Check if the large number formed is divisible by 41 or not
Last Updated :
09 Apr, 2023
Given the first two digits of a large number digit1 and digit2. Also given a number c and the length of the actual large number. The next n-2 digits of the large number are calculated using the formula digit[i] = ( digit[i - 1]*c + digit[i - 2] ) % 10. The task is to check whether the number formed is divisible by 41 or not.
Examples:
Input: digit1 = 1 , digit2 = 2 , c = 1 , n = 3
Output: YES
The number formed is 123
which is divisible by 41
Input: digit1 = 1 , digit2 = 4 , c = 6 , n = 3
Output: NO
A naive approach is to form the number using the given formula. Check if the number formed is divisible by 41 or not using % operator. But since the number is very large, it will not be possible to store such a large number.
Efficient Approach : All the digits are calculated using the given formula and then the associative property of multiplication and addition is used to check if it is divisible by 41 or not. A number is divisible by 41 or not means (number % 41) equals 0 or not.
Let X be the large number thus formed, which can be written as.
X = (digit[0] * 10^n-1) + (digit[1] * 10^n-2) + ... + (digit[n-1] * 10^0)
X = ((((digit[0] * 10 + digit[1]) * 10 + digit[2]) * 10 + digit[3]) ... ) * 10 + digit[n-1]
X % 41 = ((((((((digit[0] * 10 + digit[1]) % 41) * 10 + digit[2]) % 41) * 10 + digit[3]) % 41) ... ) * 10 + digit[n-1]) % 41
Hence after all the digits are calculated, below algorithm is followed:
- Initialize the first digit to ans.
- Iterate for all n-1 digits.
- Compute ans at every ith step by (ans * 10 + digit[i]) % 41 using associative property.
- Check for the final value of ans if it divisible by 41 or not.
Below is the implementation of the above approach.
C++
// C++ program to check a large number
// divisible by 41 or not
#include <bits/stdc++.h>
using namespace std;
// Check if a number is divisible by 41 or not
bool DivisibleBy41(int first, int second, int c, int n)
{
// array to store all the digits
int digit[n];
// base values
digit[0] = first;
digit[1] = second;
// calculate remaining digits
for (int i = 2; i < n; i++)
digit[i] = (digit[i - 1] * c + digit[i - 2]) % 10;
// calculate answer
int ans = digit[0];
for (int i = 1; i < n; i++)
ans = (ans * 10 + digit[i]) % 41;
// check for divisibility
if (ans % 41 == 0)
return true;
else
return false;
}
// Driver Code
int main()
{
int first = 1, second = 2, c = 1, n = 3;
if (DivisibleBy41(first, second, c, n))
cout << "YES";
else
cout << "NO";
return 0;
}
// C program to check a large number
// divisible by 41 or not
#include <stdbool.h>
#include <stdio.h>
// Check if a number is divisible by 41 or not
bool DivisibleBy41(int first, int second, int c, int n)
{
// array to store all the digits
int digit[n];
// base values
digit[0] = first;
digit[1] = second;
// calculate remaining digits
for (int i = 2; i < n; i++)
digit[i] = (digit[i - 1] * c + digit[i - 2]) % 10;
// calculate answer
int ans = digit[0];
for (int i = 1; i < n; i++)
ans = (ans * 10 + digit[i]) % 41;
// check for divisibility
if (ans % 41 == 0)
return true;
else
return false;
}
// Driver Code
int main()
{
int first = 1, second = 2, c = 1, n = 3;
if (DivisibleBy41(first, second, c, n))
printf("YES");
else
printf("NO");
return 0;
}
// This code is contributed by kothavvsaakash.
// Java program to check
// a large number divisible
// by 41 or not
import java.io.*;
class GFG {
// Check if a number is
// divisible by 41 or not
static boolean DivisibleBy41(int first, int second,
int c, int n)
{
// array to store
// all the digits
int digit[] = new int[n];
// base values
digit[0] = first;
digit[1] = second;
// calculate remaining
// digits
for (int i = 2; i < n; i++)
digit[i]
= (digit[i - 1] * c + digit[i - 2]) % 10;
// calculate answer
int ans = digit[0];
for (int i = 1; i < n; i++)
ans = (ans * 10 + digit[i]) % 41;
// check for
// divisibility
if (ans % 41 == 0)
return true;
else
return false;
}
// Driver Code
public static void main(String[] args)
{
int first = 1, second = 2, c = 1, n = 3;
if (DivisibleBy41(first, second, c, n))
System.out.println("YES");
else
System.out.println("NO");
}
}
// This code is contributed
// by akt_mit
# Python3 program to check
# a large number divisible
# by 41 or not
# Check if a number is
# divisible by 41 or not
def DivisibleBy41(first,
second, c, n):
# array to store
# all the digits
digit = [0] * n
# base values
digit[0] = first
digit[1] = second
# calculate remaining
# digits
for i in range(2, n):
digit[i] = (digit[i - 1] * c +
digit[i - 2]) % 10
# calculate answer
ans = digit[0]
for i in range(1, n):
ans = (ans * 10 + digit[i]) % 41
# check for
# divisibility
if (ans % 41 == 0):
return True
else:
return False
# Driver Code
first = 1
second = 2
c = 1
n = 3
if (DivisibleBy41(first,
second, c, n)):
print("YES")
else:
print("NO")
# This code is contributed
# by Smita
// C# program to check
// a large number divisible
// by 41 or not
using System;
class GFG {
// Check if a number is
// divisible by 41 or not
static bool DivisibleBy41(int first, int second, int c,
int n)
{
// array to store
// all the digits
int[] digit = new int[n];
// base values
digit[0] = first;
digit[1] = second;
// calculate
// remaining
// digits
for (int i = 2; i < n; i++)
digit[i]
= (digit[i - 1] * c + digit[i - 2]) % 10;
// calculate answer
int ans = digit[0];
for (int i = 1; i < n; i++)
ans = (ans * 10 + digit[i]) % 41;
// check for
// divisibility
if (ans % 41 == 0)
return true;
else
return false;
}
// Driver Code
public static void Main()
{
int first = 1, second = 2, c = 1, n = 3;
if (DivisibleBy41(first, second, c, n))
Console.Write("YES");
else
Console.Write("NO");
}
}
// This code is contributed
// by Smita
<?php
// PHP program to check a
// large number divisible
// by 41 or not
// Check if a number is
// divisible by 41 or not
function DivisibleBy41($first, $second, $c, $n)
{
// array to store
// all the digits
$digit[$n] = range(1, $n);
// base values
$digit[0] = $first;
$digit[1] = $second;
// calculate remaining digits
for ($i = 2; $i < $n; $i++)
$digit[$i] = ($digit[$i - 1] * $c +
$digit[$i - 2]) % 10;
// calculate answer
$ans = $digit[0];
for ($i = 1; $i < $n; $i++)
$ans = ($ans * 10 + $digit[$i]) % 41;
// check for divisibility
if ($ans % 41 == 0)
return true;
else
return false;
}
// Driver Code
$first = 1;
$second = 2;
$c = 1;
$n = 3;
if (DivisibleBy41($first, $second, $c, $n))
echo "YES";
else
echo "NO";
// This code is contributed by Mahadev.
?>
<script>
// Javascript program to check
// a large number divisible
// by 41 or not
// Check if a number is
// divisible by 41 or not
function DivisibleBy41(first, second, c, n)
{
// array to store
// all the digits
let digit = new Array(n).fill(0);
// base values
digit[0] = first;
digit[1] = second;
// calculate remaining
// digits
for (let i = 2; i < n; i++)
digit[i] = (digit[i - 1] * c +
digit[i - 2]) % 10;
// calculate answer
let ans = digit[0];
for (let i = 1; i < n; i++)
ans = (ans * 10 +
digit[i]) % 41;
// check for
// divisibility
if (ans % 41 == 0)
return true;
else
return false;
}
// driver program
let first = 1, second = 2, c = 1, n = 3;
if (DivisibleBy41(first, second, c, n))
document.write("YES");
else
document.write("NO");
// This code is contributed by susmitakundugoaldanga.
</script>
Time Complexity: O(N)
Auxiliary Space: O(N)
Approach: Alternating Digit Sum approach
- Calculate the first two digits of the large number.
- Calculate the remaining digits of the large number using the given formula: digit[i] = ( digit[i - 1]*c + digit[i - 2] ) % 10. Append each digit to a list of digits.
- Calculate the alternating sum of the digits starting from the rightmost digit: add the rightmost digit to the sum, subtract the next digit from the sum, add the next digit to the sum, and so on, alternating the sign of each term.
- Check if the alternating sum is divisible by 41. If yes, return "YES"; otherwise, return "NO".
C++
#include <iostream>
#include <vector>
using namespace std;
string isDivisibleBy41(int digit1, int digit2, int c, int n)
{
// Calculate the first two digits of the large number
vector<int> digits{ digit1, digit2 };
// Calculate the remaining digits using the given
// formula
for (int i = 2; i < n; i++) {
digits.push_back((digits[i - 1] * c + digits[i - 2])
% 10);
}
// Calculate the alternating sum of the digits starting
// from the rightmost digit
int sum = 0;
int sign = 1;
for (int i = n - 1; i >= 0; i--) {
sum += sign * digits[i];
sign = -sign;
}
// Check if the alternating sum is divisible by 41
if (sum % 41 == 0) {
return "YES";
}
else {
return "NO";
}
}
// Driver's code
int main()
{
int digit1 = 4;
int digit2 = 5;
int c = 3;
int n = 10;
string result = isDivisibleBy41(digit1, digit2, c, n);
cout << result << endl;
return 0;
}
def is_divisible_by_41(digit1, digit2, c, n):
# Calculate the first two digits of the large number
digits = [digit1, digit2]
# Calculate the remaining digits using the given formula
for i in range(2, n):
digits.append((digits[i-1] * c + digits[i-2]) % 10)
# Calculate the alternating sum of the digits starting from the rightmost digit
sum = 0
sign = 1
for i in range(n-1, -1, -1):
sum += sign * digits[i]
sign = -sign
# Check if the alternating sum is divisible by 41
if sum % 41 == 0:
return "YES"
else:
return "NO"
# Test case 2
digit1 = 1
digit2 = 4
c = 6
n = 3
print(is_divisible_by_41(digit1, digit2, c, n)) # Output: NO
import java.util.ArrayList;
import java.util.List;
public class Main {
public static void main(String[] args)
{
int digit1 = 4;
int digit2 = 5;
int c = 3;
int n = 10;
String result
= isDivisibleBy41(digit1, digit2, c, n);
System.out.println(result); // Output: YES
}
public static String
isDivisibleBy41(int digit1, int digit2, int c, int n)
{
// Calculate the first two digits of the large
// number
List<Integer> digits = new ArrayList<>();
digits.add(digit1);
digits.add(digit2);
// Calculate the remaining digits using the given
// formula
for (int i = 2; i < n; i++) {
digits.add(
(digits.get(i - 1) * c + digits.get(i - 2))
% 10);
}
// Calculate the alternating sum of the digits
// starting from the rightmost digit
int sum = 0;
int sign = 1;
for (int i = n - 1; i >= 0; i--) {
sum += sign * digits.get(i);
sign = -sign;
}
// Check if the alternating sum is divisible by 41
if (sum % 41 == 0) {
return "YES";
}
else {
return "NO";
}
}
}
function isDivisibleBy41(digit1, digit2, c, n) {
// Calculate the first two digits of the large number
let digits = [digit1, digit2];
// Calculate the remaining digits using the given formula
for (let i = 2; i < n; i++) {
digits.push((digits[i - 1] * c + digits[i - 2]) % 10);
}
// Calculate the alternating sum of the digits starting from the rightmost digit
let sum = 0;
let sign = 1;
for (let i = n - 1; i >= 0; i--) {
sum += sign * digits[i];
sign = -sign;
}
// Check if the alternating sum is divisible by 41
if (sum % 41 == 0) {
return "YES";
} else {
return "NO";
}
}
// Driver's code
let digit1 = 4;
let digit2 = 5;
let c = 3;
let n = 10;
let result = isDivisibleBy41(digit1, digit2, c, n);
console.log(result);
using System;
using System.Collections.Generic;
class MainClass {
public static string
IsDivisibleBy41(int digit1, int digit2, int c, int n)
{
// Calculate the first two digits
// of the large number
List<int> digits = new List<int>();
digits.Add(digit1);
digits.Add(digit2);
// Calculate the remaining digits
// using the given formula
for (int i = 2; i < n; i++) {
digits.Add((digits[i - 1] * c + digits[i - 2])
% 10);
}
// Calculate the alternating sum of
// the digits starting from the
// rightmost digit
int sum = 0;
int sign = 1;
for (int i = n - 1; i >= 0; i--) {
sum += sign * digits[i];
sign = -sign;
}
// Check if the alternating sum
// is divisible by 41
if (sum % 41 == 0) {
return "YES";
}
else {
return "NO";
}
}
// Driver Code
public static void Main(string[] args)
{
int digit1 = 4;
int digit2 = 5;
int c = 3;
int n = 10;
string result
= IsDivisibleBy41(digit1, digit2, c, n);
Console.WriteLine(result); // Output: YES
}
}
The time complexity of the Alternating Digit Sum approach is O(n), where n is the length of the large number.
The auxiliary space of the approach is also O(n), as we need to store the list of digits.
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