To check divisibility of any large number by 999

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.

Implementation:
 

Divisible

Time complexity : O(n) 
Auxiliary Space : O(1)

Method 2: Checking given any number is divisible by 999 or not by using modulo division operator "%".  

Implementation:

Yes

Time Complexity : O(1)

Auxiliary Space: O(1)

Method 3: The function returns a Boolean value indicating whether the input integer is divisible by 999 or not. The function calculates this by checking if the sum of the digits of num is divisible by 9 and if the last three digits of num are divisible by 27. This code then tests the function with the examples: 999, 998 and 9987 and outputs the values True, False and False respectively.

True
False
False

Time complexity: O(1),
Auxiliary Space : O(1) 

Method 4: Using string manipulation 

1. Convert the number to a string.
2. Find the sum of every third digit from the right end to the left end.
3. If the sum is divisible by 3, then the original number is divisible by 999.

False
True

The time complexity of the given function is_divisible_by_999 is O(N/3) where N is the number of digits in the input number n. This is because the loop iterates over every third digit of the input number, and performs constant time operations (such as string slicing and integer conversion) on each iteration.

The auxiliary space used by the function is O(1), since it uses a constant amount of extra space to store the loop variables and the running sum.

More Divisibility Algorithms.