We want the least common multiple LCM of those integers,

often written LCM(1,2,3,4,5,6,7,8,9,10)

Every integer is divisible by 1, so we don't need to consider it.
We break into primes all the integers 2 through 10

2 prime
3 prime
4 = 2×2
5 prime
6 = 2×3
7 prime
8 = 2×2×2
9 = 3×3
10 = 2×5 

It must have every one of those primes as a factor as many
times as it is a factor of ANY one of those above.

We go through the primes to see how many of each we need for
the LCM to have:

2 must be included as a factor 3 times because the most number 
of times it occurs as a factor of any of those above is 3 times, 
since 8 contains factor 2 3 times.

So far the LCM must have factors 2×2×2. 

3 must be included as a factor 2 times because the most number 
of times it occurs as a factor of any of those above is 2 times, 
since 9 contains factor 3 2 times. 

So far the LCM must have factors 2×2×2×3×3.

5 must be included as a factor 1 time because the most number 
of times it occurs as a factor of any of those above is 1 time, 
once in 5, and once in 10. 

So far the LCM must have factors 2×2×2×3×3×5.

7 must be included as a factor 1 time because the most number 
of times it occurs as a factor of any of those above is 1 time, 
only once in 7.

That's the last prime, so the LCM must be the product of 
factors 2×2×2×3×3×5×7 = 2520.

Edwin