Simplest Solution [C++]|| Beats 100% || 1 line of code ✅✅
Approach\n \nVery simple approach -> If n is odd, the next possible multiple of n and 2 is obviously 2*n, otherwise if n is even, it is  already a multiple of n
# Approach\n \nVery simple approach -> If n is odd, the next possible multiple of n and 2 is obviously 2*n, otherwise if n is even, it is  already a multiple of n and 2. \n# Code\n```\nclass Solution {\npublic:\n    int smallestEvenMultiple(int n) {\n        \n        if(n%2==0)return n;\n            return 2*n;\n        \n    }\n};\n```
['C++']