Evaluate $x$ if:
$$x\cdot\operatorname{lcm}{(102\ldots 200)}=\operatorname{lcm}{(1,2,\ldots 200)}$$
Here's what I have so far,
LEMMA 1: In any set of $n$ consecutive positive integers, there must be atleast one number divisible by $n$.
LEMMA 2:
$\operatorname{lcm}{(a_1,a_2\ldots)}=\operatorname{lcm}{(\operatorname{lcm}{(a_1,a_2)},a_3\ldots)}$
LEMMA 3:If $a\mid b$ then,
$\operatorname{lcm}{(a,b)}=b$.
Let
$$A=\{1,2 \ldots 200\}$$
$$B=\{102,103\ldots 200\}$$
Now, $B$ contains 99 integers.
So, there must be subsets of $B$ with $k$ consecutive integers for all $1\leq k\leq 99$.
So for each such $k$, using Lemma 1, there is a
$$l: k\mid l$$
Therefore, using Lemma 2:
$$\operatorname{lcm}{(A)}=\operatorname{lcm}{(A-\{k,l\},\operatorname{lcm}{(k,l)})}$$
Now, using Lemma 3,
$$\operatorname{lcm}{(A)}=\operatorname{lcm}{(A-\{k,l\},l)}=\operatorname{lcm}{(A-\{k\})}$$
So, doing this with all the $k$, we can conclude that,
$$\operatorname{lcm}{(1,2\ldots 200)}=\operatorname{lcm}{(100,101,102\ldots 200)}$$
Trivially, we can remove the 100 as 200 is divisible by it.
So, the original equation becomes:
$$ x=\dfrac{\operatorname{lcm}{(101,102\ldots 200)}}{\operatorname{lcm}{(102,103\ldots 200)}}$$
Thus, I conclude $\boxed{x=101}$.
Is this proof correct?
(Any proof writing tips are also appreciated. I have no experience writing number theoretic proofs)
Best Answer
This is correct. It is easier to note that all the numbers from $1$ to $100$ have a multiple in the range $102-200$ so they don't contribute to the LCM