A set of numbers where none can be made by multiplying others in the set.

Given a finite list $(N_i)_{1\leq i\leq n}$ of natural numbers $\geq1$ one may ask for the smallest number $N$ that is a multiple of all $N_i$. This $N$ is called the least common multiple (LCM) of the $N_i$ and may be obtained as follows: Each $N_i$ has a unique prime factorization of the form $$N_i=p_1^{\alpha_{i1}}\ p_2^{\alpha_{i2}}\ p_3^{\alpha_{i3}}\cdot\ldots=\prod_{k\geq1} p_k^{\alpha_{ik}}\ ,$$ where $(p_1,p_2,p_3,\ldots)$ is the sequence of all primes, the $\alpha_{ik}$ are natural numbers $\geq0$, and for each $i$ only finitely many $\alpha_{ik}$ are $\ne0$.

Consider a fixed prime $p_k$. This prime appears in the $N_i$ with various exponents $\alpha_{ik}$, and there will be a largest among them: Let $$\rho_k:=\max\{\alpha_{ik}\>|\>1\leq i\leq n\}\ .$$ This means: Among the $N_i$ there is at least one that has $p_k^{\rho_k}$ as a factor, but none containing a higher power of $p_k$. Taking again the fundamental theorem of arithmetic for granted it follows that the LCM $N$ of the $N_i$ is given by $$N=\prod_{k\geq1} p_k^{\rho_k}\ .\tag{1}$$

Formula (1) is more of a theoretical nature. In practice the LCM is determined using the so-called euclidean algorithm; and primes don't even make an appearance then.

You're looking for the least common multiple of {1, 2, ..., 10}. If you generalize to lcm({1, 2, ..., n}) this is A003418, which has a great deal of information on this problem. (You may find the OEIS a useful resource on this sort of problem.)

Here's one way to compute these numbers. For all the primes from 2 to the target number $n$, take the largest power of the prime which is less than or equal to $n$, and take the product of all these prime powers.