1 to 9 Divisibility

Firstly we can ignore 1, as every integer is divisible by 1, then we have: $\begin{aligned} 2 & = & 2 \\ 3 & = & 3 \\ 4 & = & 2\times 2 \\ 5 & = & 5 \\ 6 & = & 2\times 3 \\ 7 & = & 7 \\ 8 & = & 2\times 2\times 2 \\ 9 & = & 3\times 3 \end{aligned}$ Now, we can start cancelling numbers, if they're including already. So let's begin with the largest number, 9. As $9=3\times 3$ , we can eliminate 3. Now the next biggest is 8. So $9\times 8 = 3^2 \times 2^3$ Now we can eliminate 2, 4 and 6 as they are factors of $9\times 8$ . Now, as 5 and 7 are prime, they are unique.

This means that overall we have: $2^3\times 3^2\times 5\times 7 = \boxed {2520}$

I guess its simple as it looks...just do the LCM of these nos. to get the correct answer

8 7 9 5, because 8 = 2 4, 8 9 = 2 3 3 4, till here it contains all the factors minimum times and no factors of 5 and 7 are there , so we multiply with 5 and 7.