How can I find the quotient of the LCMs of the first 116 and first 113 positive integers? I'm trying to solve this problem: What is the quotient if we divide the LCM of the first 116 positive integers by the LCM of the first 113 positive integers? 
Actually I thought if approaching it by going through the numbers one by one to find the LCMs, but it makes the problem too  large. Will you please help me with a simpler approach?
 A: I take "natural number" in your question to mean "positive integer" (which is not the standard usage in mathematics), since the LCM of any set containing $0$ is $0$, and the quotient of $0$ by $0$ is undefined.  So you want:
$$ \frac{\operatorname{LCM}\{1,\dots,116\}}{\operatorname{LCM}\{1,\dots,113\}} $$
Let $X$ denote the denominator.  Then the numerator is $\operatorname{LCM}\{1,\dots,116\} = \operatorname{LCM}\{X,114,115,116\}$.  Since $X$ is divisible by every number less than or equal to $113$, and most numbers are products of small numbers, one should expect that $\operatorname{LCM}\{X,114,115,116\}$ is not much more than $X$.  Let's check this:
$$114 = 2\cdot 3\cdot 19, \quad 115 = 5\cdot 23, \quad 116 = 2^2 \cdot 29$$
Note that $X$ is divisible by the prime powers $3,4,5,19$ and $29$ since all of these are less than $113$.  It is therefore divisible by any product of these, and in particular by all of $114$, $115$, and $116$.  This proves that:
$$ \operatorname{LCM}\{X,114,115,116\} = X $$
and so the ratio is $\frac XX = 1$.
A: I find the answer to be 1. Since the LCM of the first 113 natural numbers is also a multiple of the remaining three: 114, 115, and 116 (why ?)
