Question

Compute LCM of (1,2,3,4........,200)/LCM of (102,103,104........,200).

Answer 1

Step-by-step explanation:

Evaluate x if:

x⋅lcm(102…200)=lcm(1,2,…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: lcm(a1,a2…)=lcm(lcm(a1,a2),a3…)

LEMMA 3:If a∣b then, lcm(a,b)=b.

Let

A={1,2…200}

B={102,103…200}

Now, B contains 99 integers.

So, there must be subsets of B with k consecutive integers for all 1≤k≤99.

So for each such k, using Lemma 1, there is a

l:k∣l

Therefore, using Lemma 2:

lcm(A)=lcm(A−{k,l},lcm(k,l))

Now, using Lemma 3,

lcm(A)=lcm(A−{k,l},l)=lcm(A−{k})

So, doing this with all the k, we can conclude that,

lcm(1,2…200)=lcm(100,101,102…200)

Trivially, we can remove the 100 as 200 is divisible by it.

So, the original equation becomes:

x=lcm(101,102…200)lcm(102,103…200)

Thus, I conclude x=101. Is this proof correct? (Any proof writing tips are also appreciated. I have no experience writing number theoretic proofs)