2018 Putnam B Problems

Revision as of 22:35, 21 April 2019 by Superram (talk | contribs) (Created page with "===Problem B1=== Let <math>\mathcal{P}</math> be the set of vectors defined by <cmath>\mathcal{P} = \left\{\begin{pmatrix} a \\ b \end{pmatrix} \, \middle\vert \, 0 \le a \le...")
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)

Problem B1

Let $\mathcal{P}$ be the set of vectors defined by \[\mathcal{P} = \left\{\begin{pmatrix} a \\ b \end{pmatrix} \, \middle\vert \, 0 \le a \le 2, 0 \le b \le 100, \, \text{and} \, a, b \in \mathbb{Z}\right\}.\]Find all $\mathbf{v} \in \mathcal{P}$ such that the set $\mathcal{P}\setminus\{\mathbf{v}\}$ obtained by omitting vector $\mathbf{v}$ from $\mathcal{P}$ can be partitioned into two sets of equal size and equal sum.

Solution

Problem B2

Let $n$ be a positive integer, and let $f_n(z) = n + (n-1)z + (n-2)z^2 + \dots + z^{n-1}$. Prove that $f_n$ has no roots in the closed unit disk $\{z \in \mathbb{C}: |z| \le 1\}$.

Solution

Problem B3

Find all positive integers $n < 10^{100}$ for which simultaneously $n$ divides $2^n$, $n-1$ divides $2^n - 1$, and $n-2$ divides $2^n - 2$.

Solution

Problem B4

Given a real number $a$, we define a sequence by $x_0 = 1$, $x_1 = x_2 = a$, and $x_{n+1} = 2x_nx_{n-1} - x_{n-2}$ for $n \ge 2$. Prove that if $x_n = 0$ for some $n$, then the sequence is periodic.

Solution

Problem B5

Let $f = (f_1, f_2)$ be a function from $\mathbb{R}^2$ to $\mathbb{R}^2$ with continuous partial derivatives $\tfrac{\partial f_i}{\partial x_j}$ that are positive everywhere. Suppose that \[\frac{\partial f_1}{\partial x_1} \frac{\partial f_2}{\partial x_2} - \frac{1}{4} \left(\frac{\partial f_1}{\partial x_2} + \frac{\partial f_2}{\partial x_1} \right)^2 > 0\]everywhere. Prove that $f$ is one-to-one.

Solution

Problem B6

Let $S$ be the set of sequences of length 2018 whose terms are in the set $\{1, 2, 3, 4, 5, 6, 10\}$ and sum to 3860. Prove that the cardinality of $S$ is at most \[2^{3860} \cdot \left(\frac{2018}{2048}\right)^{2018}.\]

Solution