<p>Could be</p>
<p>1, 11, 21, 1112, 3112, 211213, 312213, 212223, …</p>
<p>until 8 digits repeat in a loop.</p>
<p>Yep. And you get a different “limit” if you start with a different “seed”. It’s fun to play and see. I don’t know whether every seed leads to a steady-state limit though. If anyone knows of a seed that does NOT converge, please post!</p>
<p>@pckeller</p>
<p>Oh, it’s like reading the previous line out-aloud. For example, 212223 is one 1’s, 4 2’s and one 3. Then you write it like 114213. You keep going till you get 21322314- and it stops. You know, if you want to find a seed that doesn’t converge, I guess you can use C++ or something. I’ll write a program and see if there are seeds that don’t result in a converging sequence.</p>
<p>I’ll write the questions again so they won’t get lost:
(1) Show that the product of an even integer and any other integer is even.
(2) Show that the product of 2 consecutive integers is even.
(3) Show that the product of 3 consecutive integers is divisible by 6.
(4) Show that the product of 4 consecutive integers is divisible by 24.
(5) Show that the product of n consecutive integers is divisible by n!</p>
<p>1.
n = even integer
d = any other integer</p>
<p>nd be even because n can be expressed as 2k for some integer k.
2(kd) is an even integer because kd must be an integer…</p>
<ol>
<li><p>n, n+1
One of the integers must be divisible by 2. (even)</p></li>
<li><p>n, n+1, n+2
One of the integers must be divisible by 3.
One of the integers must be divisible by 2.
The prime factorization of n(n+1)(n+2) must include 2 and 3. Therefore, n(n+1)(n+2) must be divisible by 6…</p></li>
<li><p>Same reason as above…</p></li>
</ol>
<p>@Jeffery: Number 1 is essentially correct, but let me add a bit more detail:</p>
<p>Let n and d be integers with n even. Then there is an integer k such that n=2k.
nd=(2k)d=2(kd) because the set of integers is associative for multiplication. kd is an integer because the set of integers is closed for multiplication. Therefore nd is even.</p>
<p>The others require more proof. Let’s take number 2 for example? Why does one of n or n+1 have to be divisible by 2? This requires proof. Two fairly straightforward ways I see to do this are by mathematical induction, or by using the division algorithm (This will take care of the positive integers at least. You will need a bit more argument if n is nonpositive).</p>
<p>These are some awesome questions. Kudos.</p>
<p>^^ Could the answer to number 2 be that:</p>
<p>Product of 2 consecutive numbers: n(n-1) = n^2 - n</p>
<p>From this, you would need to know that odd<em>odd = odd and even</em>even=even</p>
<p>Then you would need to know that odd-odd = even and even-even = even</p>
<p>That’s just my take though</p>
<p>@Good job. </p>
<p>Yep that’s a perfectly reasonable argument. I would call that a proof by cases. I think it’s a little neater to write the 2 consecutive integers as n(n+1)=n^2+n. For a complete argument you will need 4 lemmas:</p>
<p>Lemma 1: The sum of two even integers is even.
Lemma 2: The sum of two odd integers is even.
Lemma 3: The product of an even integer and any other integer is even.
Lemma 4: The product of two odd integers is odd.</p>
<p>Lemma 3 was question number 1, and I’ve already given a complete argument (#125). Can you give complete arguments for the other 3 Lemmas?</p>
<p>Here is a complete proof by cases using these Lemmas:</p>
<p>Let n and m be consecutive integers with n<m. Then m=n+1, and nm=n(n+1)=n^2+n. If n is odd, then by Lemma 4 n^2 is odd, and by Lemma 2 n^2+n is even. If n is even, then by Lemma 3 n^2 is even, and by Lemma 1 n^2+n is even.</p>
<p>Can anyone show this using one of the other 2 methods I mentioned? (induction or division algorithm)</p>
<p>If you have n consecutive integers, then for every integer k, 1 ≤ k ≤ n, at least one of those n integers is divisible by k, by a simple Pigeonhole/mod argument.</p>
<p>Therefore, the product of n consecutive integers is divisible by 1,2,…,n, that is, it is divisible by n!.</p>
<p>Consider the sequence of numbers sqrt(2), 2sqrt(2), 3sqrt(2), 4sqrt(2), … Prove that one of these numbers gets arbitrarily close to an integer.</p>