Least common multiple of $1,2,\ldots,2n+1$ Here's a problem that I heard: Show that $lcm(1,2,\ldots,2n+1)\geq4^n$.
If you assume the prime number theorem then it's not too hard to show that $lcm(1,2,\ldots,2n+1)$ is eventually larger than $4^n$. The hard part is making this effective for small values of $n$.
Also, I've been told that you can prove this without the prime number theorem.
 A: Here's a cool way to do this. Consider the following integral 
$$\text{lcm}(1,2,3,\cdots,2n+1)\int_0^1x^n(1-x)^ndx.$$
We can expand the product $x^n(1-x)^n$ using the binomial theorem as $$x^n\sum_{k=0}^n \binom{n}{k}(-1)^kx^k = \sum_{k=0}^n \binom{n}{k}(-1)^kx^{n+k}$$
which gives the integral as 
$$\text{lcm}(1,2,3,\cdots,2n+1) \left(\sum_{k=0}^{n} \binom{n}{k}(-1)^k \frac{x^{n+k+1}}{n+k+1}\right)\Bigg{|}_{0}^1. $$ 
Now this expression is an integer since the term $\text{lcm}(1,2,3,\cdots,2n+1)$ will clear all the denominators  of the sum. This integer is also clearly positive since the function $x^n(1-x)^n$ is positive on $(0,1)$. So we conclude that 
$$\text{lcm}(1,2,3,\cdots,2n+1)\int_0^1x^n(1-x)^ndx \geq 1.$$
Now using simple calculus, we see that the maximum of the function $x^n(1-x)^n$ on $[0,1]$ occurs at $x = \frac{1}{2}$, and so $x^n(1-x)^n <\frac{1}{4^n}$. This immediately gives that 
$$1 \leq \text{lcm}(1,2,3,\cdots,2n+1)\int_0^1x^n(1-x)^ndx < \text{lcm}(1,2,3,\cdots,2n+1)\frac{1}{4^n}$$
and so $$4^n < \text{lcm}(1,2,3,\cdots,2n+1)$$
