My Favorite Problems of All Time

To know if the series converges, we use the comparison test. Let $L_k:=\lcm(1,2,...,k)$, for $k=1,2,3,...$. By definition, $L_1:=1$ and note that for $k\geq 2$, $0< L_k\leq k(k-1)$. That is, for each integer $N>0$, This means that $S=\sum_{k=1}^{\infty}\frac{1}{L_k}$ exists and is a positive real number between 1 and 2. To prove that this number is irrational, we argue by contradiction. Suppose on the contrary that $S$ is rational, and it can be written as where $(a,b)\in\Z\times\Z\backslash\{0\}$, and $a, b$ are coprimes. Let $p_1,p_2,p_3,...,p_r,...$ be the increasing sequence of prime numbers greater than $b$. Using the famous Bertrand's Postulate, we know that Change this to a non-strict inequality, we have for each $r\in\Z^+$. We will now show that $p_{r+1}-p_r$ is indeed strictly less than $p_r-1$. To do this, we need to recall a useful lemma that we will not prove formally: Lemma (1) simply states that for infinitely many primes $p_r$, it holds that $p_r\equiv 2\;(\modular 3)$. Now, note that if $p_{r+1}-p_r=p_r-1$, then $p_{r+1}-2\equiv 1\;(\modular 3)$, which means that $p_{r+1}\equiv 0\;(\modular 3)$, and is indeed true for $3$, but not true for infinitely many primes $p_r$. Hence, $p_{r+1}-p_r< p_r-1$ for infinitely many $r$. Now, notice that Notice also that $L_{p_1-1}=\lcm(1,2,...,p_1-1)$ is a multiple of all primes less than $p_1$, and $L_k=\lcm(1,2,...,k)=\lcm(1,...,p_r)$ is a multiple of all primes less than $p_r$. Therefore which is the same as saying that Recall that we defined $p_1$ to be a prime number greater than $b$, therefore, $b\in\{1,2,...,p_1-1\}$, and hence $L_{p_1-1}=\lcm(1,2,...,p_1-1)$ must be a multiple of $b$; i.e., $L_{p_1-1}=kb$ for some integer $k$. This means that Let us remind ourselves of the formula of $\lcm$ for a list of numbers. Ok, but why? This lemma is left as an exercise for the reader.

The lemma says that $L_m$ is a number divisible by $\lcm(1,2,...,k)$, and hence $\lcm(1,...,m)=h\lcm(1,...,k)$, for some $h\in\Z$. Therefore, for the term $S_{p_1-1}L_{p_1-1}$, This means that which is completely nonsensical, since this implies that there is an integer somewhere between $0$ and $1$. Therefore, the original statement must be true and $S$ is irrational.

For this inequality to make sense, $x$ and $y$ must be greater than 0. We now restate the problem more generally: if $\mathcal{S}$ has $n$ elements, then the inequality must hold for $n$ integers $x$ satisfying: Rearranging the inequality, we obtain: Let $u(x) = \log_2(4yx^6)$ and $v(x) = [\log_2(2x)]^2$. Then, based on the above, we have: $$ 2^{u(x)} + u(x) \geq 2^{v(x)} + v(x) $$ To proceed, let's introduce a useful property of monotonic functions: Returning to the inequality: Let $f(x) = 2^x + x$, a strictly increasing function. Then, by Lemma 1: Solving for $y$: Therefore: Define $f(x) = \frac{1}{2}x^{\log_2(x) - 4}$. Our goal is to find the minimum of this function. Proceeding via standard calculus: Setting $f'(x) = 0$, we have either $\frac{1}{2}x^{\log_2(x) - 5} = 0$ or $2\log_2 x - 4 = 0$. The first case is impossible since $x > 0$. Thus: $$ 2\log_2 x - 4 = 0 \Rightarrow x = 4 $$ It's easy to verify that $f'(x)$ changes sign from negative to positive at $x = 4$, so $x = 4$ is a minimum. The core result is that: is true if and only if $y \geq f(x) > 0$. For $\mathcal{S}$ to have $n$ elements, it must be true that: Plugging in $n = 64$ gives $0 < y < 2319$. Therefore, there are 2319 values of $y$ satisfying the inequality.