that adds these two numbers and checks for Russian Doll Primes that can be formed using them (which will be the same generation function we are using above).
void generationWrapperFunction(int x) {
addToRussianDollPrimesList(2);
russianDollPrimesGeneration(2);
addToRussianDollPrimesList(5);
russianDollPrimesGeneration(5);
russianDollPrimesGeneration(0);
}
END UPDATED NOTE
This little function will produce a list of russian doll prime numbers, which can then be searched for the number we are looking for.
An alternative, yet I believe will be more time-consuming, is the following recursive function:
boolean isRussianDollPrime(int n) {
if (!isPrime(n)) return false;
if (n < 10) return true;
return isRussianDollPrime(n / 10);
}
This function can be modified to work with left-truncatable primes. The generation-based solution, however, will be much difficult to implement for left-truncatable primes.

Your problem is to use this code or to solve the problem ?
if to solve it you can generate primes using Sieve algorithm then check if the element is prime or not (if it was prime then check if element/10 is also prime)

Let's start with a simple assumption that we know how to write code to detect if a value is a prime. In a coding interview, they won't likely won't expect you to pull out "Sieve of Eratosthenes". You should start with simple code that handles the special cases of x<=1 (false) and x==2(true). Then check for an even number !(x % 2)(false). Then loop on i from 3..sqrt(x) (incrementing by +2 each time) to see if there's an odd number divisor for x.
boolean isPrime(long x)
{
// your code goes here
}
And once we have a function to tell us if a value is prime, we can easily build the function to detect if a value is a Russian Prime. Therefore we just need to loop on our value, each time check for prime, and then chop off the right hand side. And the easiest way to remove the right-most digit from a number is to simply divide it by 10.
boolean isRussianPrime(long x)
{
boolean result = isPrime(x);
while ((x != 0) && result)
{
// chop off the right digit of x
x = x / 10;
if (x != 0)
{
result = isPrime(x);
}
}
return result;
}
And that's really all there is to it.

package com.example.tests;
public class RussianDollPrimeNumber {
public static void main(String[] args) {
int x= 373;
int k;
int n =x;
for ( k= String.valueOf(x).length()-1;k>0;k--){
System.out.println(n);
if (isPrime(n)){
String m=String.valueOf(n).substring(0, k);
n=Integer.parseInt(m);
continue;
}else {
break;
}
}
if( k==0){
System.out.println("Number is Russianl Doll Number "+x);
}else {
System.out.println("Number is not Russianl Doll Number "+x);
}
}
private static boolean isPrime(int x) {
boolean check=true;
for (int i=2;i<x/2;i++){
if( (x%i)==0){
check=false;
}
}
return check;
}
}