Looking for the lowest number divisible by 1 to A. What would the math equation be for finding the lowest number divisible by 1 to A? I know factorial can make numbers divisible by 1 to A but that dosn't give me the lowest number.
Example of what I'm talking about:
the lowest number divisible by 1,2,3,4,5,6,7,8 = 840
Example of factorial (What I'm not talking about):
8! = 40,320
Note:
A is anything that's more then 1 and is whole.
 A: The problem with taking the factorial is that it repeats prime powers too many times: For example, $8!$ contains $6$ powers of $2$, while $8$ is only divisble by $2^3$. 
You're looking for the number that's divisible by all the primes at most $A$, and powers of primes that are still less than $A$; to this end, let $P_A = \{p_1, p_2,\dots p_n\}$ be the collection of primes in $\{1, \dots, A\}$. Then for primes $p$, 
$$p^k \le A \iff k \le \frac{\log A}{\log p}$$
Thus, the desired number is
$$\operatorname{lcm}(1,2,\dots,A) = \prod_{k = 1}^n p_k^{\lfloor\log A / \log p_k\rfloor}$$
where $\lfloor \cdot \rfloor$ denotes the floor function.
As an example, for $A = 8$, we have $P_A = \{2,3,5,7\}$. Then the exponent for $2$ is
$$\left\lfloor \frac{\log 8}{\log 2} \right\rfloor = 3$$
as expected. Similarly, the exponent for $3$ is $\lfloor \log 8 / \log 3\rfloor = 1$, and likewise for $5$ and $7$. Thus the desired number is
$$2^3 \cdot 3 \cdot 5 \cdot 7  = 840$$
A: You can write it as $LCM(1,2,3,4,\dots A)$  To figure it out, you take the highest power of each prime less than $A$ and multiply them.  For $A=37$, we have $2^5,3^3,5^2$ and all other primes less than $37$, so it is $2^5\cdot 3^3\cdot 5^2\cdot 7 \cdot 11 \dots \cdot 37$  They are given in OEIS A003418
A: Look at the LCM. Note $lcm\{a,b\} = \frac{ab}{gcd(a, b)}$. 
If you have a prime $p$ and a power of $p$, $p^{n}$, you will simply discard $p$ and keep $p^{n}$ instead. As $p^{n}|x \implies p|x$, but the converse is not true.
Note as well you can ignore $1$ in your $gcd$ calculations.
