4
$\begingroup$

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)

$\endgroup$
1
  • $\begingroup$ Looks good to me. The only minor thing is that most people prefer to use \setminus ($A\setminus B$) to - ($A-B$) for the set difference in $\LaTeX$. $\endgroup$
    – fedja
    Commented May 27, 2017 at 16:17

1 Answer 1

1
$\begingroup$

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

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .