Given a large number n, find if the number is divisible by 29.
Examples :
Input : 363927598 Output : No Input : 292929002929 Output : Yes
A quick solution to check if a number is divisible by 29 or not is to add 3 times of last digit to rest number and repeat this process until number comes 2 digit. The given number is divisible by 29 if the obtained two digit number is divisible by 29.
Number is 348,
Three times of last digit + Rest of the number = 8*3 + 34 = 58
Since 58 is divisible by 29, 348 is also divisible by 29.
C++
// CPP program to demonstrate above method// to check divisibility by 29.#include <iostream>using namespace std;// Returns true if n is divisible by 29// else returns false.bool isDivisible(long long int n){ // add the lastdigit*3 to renaming // number until number comes only // 2 digit while (n / 100) { int last_digit = n % 10; n /= 10; n += last_digit * 3; } // return true if number is // divisible by 29 another return (n % 29 == 0);}// Driver Codeint main(){ long long int n = 348; if (isDivisible(n)) cout << "Yes" << endl; else cout << "No" << endl; return 0;} |
Java
// Java program to demonstrate above method// to check divisibility by 29.import java.io.*;class GFG { // Returns true if n is divisible by 29 // else returns false. static boolean isDivisible(long n) { // add the lastdigit*3 to renaming // number until number comes only // 2 digit while (n / 100 > 0) { int last_digit = (int)n % 10; n /= 10; n += last_digit * 3; } // return true if number is // divisible by 29 another return (n % 29 == 0); } // Driver code public static void main(String[] args) { long n = 348; if (isDivisible(n)) System.out.println("Yes"); else System.out.println("No"); }}// This code is contributed by vt_m. |
Python3
# Python3 program to demonstrate above # method to check divisibility by 29.# Returns true if n is divisible # by 29 else returns false.def isDivisible(n): # add the lastdigit*3 to renaming # number until number comes only # 2 digit while (int(n / 100)) : last_digit = int(n % 10) n = int(n / 10) n += last_digit * 3 # return true if number is # divisible by 29 another return (n % 29 == 0)# Driver Coden = 348if(isDivisible(n) != 0): print("Yes")else: print("No")# This code is contributed by Smitha Dinesh Semwal. |
C#
// C# program to demonstrate above method// to check divisibility by 29.using System;class GFG { // Returns true if n is divisible by 29 // else returns false. static bool isDivisible(long n) { // add the lastdigit*3 to renaming // number until number comes only // 2 digit while (n / 100 > 0) { int last_digit = (int)n % 10; n /= 10; n += last_digit * 3; } // return true if number is // divisible by 29 another return (n % 29 == 0); } // Driver code public static void Main() { long n = 348; if (isDivisible(n)) Console.Write("Yes"); else Console.Write("No"); }}// This code is contributed by nitin mittal |
PHP
<?php// PHP program to demonstrate // above method to check // divisibility by 29.// Returns true if n is // divisible by 29// else returns false.function isDivisible($n){ // add the lastdigit*3 to // remaining number until // number becomes of only // 2 digit while (intval($n / 100)) { $last_digit = $n % 10; $n = intval($n / 10); $n += $last_digit * 3; } // return true if number is // divisible by 29 another return ($n % 29 == 0);}// Driver Code$n = 348;if (isDivisible($n)) echo "Yes";else echo "No" ;// This code is contributed by Sam007?> |
Javascript
<script>// Javascript program to demonstrate // above method to check // divisibility by 29.// Returns true if n is // divisible by 29// else returns false.function isDivisible(n){ // add the lastdigit*3 to // remaining number until // number becomes of only // 2 digit while (parseInt(n / 100)) { let last_digit = n % 10; n = parseInt(n / 10); n += last_digit * 3; } // return true if number is // divisible by 29 another return (n % 29 == 0);}// Driver Codelet n = 348;if (isDivisible(n)) document.write("Yes");else document.write("No") ;// This code is contributed by _saurabh_jaiswal</script> |
Output :
Yes
Time Complexity: O(n) where n is given number.
Space Complexity: O(1) as we are not using any extra space.
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