Part (i)

There are many examples but n = 5 and n = 9 are the smallest, and can be found easily by working out a list of the values of L(n) for n = 1, 2, 3, …

For example,

L(1) = 1

L(2) = 2

L(3) = 2 x 3 = 6

L(4) = 2 x 2 x 3 = 12

L(5) = 2 x 2 x 3 x 5 = 60

As L(5) already has a factor of 6, L(6) = L(5).

Part (ii)

For L(n) to equal L(n+3) it is necessary to find a run of three integers that do not have any “new” prime numbers in their prime factorisation

For example if we try out L(n) for n =7, then none of L(n) for n = 8, 9 or 10 will use a “new” prime number, as their prime factors are 2, 3 and 5, all of which will form part of L(7).

However, when we look more closely, this condition is not sufficient, as L(8) requires an extra factor of 2 and L(9) requires an extra factor of 3.

L(7) = L(6) x 7 = 2 x 2 x 3 x 5 x 7 = 420

L(8) = L(8) x 2 = 2 x 2 x 3 x 5 x 7 x 2 = 840

L(9) = L(8) x 3 = 2 x 2 x 2 x 3 x 5 x 7 x 3 = 2520

L(10 ) = L(9) = 2520

This “extra” prime factor will be needed every time there is a power of a prime in the run of integers examined, so the original condition we considered is not sufficient to find a suitable n.

So now let’s look for a run of 3 integers with no “new” prime numbers and no powers of primes.

The smallest number with this property is n = 19, with the run of integers 20 (= 2 x 2 x 5), 21 (= 3 x 7) and 22 (=2 x 11) following.

The prime factorisations of these three integers are already contained in L(19), so the answer to the problem is n = 19.

(For reference, L(19) = 2 x 2 x 2 x 2 x 3 x 3 x 5 x 7 x 11 x 13 x 17 x 19)

Challenge 311: Large LCMs