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Fermat's little theorem
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Rosser's Theorem

Last Updated : 08 Oct, 2023
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In mathematics, Rosser's Theorem states that the nth prime number is greater than the product of n and natural logarithm of n for all n greater than 1. 
Mathematically, 
For n >= 1, if pn is the nth prime number, then 
pn > n * (ln n)

Illustrative Examples:
      
       For n = 1, nth prime number = 2
    2 > 1 * ln(1)
Explanations, the above relation inherently holds true and it verifies the 
statement of Rosser's Theorem clearly.
Some more examples are, 
       For n = 2, nth prime number = 3
    3 > 2 * ln(2)
       For n = 3, nth prime number = 5
    5 > 3 * ln(3)
       For n = 4, nth prime number = 7
    7 > 4 * ln(4)
       For n = 5, nth prime number = 11
    11 > 5 * ln(5)
       For n = 6, nth prime number = 13
    13 > 6 * ln(6)

Approach for code: 
Efficient generation of prime numbers using prime sieve and checking the condition of Rosser's Theorem for each prime number individually. 

C++ 
// CPP code to verify Rosser's Theorem #include <bits/stdc++.h> using namespace std;  #define show(x) cout << #x << " = " << x << "\n";  vector<int> prime;  // Sieve of Eratosthenes void sieve() {     // prime sieve to generate prime numbers efficiently     int n = 1e5;     vector<bool> isprime(n + 2, true);     isprime[0] = isprime[1] = false;     for (long long i = 2; i <= n; i++) {         if (isprime[i]) {             for (long long j = i * i; j <= n; j += i)                 isprime[j] = false;         }     }     // store primes in prime[] vector     for (int i = 0; i <= n; i++)         if (isprime[i])             prime.push_back(i); }  // Verifies ROSSER'S THEOREM for all numbers  // smaller than n. void verifyRosser(int n) {     cout << "ROSSER'S THEOREM: nth prime "             "number > n * (ln n)\n";     for (int i = 0; i < n; i++)         if (prime[i] > (i + 1) * log(i + 1)) {             cout << "For n = " << i + 1                  << ", nth prime number = "                  << prime[i] << "\n\t"                  << prime[i] << " > " << i + 1                   << " * ln(" << i + 1 << ")\n";         } }   int main() {     sieve();     verifyRosser(20);     return 0; }
Java 
// Java code to verify Rosser's Theorem  import java.util.*; class GFG {     static Vector<Integer> prime=new Vector<Integer>();      // Sieve of Eratosthenes     static void sieve()     {         // prime sieve to generate prime numbers efficiently         int n = 10000;         boolean []isprime=new boolean[n+2];         for(int i=0;i<n;i++)         isprime[i]=true;                  isprime[0]=false;         isprime[1] =false;         for (int i = 2; i <= n; i++) {             if (isprime[i]) {                 for (int j = i * i; j <= n; j += i)                     isprime[j] =false;             }         }         // store primes in prime[] vector         for (int i = 0; i <= n; i++)             if (isprime[i])                 prime.add(i);     }          // Verifies ROSSER'S THEOREM for all numbers      // smaller than n.     static void verifyRosser(int n)     {         System.out.println("ROSSER'S THEOREM: nth prime number > n * (ln n)");                          for (int i = 0; i < n; i++)             if (prime.get(i) > (i + 1) * Math.log(i + 1)) {                 System.out.println( "For n = " + (i+1)                      + ", nth prime number = "                     + prime.get(i) + "\n\t"                     + prime.get(i) + " > " + (i + 1)                     + " * ln(" + (i + 1) + ")");             }     }          // Driver code     public static void main(String [] args)     {         sieve();         verifyRosser(20);              } }  // This code is contributed by ihritik
Python3 
# Python3 code to verify Rosser's Theorem import math prime = [];  # Sieve of Eratosthenes def sieve():          # prime sieve to generate      # prime numbers efficiently     n = 100001;     isprime = [True] * (n + 2);     isprime[0] = False;     isprime[1] = False;     for i in range(2, n + 1):         if(isprime[i]):             j = i * i;             while (j <= n):                 isprime[j] = False;                 j += i;          # store primes in      # prime[] vector     for i in range(n + 1):         if (isprime[i]):             prime.append(i);  # Verifies ROSSER'S THEOREM  # for all numbers smaller than n. def verifyRosser(n):          print("ROSSER'S THEOREM: nth",            "prime number > n * (ln n)");     for i in range(n):         if (prime[i] > int((i + 1) * math.log(i + 1))):             print("For n =", (i + 1), ", nth prime number =",                     prime[i], "\n\t", prime[i], " >", (i + 1),                                        "* ln(", (i + 1), ")");  # Driver Code sieve(); verifyRosser(20);  # This code is contributed  # by mits
C# 
// C# code to verify Rosser's Theorem using System; using System.Collections.Generic;      class GFG {     static List<int> prime = new List<int>();           // Sieve of Eratosthenes     static void sieve()     {         // prime sieve to generate          // prime numbers efficiently         int n = 10000;         bool []isprime = new bool[n + 2];         for(int i = 0; i < n; i++)         isprime[i] = true;                  isprime[0] = false;         isprime[1] = false;         for (int i = 2; i <= n; i++)          {             if (isprime[i])             {                 for (int j = i * i;                           j <= n; j += i)                     isprime[j] = false;             }         }                  // store primes in prime[] vector         for (int i = 0; i <= n; i++)             if (isprime[i])                 prime.Add(i);     }          // Verifies ROSSER'S THEOREM for      // all numbers smaller than n.     static void verifyRosser(int n)     {         Console.WriteLine("ROSSER'S THEOREM: " +                               "nth prime number > n * (ln n)");                          for (int i = 0; i < n; i++)             if (prime[i] > (i + 1) * Math.Log(i + 1))              {                 Console.WriteLine( "For n = " + (i + 1) +                                  ", nth prime number = " +                             prime[i] + "\n\t" + prime[i] +                               " > " + (i + 1) + " * ln(" +                                            (i + 1) + ")");             }     }          // Driver code     public static void Main(String [] args)     {         sieve();         verifyRosser(20);     } }  // This code is contributed by PrinciRaj1992
JavaScript 
<script>  // JavaScript code to verify // Rosser's Theorem let prime = new Array(); let y = 0;  // function show(x) // {echo x" = ".x."\n";}  // Sieve of Eratosthenes function sieve() {     // prime sieve to generate     // prime numbers efficiently     let n = 100001;     let isprime = new Array(n + 2).fill(true);     isprime[0] = false;     isprime[1] = false;     for (let i = 2; i <= n; i++)     {         if (isprime[i])         {             for (let j = i * i;                 j <= n; j += i)                 isprime[j] = false;         }     }          // store primes in     // prime[] vector     for (let i = 0; i <= n; i++)         if (isprime[i])             prime[y++] = i; }  // Verifies ROSSER'S THEOREM // for all numbers smaller than n. function verifyRosser(n) {         document.write(     "ROSSER'S THEOREM: nth prime number > n * (ln n) <br>"     );     for (let i = 0; i < n; i++)         if (prime[i] > Math.floor((i + 1) * Math.log(i + 1)))         {             document.write("For n = " + (i + 1) +                 ", nth prime number = " +                 prime[i] + "<br>" +                 prime[i] + " > " +                 (i + 1) + " * ln(" +                 (i + 1) + ")<br>");         } }  // Driver Code sieve(); verifyRosser(20);  // This code is contributed by gfgking  </script>
PHP 
<?php // PHP code to verify  // Rosser's Theorem $prime; $y = 0;  // function show($x) // {echo $x" = ".$x."\n";}  // Sieve of Eratosthenes function sieve() {     global $prime,$y;          // prime sieve to generate      // prime numbers efficiently     $n = 100001;     $isprime = array_fill(0, ($n + 2), true);     $isprime[0] = false;     $isprime[1] = false;     for ($i = 2; $i <= $n; $i++)     {         if ($isprime[$i])          {             for ($j = $i * $i;                   $j <= $n; $j += $i)                 $isprime[$j] = false;         }     }          // store primes in      // prime[] vector     for ($i = 0; $i <= $n; $i++)         if ($isprime[$i])             $prime[$y++]=$i; }  // Verifies ROSSER'S THEOREM  // for all numbers smaller than n. function verifyRosser($n) {     global $prime;          echo "ROSSER'S THEOREM: nth " .           "prime number > n * (ln n)\n";     for ($i = 0; $i < $n; $i++)         if ($prime[$i] > (int)(($i + 1) * log($i + 1)))         {             echo "For n = " . ($i + 1).                  ", nth prime number = " .                    $prime[$i] . "\n\t" .                    $prime[$i] . " > " .                    ($i + 1) . " * ln(" .                    ($i + 1) . ")\n";         } }  // Driver Code sieve(); verifyRosser(20);  // This code is contributed  // by mits ?>

Output
ROSSER'S THEOREM: nth prime number > n * (ln n) For n = 1, nth prime number = 2     2 > 1 * ln(1) For n = 2, nth prime number = 3     3 > 2 * ln(2) For n = 3, nth prime number = 5     5 > 3 * ln(3) For n = 4, nth prime number = 7     7 > 4 * ln(4) For n = 5, nth prime number = 11     11 > 5 * ln(5) For n = 6, nth prime number = 13     13 > 6 * ln(6) For n = 7, nth prime number = 17     17 > 7 * ln(7) For n = 8, nth prime number = 19     19 > 8 * ln(8) For n = 9, nth prime number = 23     23 > 9 * ln(9) For n = 10, nth prime number = 29     29 > 10 * ln(10) For n = 11, nth prime number = 31     31 > 11 * ln(11) For n = 12, nth prime number = 37     37 > 12 * ln(12) For n = 13, nth prime number = 41     41 > 13 * ln(13) For n = 14, nth prime number = 43     43 > 14 * ln(14) For n = 15, nth prime number = 47     47 > 15 * ln(15) For n = 16, nth prime number = 53     53 > 16 * ln(16) For n = 17, nth prime number = 59     59 > 17 * ln(17) For n = 18, nth prime number = 61     61 > 18 * ln(18) For n = 19, nth prime number = 67     67 > 19 * ln(19) For n = 20, nth prime number = 71     71 > 20 * ln(20)

Time Complexity: O(n*log(log(n)))
Auxiliary Space: O(n)

Another Approach :

C++ 
#include <bits/stdc++.h> using namespace std;  #define show(x) cout << #x << " = " << x << "\n";  // Computes the nth prime number using the sieve of Eratosthenes int nthPrime(int n) {     int maxN = max((int)(1.2 * n * log(n)), 20); // maxN is an estimate of the nth prime     vector<bool> isprime(maxN + 1, true);     int count = 0;     for (int i = 2; i <= maxN; i++) {         if (isprime[i]) {             count++;             if (count == n) {                 return i;             }             for (int j = i * i; j <= maxN; j += i) {                 isprime[j] = false;             }         }     }     return -1; // nth prime not found }  // Verifies ROSSER'S THEOREM for all numbers  // smaller than n. void verifyRosser(int n) {     cout << "ROSSER'S THEOREM: nth prime "             "number > n * (ln n)\n";     for (int i = 1; i <= n; i++) {         int nth = nthPrime(i);         if (nth > i * log(i)) {             cout << "For n = " << i << ", nth prime number = "                  << nth << "\n\t"                  << nth << " > " << i << " * ln(" << i << ")\n";         }     } }  int main() {     verifyRosser(20);     return 0; }
Java 
import java.util.Arrays;  public class GFG {      // Computes the nth prime number using the sieve of Eratosthenes     static int nthPrime(int n) {         int maxN = Math.max((int) (1.2 * n * Math.log(n)), 20); // maxN is an estimate of the nth prime         boolean[] isPrime = new boolean[maxN + 1];         Arrays.fill(isPrime, true);          int count = 0;         for (int i = 2; i <= maxN; i++) {             if (isPrime[i]) {                 count++;                 if (count == n) {                     return i;                 }                 for (int j = i * i; j <= maxN; j += i) {                     isPrime[j] = false;                 }             }         }         return -1; // nth prime not found     }      // Verifies ROSSER'S THEOREM for all numbers      // smaller than n.     static void verifyRosser(int n) {         System.out.println("ROSSER'S THEOREM: nth prime number > n * (ln n)");         for (int i = 1; i <= n; i++) {             int nth = nthPrime(i);             if (nth > i * Math.log(i)) {                 System.out.println("For n = " + i + ", nth prime number = " + nth);                 System.out.println("\t" + nth + " > " + i + " * ln(" + i + ")");             }         }     }      public static void main(String[] args) {         verifyRosser(20);     } }
Python3 
import math  # Computes the nth prime number using the sieve of Eratosthenes def nth_prime(n):     maxN = max(int(1.2 * n * math.log(n)), 20)  # maxN is an estimate of the nth prime     isprime = [True] * (maxN + 1)     count = 0     for i in range(2, maxN + 1):         if isprime[i]:             count += 1             if count == n:                 return i             for j in range(i * i, maxN + 1, i):                 isprime[j] = False     return -1  # nth prime not found  # Verifies ROSSER'S THEOREM for all numbers smaller than n. def verify_rosser(n):     print("ROSSER'S THEOREM: nth prime number > n * ln n")     for i in range(1, n + 1):         nth = nth_prime(i)         if nth > i * math.log(i):             print(f"For n = {i}, nth prime number = {nth}")             print(f"{nth} > {i} * ln({i})")  if __name__ == "__main__":     verify_rosser(20)
C# 
using System; using System.Collections.Generic;  namespace RosserTheoremVerification {     class GFG     {         // Computes the nth prime number using the sieve of Eratosthenes         static int NthPrime(int n)         {             int maxN = (int)(1.2 * n * Math.Log(n)); // maxN is an estimate of the nth prime             List<bool> isprime = new List<bool>(new bool[maxN + 1]);             int count = 0;             for (int i = 2; i <= maxN; i++)             {                 if (isprime[i])                 {                     count++;                     if (count == n)                     {                         return i;                     }                     for (int j = i * i; j <= maxN; j += i)                     {                         isprime[j] = false;                     }                 }             }             return -1; // nth prime not found         }          // Verifies ROSSER'S THEOREM for all numbers          // smaller than n.         static void VerifyRosser(int n)         {             Console.WriteLine("ROSSER'S THEOREM: nth prime number > n * (ln n)");             for (int i = 1; i <= n; i++)             {                 int nth = NthPrime(i);                 if (nth > i * Math.Log(i))                 {                     Console.WriteLine($"For n = {i}, nth prime number = {nth}");                     Console.WriteLine($"\t{nth} > {i} * ln({i})");                 }             }         }          static void Main(string[] args)         {             VerifyRosser(20);         }     } }
JavaScript 
// Computes the nth prime number using the sieve of Eratosthenes function nthPrime(n) {     let maxN = Math.max(Math.floor((1.2 * n * Math.log(n))), 20); // maxN is an estimate of the nth prime     let isprime=new Array(maxN + 1).fill(true);     let count = 0;     for (let i = 2; i <= maxN; i++) {         if (isprime[i]) {             count++;             if (count == n) {                 return i;             }             for (let j = i * i; j <= maxN; j += i) {                 isprime[j] = false;             }         }     }     return -1; // nth prime not found }  // Verifies ROSSER'S THEOREM for all numbers  // smaller than n. function verifyRosser( n) {     document.write("ROSSER'S THEOREM: nth prime " +             "number > n * (ln n)");     for (let i = 1; i <= n; i++) {         let nth = nthPrime(i);         if (nth > i * Math.floor(Math.log(i))) {             document.write("For n = " + i + ", nth prime number = "                  + nth + "\n\t"                  + nth + " > " + i + " * ln(" + i + ")");         }     } }     verifyRosser(20);

Output :

ROSSER'S THEOREM : nth prime number >n *(ln n) 
for n = 1, nth prime number = 2
    2 > 1 * ln(1)
For n = 2, nth prime number = 3
    3 > 2 * ln(2)
For n = 3, nth prime number = 5
    5 > 3 * ln(3)
For n = 4, nth prime number = 7
    7 > 4 * ln(4)
For n = 5, nth prime number = 11
    11 > 5 * ln(5)
For n = 6, nth prime number = 13
    13 > 6 * ln(6)
For n = 7, nth prime number = 17
    17 > 7 * ln(7)
For n = 8, nth prime number = 19
    19 > 8 * ln(8)
For n = 9, nth prime number = 23
    23 > 9 * ln(9)
For n = 10, nth prime number = 29
    29 > 10 * ln(10)
For n = 11, nth prime number = 31
    31 > 11 * ln(11)
For n = 12, nth prime number = 37
    37 > 12 * ln(12)
For n = 13, nth prime number = 41
    41 > 13 * ln(13)
For n = 14, nth prime number = 43
    43 > 14 * ln(14)
For n = 15, nth prime number = 47
    47 > 15 * ln(15)
For n = 16, nth prime number = 53
    53 > 16 * ln(16)
For n = 17, nth prime number = 59
    59 > 17 * ln(17)
For n = 18, nth prime number = 61
    61 > 18 * ln(18)
For n = 19, nth prime number = 67
    67 > 19 * ln(19)
For n = 20, nth prime number = 71
    71 > 20 * ln(20)

Time Complexity: O(n*log(log(n)))
Auxiliary Space: O(n)


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    Given integers n and m, construct a strictly increasing sequence of m positive integers with sum exactly equal to n such that the GCD of the sequence is maximized. If multiple sequences have the same maximum GCD, return the lexicographically smallest one. If not possible, return -1.Examples : Input
    7 min read
    Largest Subset with GCD 1
    Given n integers, we need to find size of the largest subset with GCD equal to 1. Input Constraint : n <= 10^5, A[i] <= 10^5Examples: Input : A = {2, 3, 5}Output : 3Explanation: The largest subset with a GCD greater than 1 is {2, 3, 5}, and the GCD of all the elements in the subset is 3.Input
    6 min read
    Summation of GCD of all the pairs up to n
    Given a number n, find sum of all GCDs that can be formed by selecting all the pairs from 1 to n. Examples: Input : n = 4Output : 7Explanation: Numbers from 1 to 4 are: 1, 2, 3, 4Result = gcd(1,2) + gcd(1,3) + gcd(1,4) + gcd(2,3) + gcd(2,4) + gcd(3,4) = 1 + 1 + 1 + 1 + 2 + 1 = 7Input : n = 12Output
    10 min read

    Series

    Juggler Sequence
    Juggler Sequence is a series of integer number in which the first term starts with a positive integer number a and the remaining terms are generated from the immediate previous term using the below recurrence relation : a_{k+1}=\begin{Bmatrix} \lfloor a_{k}^{1/2} \rfloor & for \quad even \quad a
    5 min read
    Padovan Sequence
    Padovan Sequence similar to Fibonacci sequence with similar recursive structure. The recursive formula is, P(n) = P(n-2) + P(n-3) P(0) = P(1) = P(2) = 1 Fibonacci Sequence: 0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55...... Spiral of squares with side lengths which follow the Fibonacci sequence. Padovan Sequ
    4 min read
    Aliquot Sequence
    Given a number n, the task is to print its Aliquot Sequence. Aliquot Sequence of a number starts with itself, remaining terms of the sequence are sum of proper divisors of immediate previous term. For example, Aliquot Sequence for 10 is 10, 8, 7, 1, 0. The sequence may repeat. For example, for 6, we
    8 min read
    Moser-de Bruijn Sequence
    Given an integer 'n', print the first 'n' terms of the Moser-de Bruijn Sequence. Moser-de Bruijn sequence is the sequence obtained by adding up the distinct powers of the number 4 (For example, 1, 4, 16, 64, etc). Examples: Input : 5 Output : 0 1 4 5 16 Input : 10 Output : 0 1 4 5 16 17 20 21 64 65
    12 min read
    Stern-Brocot Sequence
    Stern Brocot sequence is similar to Fibonacci sequence but it is different in the way fibonacci sequence is generated . Generation of Stern Brocot sequence : 1. First and second element of the sequence is 1 and 1.2. Consider the second member of the sequence . Then, sum the considered member of the
    5 min read
    Newman-Conway Sequence
    Newman-Conway Sequence is the one that generates the following integer sequence. 1 1 2 2 3 4 4 4 5 6 7 7... In mathematical terms, the sequence P(n) of Newman-Conway numbers is defined by the recurrence relation P(n) = P(P(n - 1)) + P(n - P(n - 1)) with seed values P(1) = 1 and P(2) = 1 Given a numb
    6 min read
    Sylvester's sequence
    In number system, Sylvester's sequence is an integer sequence in which each member of the sequence is the product of the previous members, plus one. Given a positive integer N. The task is to print the first N member of the sequence. Since numbers can be very big, use %10^9 + 7.Examples: Input : N =
    4 min read
    Recaman's sequence
    Given an integer n. Print first n elements of Recaman’s sequence. Recaman's Sequence starts with 0 as the first term. For each next term, calculate previous term - index (if positive and not already in sequence); otherwise, use previous term + index.Examples: Input: n = 6Output: 0, 1, 3, 6, 2, 7Expl
    8 min read
    Sum of the sequence 2, 22, 222, .........
    Given an integer n. The task is to find the sum of the following sequence: 2, 22, 222, ......... to n terms. Examples : Input: n = 2Output: 24Explanation: For n = 2, the sum of first 2 terms are 2 + 22 = 24Input: 3Output: 246Explanation: For n = 3, the sum of first 3 terms are 2 + 22 + 222 = 246Usin
    8 min read
    Sum of series 1^2 + 3^2 + 5^2 + . . . + (2*n - 1)^2
    Given a series 12 + 32 + 52 + 72 + . . . + (2*n - 1)2, find the sum of the series.Examples: Input: n = 4Output: 84Explanation: sum = 12 + 32 + 52 + 72 = 1 + 9 + 25 + 49 = 84Input: n = 10 Output: 1330Explanation: sum = 12 + 32 + 52 + 72 + 92 + 112 + 132 + 152 + 172 + 192 = 1 + 9 + 24 + 49 + . . . + 3
    3 min read
    Sum of the series 0.6, 0.06, 0.006, 0.0006, ...to n terms
    Given the number of terms i.e. n. Find the sum of the series 0.6, 0.06, 0.006, 0.0006, ...to n terms.Examples: Input: 2Output: 0.66Explanation: sum of the series upto 2 terms: 0.6 + 0.06 = 0.66.Input: 3Output: 0.666Explanation: sum of the series upto 3 terms: 0.6 + 0.06 + 0.006 = 0.666.Table of Cont
    7 min read
    n-th term in series 2, 12, 36, 80, 150....
    Given a series 2, 12, 36, 80, 150.. Find the n-th term of the series.Examples : Input : 2 Output : 12 Input : 4 Output : 80 If we take a closer look, we can notice that series is sum of squares and cubes of natural numbers (1, 4, 9, 16, 25, .....) + (1, 8, 27, 64, 125, ....).Therefore n-th number of
    3 min read

    Number Digits

    Minimum digits to remove to make a number Perfect Square
    Given an integer n, we need to find how many digits remove from the number to make it a perfect square. Examples : Input : 8314 Output: 81 2 Explanation: If we remove 3 and 4 number becomes 81 which is a perfect square. Input : 57 Output : -1 The idea is to generate all possible subsequences and ret
    9 min read
    Print first k digits of 1/n where n is a positive integer
    Given a positive integer n, print first k digits after point in value of 1/n. Your program should avoid overflow and floating point arithmetic.Examples : Input: n = 3, k = 3 Output: 333 Input: n = 50, k = 4 Output: 0200 We strongly recommend to minimize the browser and try this yourself first.Let us
    5 min read
    Check if a given number can be represented in given a no. of digits in any base
    Given a number and no. of digits to represent the number, find if the given number can be represented in given no. of digits in any base from 2 to 32.Examples : Input: 8 4 Output: Yes Possible in base 2 as 8 in base 2 is 1000 Input: 8 2 Output: Yes Possible in base 3 as 8 in base 3 is 22 Input: 8 3
    12 min read
    Minimum Segments in Seven Segment Display
    A seven-segment display can be used to display numbers. Given an array of n natural numbers. The task is to find the number in the array that uses the minimum number of segments to display the number. If multiple numbers have a minimum number of segments, output the number having the smallest index.
    6 min read
    Find next greater number with same set of digits
    Given a number N as string, find the smallest number that has same set of digits as N and is greater than N. If N is the greatest possible number with its set of digits, then print "Not Possible".Examples: Input: N = "218765"Output: "251678"Explanation: The next number greater than 218765 with same
    9 min read
    Check if a number is jumbled or not
    Write a program to check if a given integer is jumbled or not. A number is said to be Jumbled if for every digit, its neighbours digit differs by max 1. Examples : Input : 6765Output : TrueAll neighbour digits differ by atmost 1. Input : 1223Output : True Input : 1235Output : False Approach: Find th
    6 min read
    Numbers having difference with digit sum more than s
    You are given two positive integer value n and s. You have to find the total number of such integer from 1 to n such that the difference of integer and its digit sum is greater than given s.Examples : Input : n = 20, s = 5 Output :11 Explanation : Integer from 1 to 9 have diff(integer - digitSum) =
    7 min read
    Total numbers with no repeated digits in a range
    Given a range L, R find total such numbers in the given range such that they have no repeated digits. For example: 12 has no repeated digit. 22 has repeated digit. 102, 194 and 213 have no repeated digit. 212, 171 and 4004 have repeated digits. Examples: Input : 10 12 Output : 2 Explanation : In the
    15+ min read
    K-th digit in 'a' raised to power 'b'
    Given three numbers a, b and k, find k-th digit in ab from right sideExamples: Input : a = 3, b = 3, k = 1Output : 7Explanation: 3^3 = 27 for k = 1. First digit is 7 in 27Input : a = 5, b = 2, k = 2Output : 2Explanation: 5^2 = 25 for k = 2. First digit is 2 in 25The approach is simple. Computes the
    3 min read

    Algebra

    Program to add two polynomials
    Given two polynomials represented by two arrays, write a function that adds given two polynomials. Example: Input: A[] = {5, 0, 10, 6} B[] = {1, 2, 4} Output: sum[] = {6, 2, 14, 6} The first input array represents "5 + 0x^1 + 10x^2 + 6x^3" The second array represents "1 + 2x^1 + 4x^2" And Output is
    15+ min read
    Multiply two polynomials
    Given two polynomials represented by two arrays, write a function that multiplies the given two polynomials. In this representation, each index of the array corresponds to the exponent of the variable(e.g. x), and the value at that index represents the coefficient of the term. For example, the array
    15+ min read
    Find number of solutions of a linear equation of n variables
    Given a linear equation of n variables, find number of non-negative integer solutions of it. For example, let the given equation be "x + 2y = 5", solutions of this equation are "x = 1, y = 2", "x = 5, y = 0" and "x = 3, y = 1". It may be assumed that all coefficients in given equation are positive i
    10 min read
    Calculate the Discriminant Value
    In algebra, Discriminant helps us deduce various properties of the roots of a polynomial or polynomial function without even computing them. Let's look at this general quadratic polynomial of degree two: ax^2+bx+c Here the discriminant of the equation is calculated using the formula: b^2-4ac Now we
    5 min read
    Program for dot product and cross product of two vectors
    There are two vector A and B and we have to find the dot product and cross product of two vector array. Dot product is also known as scalar product and cross product also known as vector product.Dot Product - Let we have given two vector A = a1 * i + a2 * j + a3 * k and B = b1 * i + b2 * j + b3 * k.
    8 min read
    Iterated Logarithm log*(n)
    Iterated Logarithm or Log*(n) is the number of times the logarithm function must be iteratively applied before the result is less than or equal to 1. \log ^{*}n:=\begin{cases}0n\leq 1;\\1+\log ^{*}(\log n)n>1\end{cases} Applications: It is used in the analysis of algorithms (Refer Wiki for detail
    4 min read
    Program to Find Correlation Coefficient
    The correlation coefficient is a statistical measure that helps determine the strength and direction of the relationship between two variables. It quantifies how changes in one variable correspond to changes in another. This coefficient, sometimes referred to as the cross-correlation coefficient, al
    8 min read
    Program for Muller Method
    Given a function f(x) on floating number x and three initial distinct guesses for root of the function, find the root of function. Here, f(x) can be an algebraic or transcendental function.Examples: Input : A function f(x) = x^3 + 2x^2 + 10x - 20 and three initial guesses - 0, 1 and 2 .Output : The
    13 min read
    Number of non-negative integral solutions of a + b + c = n
    Given a number n, find the number of ways in which we can add 3 non-negative integers so that their sum is n.Examples : Input : n = 1 Output : 3 There are three ways to get sum 1. (1, 0, 0), (0, 1, 0) and (0, 0, 1) Input : n = 2 Output : 6 There are six ways to get sum 2. (2, 0, 0), (0, 2, 0), (0, 0
    7 min read
    Generate Pythagorean Triplets
    Given a positive integer limit, your task is to find all possible Pythagorean Triplet (a, b, c), such that a <= b <= c <= limit.Note: A Pythagorean triplet is a set of three positive integers a, b, and c such that a2 + b2 = c2. Input: limit = 20Output: 3 4 5 5 12 13 6 8 10 8 15 17 9 12 15 1
    14 min read

    Number System

    Exponential notation of a decimal number
    Given a positive decimal number, find the simple exponential notation (x = a·10^b) of the given number. Examples: Input : 100.0 Output : 1E2 Explanation: The exponential notation of 100.0 is 1E2. Input :19 Output :1.9E1 Explanation: The exponential notation of 16 is 1.6E1. Approach: The simplest way
    5 min read
    Check if a number is power of k using base changing method
    This program checks whether a number n can be expressed as power of k and if yes, then to what power should k be raised to make it n. Following example will clarify : Examples: Input : n = 16, k = 2 Output : yes : 4 Explanation : Answer is yes because 16 can be expressed as power of 2. Input : n = 2
    8 min read
    Program to convert a binary number to hexadecimal number
    Given a Binary Number, the task is to convert the given binary number to its equivalent hexadecimal number. The input could be very large and may not fit even into an unsigned long long int.Examples: Input: 110001110Output: 18EInput: 1111001010010100001.010110110011011Output: 794A1.5B36 794A1D9B App
    13 min read
    Program for decimal to hexadecimal conversion
    Given a decimal number as input, we need to write a program to convert the given decimal number into an equivalent hexadecimal number. i.e. convert the number with base value 10 to base value 16.Hexadecimal numbers use 16 values to represent a number. Numbers from 0-9 are expressed by digits 0-9 and
    8 min read
    Converting a Real Number (between 0 and 1) to Binary String
    Given a real number between 0 and 1 (e.g., 0.72) that is passed in as a double, print the binary representation. If the number cannot be represented accurately in binary with at most 32 characters, print" ERROR:' Examples: Input : (0.625)10 Output : (0.101)2 Input : (0.72)10 Output : ERROR Solution:
    12 min read
    Convert from any base to decimal and vice versa
    Given a number and its base, convert it to decimal. The base of number can be anything such that all digits can be represented using 0 to 9 and A to Z. The value of A is 10, the value of B is 11 and so on. Write a function to convert the number to decimal. Examples: Input number is given as string a
    15+ min read
    Decimal to binary conversion without using arithmetic operators
    Find the binary equivalent of the given non-negative number n without using arithmetic operators. Examples: Input : n = 10Output : 1010 Input : n = 38Output : 100110 Note that + in below algorithm/program is used for concatenation purpose. Algorithm: decToBin(n) if n == 0 return "0" Declare bin = ""
    8 min read

    Prime Numbers & Primality Tests

    Prime Numbers | Meaning | List 1 to 100 | Examples
    Prime numbers are those natural numbers that are divisible by only 1 and the number itself. Numbers that have more than two divisors are called composite numbers All primes are odd, except for 2.Here, we will discuss prime numbers, the list of prime numbers from 1 to 100, various methods to find pri
    12 min read
    Left-Truncatable Prime
    A Left-truncatable prime is a prime which in a given base (say 10) does not contain 0 and which remains prime when the leading ("left") digit is successively removed. For example, 317 is left-truncatable prime since 317, 17 and 7 are all prime. There are total 4260 left-truncatable primes.The task i
    13 min read
    Program to Find All Mersenne Primes till N
    Mersenne Prime is a prime number that is one less than a power of two. In other words, any prime is Mersenne Prime if it is of the form 2k-1 where k is an integer greater than or equal to 2. First few Mersenne Primes are 3, 7, 31 and 127.The task is print all Mersenne Primes smaller than an input po
    9 min read
    Super Prime
    Given a positive integer n and the task is to print all the Super-Primes less than or equal to n. Super-prime numbers (also known as higher order primes) are the subsequence of prime number sequence that occupy prime-numbered positions within the sequence of all prime numbers. The first few super pr
    6 min read
    Hardy-Ramanujan Theorem
    Hardy Ramanujam theorem states that the number of prime factors of n will approximately be log(log(n)) for most natural numbers nExamples : 5192 has 2 distinct prime factors and log(log(5192)) = 2.1615 51242183 has 3 distinct prime facts and log(log(51242183)) = 2.8765 As the statement quotes, it is
    6 min read
    Rosser's Theorem
    In mathematics, Rosser's Theorem states that the nth prime number is greater than the product of n and natural logarithm of n for all n greater than 1. Mathematically, For n >= 1, if pn is the nth prime number, then pn > n * (ln n) Illustrative Examples: For n = 1, nth prime number = 2 2 >
    15 min read
    Fermat's little theorem
    Fermat's little theorem states that if p is a prime number, then for any integer a, the number a p - a is an integer multiple of p. Here p is a prime number ap ≡ a (mod p).Special Case: If a is not divisible by p, Fermat's little theorem is equivalent to the statement that a p-1-1 is an integer mult
    8 min read
    Introduction to Primality Test and School Method
    Given a positive integer, check if the number is prime or not. A prime is a natural number greater than 1 that has no positive divisors other than 1 and itself. Examples of the first few prime numbers are {2, 3, 5, ...}Examples : Input: n = 11Output: trueInput: n = 15Output: falseInput: n = 1Output:
    10 min read
    Vantieghems Theorem for Primality Test
    Vantieghems Theorem is a necessary and sufficient condition for a number to be prime. It states that for a natural number n to be prime, the product of 2^i - 1 where 0 < i < n , is congruent to n~(mod~(2^n - 1)) . In other words, a number n is prime if and only if.{\displaystyle \prod _{1\leq
    4 min read
    AKS Primality Test
    There are several primality test available to check whether the number is prime or not like Fermat's Theorem, Miller-Rabin Primality test and alot more. But problem with all of them is that they all are probabilistic in nature. So, here comes one another method i.e AKS primality test (Agrawal–Kayal–
    11 min read
    Lucas Primality Test
    A number p greater than one is prime if and only if the only divisors of p are 1 and p. First few prime numbers are 2, 3, 5, 7, 11, 13, ...The Lucas test is a primality test for a natural number n, it can test primality of any kind of number.It follows from Fermat’s Little Theorem: If p is prime and
    12 min read

    Prime Factorization & Divisors

    Print all prime factors of a given number
    Given a number n, the task is to find all prime factors of n.Examples:Input: n = 24Output: 2 2 2 3Explanation: The prime factorization of 24 is 23×3.Input: n = 13195Output: 5 7 13 29Explanation: The prime factorization of 13195 is 5×7×13×29.Approach:Every composite number has at least one prime fact
    6 min read
    Smith Number
    Given a number n, the task is to find out whether this number is smith or not. A Smith Number is a composite number whose sum of digits is equal to the sum of digits in its prime factorization. Examples: Input : n = 4Output : YesPrime factorization = 2, 2 and 2 + 2 = 4Therefore, 4 is a smith numberI
    15 min read
    Sphenic Number
    A Sphenic Number is a positive integer n which is a product of exactly three distinct primes. The first few sphenic numbers are 30, 42, 66, 70, 78, 102, 105, 110, 114, ... Given a number n, determine whether it is a Sphenic Number or not. Examples: Input: 30Output : YesExplanation: 30 is the smalles
    8 min read
    Hoax Number
    Given a number 'n', check whether it is a hoax number or not. A Hoax Number is defined as a composite number, whose sum of digits is equal to the sum of digits of its distinct prime factors. It may be noted here that, 1 is not considered a prime number, hence it is not included in the sum of digits
    13 min read
    k-th prime factor of a given number
    Given two numbers n and k, print k-th prime factor among all prime factors of n. For example, if the input number is 15 and k is 2, then output should be "5". And if the k is 3, then output should be "-1" (there are less than k prime factors). Examples: Input : n = 225, k = 2 Output : 3 Prime factor
    15+ min read
    Pollard's Rho Algorithm for Prime Factorization
    Given a positive integer n, and that it is composite, find a divisor of it.Example:Input: n = 12;Output: 2 [OR 3 OR 4]Input: n = 187;Output: 11 [OR 17]Brute approach: Test all integers less than n until a divisor is found. Improvisation: Test all integers less than ?nA large enough number will still
    14 min read
    Finding power of prime number p in n!
    Given a number 'n' and a prime number 'p'. We need to find out the power of 'p' in the prime factorization of n!Examples: Input : n = 4, p = 2 Output : 3 Power of 2 in the prime factorization of 2 in 4! = 24 is 3 Input : n = 24, p = 2 Output : 22 Naive approach The naive approach is to find the powe
    8 min read
    Find all factors of a Positive Number
    Given a positive integer n, find all the distinct divisors of n.Examples:Input: n = 10 Output: [1, 2, 5, 10]Explanation: 1, 2, 5 and 10 are the divisors of 10. Input: n = 100Output: [1, 2, 4, 5, 10, 20, 25, 50, 100]Explanation: 1, 2, 4, 5, 10, 20, 25, 50 and 100 are divisors of 100.Table of Content[
    7 min read
    Find numbers with n-divisors in a given range
    Given three integers a, b, n .Your task is to print number of numbers between a and b including them also which have n-divisors. A number is called n-divisor if it has total n divisors including 1 and itself. Examples: Input : a = 1, b = 7, n = 2 Output : 4 There are four numbers with 2 divisors in
    14 min read

    Modular Arithmetic

    Modular Exponentiation (Power in Modular Arithmetic)
    Given three integers x, n, and M, compute (x^n) % M (remainder when x raised to the power n is divided by M).Examples : Input: x = 3, n = 2, M = 4Output: 1Explanation: 32 % 4 = 9 % 4 = 1.Input: x = 2, n = 6, M = 10Output: 4Explanation: 26 % 10 = 64 % 10 = 4.Table of Content[Naive Approach] Repeated
    6 min read
    Modular multiplicative inverse
    Given two integers A and M, find the modular multiplicative inverse of A under modulo M.The modular multiplicative inverse is an integer X such that:A X ≡ 1 (mod M) Note: The value of X should be in the range {1, 2, ... M-1}, i.e., in the range of integer modulo M. ( Note that X cannot be 0 as A*0 m
    15+ min read
    Modular Division
    Given three positive integers a, b, and M, the objective is to find (a/b) % M i.e., find the value of (a × b-1 ) % M, where b-1 is the modular inverse of b modulo M.Examples: Input: a = 10, b = 2, M = 13Output: 5Explanation: The modular inverse of 2 modulo 13 is 7, so (10 / 2) % 13 = (10 × 7) % 13 =
    13 min read
    Euler's criterion (Check if square root under modulo p exists)
    Given a number 'n' and a prime p, find if square root of n under modulo p exists or not. A number x is square root of n under modulo p if (x*x)%p = n%p. Examples : Input: n = 2, p = 5 Output: false There doesn't exist a number x such that (x*x)%5 is 2 Input: n = 2, p = 7 Output: true There exists a
    11 min read
    Find sum of modulo K of first N natural number
    Given two integer N ans K, the task is to find sum of modulo K of first N natural numbers i.e 1%K + 2%K + ..... + N%K. Examples : Input : N = 10 and K = 2. Output : 5 Sum = 1%2 + 2%2 + 3%2 + 4%2 + 5%2 + 6%2 + 7%2 + 8%2 + 9%2 + 10%2 = 1 + 0 + 1 + 0 + 1 + 0 + 1 + 0 + 1 + 0 = 5.Recommended PracticeReve
    9 min read
    How to compute mod of a big number?
    Given a big number 'num' represented as string and an integer x, find value of "num % a" or "num mod a". Output is expected as an integer. Examples : Input: num = "12316767678678", a = 10 Output: num (mod a) ? 8 The idea is to process all digits one by one and use the property that xy (mod a) ? ((x
    4 min read
    Exponential Squaring (Fast Modulo Multiplication)
    Given two numbers base and exp, we need to compute baseexp under Modulo 10^9+7 Examples: Input : base = 2, exp = 2Output : 4Input : base = 5, exp = 100000Output : 754573817In competitions, for calculating large powers of a number we are given a modulus value(a large prime number) because as the valu
    12 min read
    Trick for modular division ( (x1 * x2 .... xn) / b ) mod (m)
    Given integers x1, x2, x3......xn, b, and m, we are supposed to find the result of ((x1*x2....xn)/b)mod(m). Example 1: Suppose that we are required to find (55C5)%(1000000007) i.e ((55*54*53*52*51)/120)%1000000007 Naive Method :  Simply calculate the product (55*54*53*52*51)= say x,Divide x by 120 a
    9 min read
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