Mind-boggling Factors

Relevant wiki: Lowest Common Multiple

This problem is to find the lowest common multiplier of 1, 2, 3, 4, 5, 6, 7, 8 and 9: $\text{lcm } (1, 2, 3, 4, {\color{#624F41}5}, 6, {\color{#20A900}7}, {\color{#3D99F6}8}, {\color{#D61F06}9}) = {\color{#3D99F6}2^3} \times {\color{#D61F06}3^2} \times {\color{#624F41}5} \times {\color{#20A900}7} = \boxed{2520}$ .

1. To do this it’s useful to think of the prime factorisations of all the numbers from 2 to 9. These are listed below. Number Prime factorisation Primes in the prime factorisation, with repeats 2 2 2 3 3 3 4 2 x 2 2, 2 5 5 5 6 2 x 3 2, 3 7 7 7 8 2 x 2 x 2 2, 2, 2 9 3 x 3 3, 3
2. Now think about the prime factorisation of the smallest number which is a multiple of all the numbers from 1 to 9. It needs to include all the primes which are listed in the right hand column above. The number of times we need each prime is given by the highest number of occurences of it in the right hand column above. The primes appearing are 2, 3, 5, 7. The highest number of occurences of 2 is 3 (in 8 = 2 x 2 x 2). If each prime appears that many times we will have the smallest number that is divisible by all of the above.
3. This gives us the number 2 x 2 x 2 x 3 x 3 x 5 x 7 = 2520. You can see from the prime factorisation that it is divisible by all the numbers from 1 to 9. Number Appearance as a divisor 2 2 x 2 x 2 x 3 x 3 x 5 x 7 3 2 x 2 x 2 x 3 x 3 x 5 x 7 4 2 x 2 x 2 x 3 x 3 x 5 x 7 5 2 x 2 x 2 x 3 x 3 x 5 x 7 6 2 x 2 x 2 x 3 x 3 x 5 x 7 7 2 x 2 x 2 x 3 x 3 x 5 x 7 8 2 x 2 x 2 x 3 x 3 x 5 x 7 9 2 x 2 x 2 x 3 x 3 x 5 x 7