LCM Dilemma

What I'm going to show here is that $\text{lcm}(1,\ldots,2018)=\text{lcm}(1010,\ldots,2018)$ because the numbers $\{1,\ldots,1009\}$ don't contribute any factors to the LCM.

Consider the sets $C=\{1,2,3, \ldots, 1009\}$ and $D=\{1010,1011,1012, \ldots, 2018\}.$

Claim : Every integer in set $C$ divides some integer in set $D.$

Proof : Partition set $C$ as follows:

$\begin{aligned} E &= \{1,2,3, \ldots, 504\} \\ F &= \{505,506,507, \ldots, 1009\} \\ \end{aligned}$

Note that every element in set $F$ is exactly $\frac{1}{2}$ of some element in set $D.$ Now partition set $E$ as follows:

$\begin{aligned} G &= \{1,2,3, \ldots, 252\} \\ H &= \{253,254,255, \ldots, 504\} \\ \end{aligned}$

Note that every element in set $H$ is exactly $\frac{1}{4}$ of some element in set $D.$

We can continue partitioning the sets as above indefinitely. However, since there exists a smallest positive integer, $1,$ we will eventually have to stop. At that point, we will have established that every integer in set $C$ divides some integer in set $D.$ It then follows that the least common multiples given in the problem are equal.

By definition, $A|n \ \ \ \text{if and only if}\ \ \ \ x|n\ \text{for all}\ x = 1,2,\dots,2018; \\ B|n \ \ \ \text{if and only if}\ \ \ \ x|n\ \text{for all}\ x = 1010,\dots,2018.$

Since $x|A$ for all $x = 1, 2, \dots, 2018$ , we certainly have that $x|A$ for all $x = 1010,\dots, 2018$ , so that $\boxed{B|A}$ .

Now let $1 \leq x \leq 2018$ . There exist a unique integer $n \geq 0$ such that $1010 \leq 2^nx \leq 2018$ . Since $x | 2^nx$ and $2^nx|B$ we see that $x|B$ . Since this is true for all $x = 1, 2,\dots, 2018$ , we conclude that $\boxed{A|B}$ .

From $A|B$ and $B|A$ it follows that $\boxed{A = B}$ .

Let $C = \{1, 2, 3, \dots, 1009\}$ and $D = \{1010, 1011, 1012, \dots, 2018\}$ . Then $A = \text{lcm}(C \cup D)$ and $B = \text{lcm}(D)$ .

Since each element in $C$ multiplied by some power of $2$ is already found in $D$ , and $D$ has at least one even number, $C$ has no unique factors to $D$ . Therefore, $\text{lcm}(C \cup D) = \text{lcm}(D)$ , and so $A = B$ .

It should be intuitively clear that any $n$ consecutive positive integers must have a multiple of $n$ among them. However, as a formality, a proof is included at the bottom.

Using this idea, it is immediate that the set $B^{*}=\{1010,\,1011,\,1012,\,\dots,\,2018\}$ , which contains 1009 consecutive positive integers, must contain a multiple of $a$ for every $a \in A^{*}=\{1,\,2,\,3,\dots,1009\}$ . Therefore, $\text{lcm}\,(1,\,2,\,3,\dots,2018) = \text{lcm}\,(1010,\,1011,\,1012,\dots,2018)$ .

Here's the promised proof.

Claim: Every sequence of $n$ consecutive positive integers contains a multiple of $n$ .

Proof: Let $t$ be the largest term in the sequence. If $n \mid t$ then we're done.

If not, let $t \equiv r \pmod{n}$ for some $r \in \{1,\,2,\,\dots, n-1\}$ . Then $t-r$ must be a member of the sequence, and clearly $n \mid t-r$ .

Mathematica

LCM @@ Range@2018 == LCM @@ Range[1010, 2018]

yelds True

The greatest prime number with 2018 in both sets is 2017 so $\boxed{A=2017 \times 2018=B}$

The difference in the LCM's are caused by the prime numbers in the list. B contains the same prime numbers as A from 1013 to 2017. And if not, the prime numbers appear as factors of other numbers in B. The largest prime number less than 1010 is 1009, and 2 x 1009 = 2018. Therefore, A = B.

The extra numbers of the first set don't put in any extra factors to the LCM. For any number n, there is at least one multiple of n every n numbers. There are 1009 numbers from 1010 to 2018, enough for at least one multiple of each number from 1 to 1009.

Quick Python solution

 from functools import reduce
 from fractions import gcd

 lcm = lambda a, b: a * b // gcd(a, b)

 A = reduce(lcm, range(1,    2019))
 B = reduce(lcm, range(1010, 2019))

 print(A > B)
 print(B > A)
 print(A == B)