Given a number, the task is to find if the number is divisible by 9 or not. The input number may be large and it may not be possible to store even if we use long long int.
Input : n = 69354 Output : Yes Input : n = 234567876799333 Output : No Input : n = 3635883959606670431112222 Output : No
Since input number may be very large, we cannot use n % 9 to check if a number is divisible by 9 or not, especially in languages like C/C++. The idea is based on following fact.
A number is divisible by 9 if sum of its digits is divisible by 9.
For example n = 9432 Sum of digits = 9 + 4 + 3 + 2 = 18 Since sum is divisible by 9, answer is Yes.
How does this work?
Let us consider 1332, we can write it as 1332 = 1*1000 + 3*100 + 3*10 + 2 The proof is based on below observation: Remainder of 10i divided by 9 is 1 So powers of 10 only results in remainder 1 when divided by 9. Remainder of "1*1000 + 3*100 + 3*10 + 2" divided by 9 can be written as : 1*1 + 3*1 + 3*1 + 2 = 9 The above expression is basically sum of all digits. Since 9 is divisible by 9, answer is yes.
Below is the implementation of above idea.
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