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:


Now, using Lemma 3,


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)

  • $\begingroup$ 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 May 27 '17 at 16:17

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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.