Editorial

Save Money

• Bruteforce approach
  • You can simply run a loop from 1 to n and add all ODD numbers. Then print the sum.
  • The time complexity will be O(n). But this will give you TLE.
• Optimized Approach
  • If you observe you will notice a pattern.
  • The sum of first x ODD numbers is x².
    • Example: 5 is the 3^rd ODD number (Before that we have 1 and 3).So here x = 3.The sum of 1+3+5 = 9 which is equal to 3² or x².
  • So you just have to identify which ODD number it is and square that to get the answer.
    • (7 is 4^th ODD number so square of 4 will give me the sum of all ODD numbers from 1 to 7, 9 is 5^th ODD number so square of 5 will give me the sum of all ODD numbers from 1 to 9)
• Solution

Evenly Divisible

• If we want to get a number which will be divisible by all the given N numbers then we just have to calculate the LCM (Least Common Multiple) (লসাগু) of those N numbers.

• Solution