Given a positive integer N > 1, the task is to find the maximum LCM among all the pairs (i, j) such that i < j ≤ N.
Input: N = 3
LCM(1, 2) = 2
LCM(1, 3) = 3
LCM(2, 3) = 6
Input: N = 4
Approach: Since the LCM of two consecutive elements is equal to their multiples then it is obvious that the maximum LCM will be of the pair (N, N – 1) i.e. (N * (N – 1)).
Below is the implementation of the above approach:
Time Complexity: O(1)
- Number of pairs from the first N natural numbers whose sum is divisible by K
- Count pairs of natural numbers with GCD equal to given number
- Find the permutation of first N natural numbers such that sum of i % Pi is maximum possible
- Find maximum N such that the sum of square of first N natural numbers is not more than X
- Sum of first n natural numbers
- LCM of First n Natural Numbers
- Natural Numbers
- Sum of fifth powers of the first n natural numbers
- Sum of cubes of first n odd natural numbers
- Sum of first N natural numbers which are divisible by 2 and 7
- Sum of all natural numbers in range L to R
- Sum of all odd natural numbers in range L and R
- Sum of first N natural numbers which are not powers of K
- Sum of squares of first n natural numbers
- Average of first n even natural numbers
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