Question 505530
2520

We want  LCM(1,2,3,4,5,6,7,8,9,10)
We can ignore 1, since any counting number is divisible by 1.
We 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

LCM = 2*2*2*3*3*5*7 = 2520

Edwin</pre>