The question rephrased:
How does one determine $m_k$, the smallest number which is divisible by all the positive integers from 1 to $k$?
As a product of primes
Well, if one starts with the product $\tilde m_k = 1 \cdot 2 \cdot \dots \cdot k$, this is divisble by $1, 2, \dots, k$, but it's not the smallest. For example $1 \cdot 2 \cdot 3 \cdot 4 = 24$, but the smallest is 12, since 4 is divisble by 2. If we remove the 2 we get the smallest number. This suggest a general procedure.
Take all the prime numbers $p_1, \dots, p_l$ from $\{1, 2, \dots, k\}$ and let $q_i$ be the largest integer such that $p_i^{q_i} \leq k$ and set
$m_k = p_1^{q_1} p_2^{q_2} \cdots p_l^{q_l}$
which should be the smallest number which is divisble by $1, \dots, k$.
Recursively
Calculate $m_k$, let $m_{k-1}$ be given.
- Is $m_{k-1}$ divisble by $k$? If so, set $m_k = m_{k-1}$.
- Otherwise, set $m_k = \operatorname{lcm}(m_{k-1}, k)$, where $\operatorname{lcm}$ is the least common multiple.
This gives a nice implementation for computer using the following formula:
$\operatorname{lcm}(a,b) = \frac{|a b|}{\operatorname{gcd}(a,b)}$
Least common multiple
Of course, you can just set:
$m_k = \operatorname{lcm}(2, 3, \dots, k)$.