You are given an n-digit large number, you have to check whether it is divisible by 999 without dividing or finding modulo of number by 999.
Examples:
Input : 235764 Output : Yes Input : 23576 Output : No
Since input number may be very large, we cannot use n % 999 to check if a number is divisible by 999 or not, especially in languages like C/C++. The idea is based on following fact.
The solutions is based on below fact.
A number is divisible by 999 if sum of its 3-digit-groups (if required groups are formed by appending a 0s at the beginning) is divisible by 999.
Illustration:
Input : 235764
Output : Yes
Explanation : Step I - read input : 235, 764
Step II - 235 + 764 = 999
As result is 999 then we can
conclude that it is divisible by 999.
Input : 1244633121
Output : Yes
Explanation : Step I - read input : 1, 244, 633, 121
Step II - 001 + 244 + 633 + 121 = 999
As result is 999 then we can conclude
that it is divisible by 999.
Input : 999999999
Output : Yes
Explanation : Step I - read input : 999, 999, 999
Step II - 999 + 999 + 999 = 2997
Step III - 997 + 002 = 999
As result is 999 then we can conclude
that it is divisible by 999.
How does this work?
Let us consider 235764, we can write it as
235764 = 2*105 + 3*104 + 5*103 +
7*102 + 6*10 + 4
The idea is based on below observation:
Remainder of 103 divided by 999 is 1
For i > 3, 10i % 999 = 10i-3 % 999
Let us see how we use above fact.
Remainder of 2*105 + 3*104 + 5*103 +
7*102 + 6*10 + 4
Remainder with 999 can be written as :
2*100 + 3*10 + 5*1 + 7*100 + 6*10 + 4
The above expression is basically sum of
groups of size 3.
Since the sum is divisible by 999, answer is yes.
A simple and efficient method is to take input in form of string (make its length in form of 3*m by adding 0 to left of number if required) and then you have to add the digits in blocks of three from right to left until it become a 3 digit number and if that result is 999 we can say that number is divisible by 999.
As in the case of “divisibility by 9” we check that sum of all digit is divisible by 9 or not, the same thing follows within the case of divisibility by 999. We sum up all 3-digits group from right to left and check whether the final result is 999 or not.
C++
// CPP for divisibility of number by 999 #include<bits/stdc++.h> using namespace std; // function to check divisibility bool isDivisible999(string num) { int n = num.length(); if (n == 0 && num[0] == '0') return true; // Append required 0s at the beginning. if (n % 3 == 1) num = "00" + num; if (n % 3 == 2) num = "0" + num; // add digits in group of three in gSum int gSum = 0; for (int i = 0; i<n; i++) { // group saves 3-digit group int group = 0; group += (num[i++] - '0') * 100; group += (num[i++] - '0') * 10; group += num[i] - '0'; gSum += group; } // calculate result till 3 digit sum if (gSum > 1000) { num = to_string(gSum); n = num.length(); gSum = isDivisible999(num); } return (gSum == 999); } // driver program int main() { string num = "1998"; int n = num.length(); if (isDivisible999(num)) cout << "Divisible"; else cout << "Not divisible"; return 0; } |
Java
//Java for divisibility of number by 999 class Test { // Method to check divisibility static boolean isDivisible999(String num) { int n = num.length(); if (n == 0 && num.charAt(0) == '0') return true; // Append required 0s at the beginning. if (n % 3 == 1) num = "00" + num; if (n % 3 == 2) num = "0" + num; // add digits in group of three in gSum int gSum = 0; for (int i = 0; i<n; i++) { // group saves 3-digit group int group = 0; group += (num.charAt(i++) - '0') * 100; group += (num.charAt(i++) - '0') * 10; group += num.charAt(i) - '0'; gSum += group; } // calculate result till 3 digit sum if (gSum > 1000) { num = Integer.toString(gSum); n = num.length(); gSum = isDivisible999(num) ? 1 : 0; } return (gSum == 999); } // Driver method public static void main(String args[]) { String num = "1998"; System.out.println(isDivisible999(num) ? "Divisible" : "Not divisible"); } } |
Python 3
# Python3 program for divisibility # of number by 999 # function to check divisibility def isDivisible999(num): n = len(num); if(n == 0 or num[0] == '0'): return true # Append required 0s at the beginning. if((n % 3) == 1): num = "00" + num if((n % 3) == 2): num = "0" + num # add digits in group of three in gSum gSum = 0 for i in range(0, n, 3): # group saves 3-digit group group = 0 group += (ord(num[i]) - 48) * 100 group += (ord(num[i + 1]) - 48) * 10 group += (ord(num[i + 2]) - 48) gSum += group # calculate result till 3 digit sum if(gSum > 1000): num = str(gSum) n = len(num) gSum = isDivisible999(num) return (gSum == 999) # Driver code if __name__=="__main__": num = "1998" n = len(num) if(isDivisible999(num)): print("Divisible") else: print("Not divisible") # This code is contributed # by Sairahul Jella |
C#
// C# code for divisibility of number by 999 using System; class Test { // Method to check divisibility static bool isDivisible999(String num) { int n = num.Length; if (n == 0 && num[0] == '0') return true; // Append required 0s at the beginning. if (n % 3 == 1) num = "00" + num; if (n % 3 == 2) num = "0" + num; // add digits in group of three in gSum int gSum = 0; for (int i = 0; i<n; i++) { // group saves 3-digit group int group = 0; group += (num[i++] - '0') * 100; group += (num[i++] - '0') * 10; group += num[i] - '0'; gSum += group; } // calculate result till 3 digit sum if (gSum > 1000) { num = Convert.ToString(gSum); n = num.Length ; gSum = isDivisible999(num) ? 1 : 0; } return (gSum == 999); } // Driver method public static void Main() { String num = "1998"; Console.WriteLine(isDivisible999(num) ? "Divisible" : "Not divisible"); } // This code is contributed by Ryuga } |
PHP
<?php // PHP for divisibility of number by 999 // function to check divisibility function isDivisible999($num) { $n = strlen($num); if ($n == 0 && $num[0] == '0') return true; // Append required 0s at the beginning. if ($n % 3 == 1) $num = "00" . $num; if ($n % 3 == 2) $num = "0" . $num; // add digits in group of three in gSum $gSum = 0; for ($i = 0; $i < $n; $i += 3) { // group saves 3-digit group $group = 0; $group += (ord($num[$i]) - 48) * 100; $group += (ord($num[$i + 1]) - 48) * 10; $group += (ord($num[$i + 2]) - 48); $gSum += $group; } // calculate result till 3 digit sum if ($gSum > 1000) { $num = strval($gSum); $n = strlen($num); $gSum = isDivisible999($num); } return ($gSum == 999); } // Driver Code $num = "1998"; if (isDivisible999($num)) echo "Divisible"; else echo "Not divisible"; // This code is contributed by mits ?> |
Output:
Divisible
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