for any positive integer n find the lcmnn1Solutionn1n This i

for any positive integer n, find the lcm(n,n+1)

Solution

(n+1)×n.

This is so because for any two numbers of which one is not the multiple of other, their Least Common Multiple or the LCM has to be a number that is divisible by both, this implies that the LCM should have both the numbers as a factor.

Now, since we are talking about the \"Least\" common multiple, one factor of both the numbers is what we need at least for any third number to be called the common multiple of first two.

Thus for all n except 1, the above explanation holds( as for n=1, n+1 (=2) becomes a multiple of n (=1)) and the LCM is (n+1)×n.

For n=1, LCM of n (=1) and n+1 (=2) is 2 which coincidentally, is (n+1)×n.

Thus the general solution of this problem comes out to be-

(n+1)×n