Check if a large number is divisible by 9 or not
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.
Attention reader! Don’t stop learning now. Get hold of all the important mathematical concepts for competitive programming with the Essential Maths for CP Course at a student-friendly price. To complete your preparation from learning a language to DS Algo and many more, please refer Complete Interview Preparation Course.
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.
This article is contributed by DANISH_RAZA . If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to email@example.com. See your article appearing on the GeeksforGeeks main page and help other Geeks.
Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.