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In this section we will see, whether a number can be represented as tree consecutive numbers or not. Suppose a number is 27. This can be represented as 8 + 9 + 10.

This can be solved in two different approach. The first approach is Naïve approach. In that approach, we have to check i + (i + 1) + (i + 2) is same as number or not. Another efficient approach is by checking whether the number is divisible by 3 or not. Suppose a number x can be represented by three consecutive 1s, then x = (y - 1) + y + (y + 1) = 3y. So the number must be divisible by 3.

#include <iostream> using namespace std; bool hasThreeNums(int n) { if(n % 3 == 0){ return true; } return false; } int main() { int num = 27; if(hasThreeNums(num)){ cout << "Can be represented"; }else{ cout << "Cannot be presented"; } }

Can be represented

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