We can ignore 1, since any counting number is divisible by 1. So our prime factor each of the counting numbers from 2 to 10 2 = 2 3 = 3 4 = 2*2 5 = 5 6 = 2*3 7 = 7 8 = 2*2*2 9 = 3*3

10 = 2*5

The LCM of all those must have as many factors of each prime that appears in any factorization 2 appears at most 3 times as a factor of 8 3 appears at most 2 times as a factor of 9 5 appears at most 1 time as a factor if 5 and 10 7 appears at most 1 time as a factor of 7

So the LCM has 3 factors of 2, 2 factors of 3, and 1 facor each of 5 and 7