Check if a large number is divisible by 4 or not

Given a number, the task is to check if a number is divisible by 4 or not. The input number may be large and it may not be possible to store even if we use long long int.

Examples:

Input : n = 1124
Output : Yes

Input : n = 1234567589333862
Output : No

Input : n = 363588395960667043875487
Output : No

Using the modulo division operator "%"

This approach directly checks divisibility by 4 using the modulo operator (%).

• Compute n % 4.
• If the remainder is 0, the number is divisible by 4; otherwise, it is not.

No

Time Complexity - O(1)
Auxiliary Space - O(1)

Checking Divisibility of the Last 2 Digits

Since input number may be very large, we cannot use n % 4 to check if a number is divisible by 4 or not, especially in languages like C/C++. The idea is based on following fact.

A number is divisible by 4 if number formed by last two digits of it is divisible by 4. For example, let us consider 76952. Number formed by last two digits = 52. Since 52 is divisible by 4, answer is YES.

How does this work? Let us consider 76952, we can write it as 76952 = 7*10000 + 6*1000 + 9*100 + 5*10 + 2
The proof is based on below observation:
Remainder of 10ⁱ divided by 4 is 0 if i greater than or equal to two. Note than 100, 1000, .. etc lead to remainder 0 when divided by 4. So remainder of "7*10000 + 6*1000 + 9*100 + 5*10 + 2" divided by 4 is equivalent to remainder
of following :
0 + 0 + 0 + 5*10 + 2 = 52
Therefore we can say that the whole number is divisible by 4 if 52 is divisible by 4.

Yes

Time Complexity - O(1)
Auxiliary Space - O(1)

Alternate Implementation - Substring to Integer Conversion

• Use substring function to get the last two characters of the string.
• Convert the string to integer
• Check if it is divisible by 4 or not, using (number%4 == 0).

Yes

Time Complexity - O(1)
Auxiliary Space - O(1)