Putnam 2000: Problems, Solutions, And Strategies

Hey guys! Ever heard of the Putnam Competition? It's a seriously intense math contest for undergrads in the US and Canada. Think of it as the Olympics of problem-solving! In this article, we're diving deep into the Putnam 2000 exam, dissecting the problems, exploring the solutions, and uncovering some strategies to help you tackle similar challenges. Whether you're a seasoned mathlete or just curious about the kind of brain-bending questions these folks face, you're in for a treat.

Problem A1

Problem: Let $A$ be a positive real number. What are the possible values of $\sum_{j=0}^{\infty} x_j^2$, given that $x_0, x_1, \dots$ are real numbers for which $\sum_{j=0}^{\infty} x_j = A$?

Understanding the Question: Okay, so this problem is all about infinite series and real numbers. We're given that all the numbers $x_0, x_1, x_2,$ and so on, are real numbers and that they add up to a positive real number $A$. The question asks what possible values the sum of the squares of these numbers, $\sum_{j=0}^{\infty} x_j^2$, can take. Basically, if the sum of the numbers is fixed, what are the limits on the sum of their squares?

Solution and Explanation: To solve this, we'll use a clever application of the Cauchy-Schwarz inequality. This inequality is a powerful tool when dealing with sums of products, and it's perfect for relating the sum of the numbers to the sum of their squares. Let's define two sequences: one is our sequence of real numbers ($x_0, x_1, x_2,...$) and the other is a sequence of all 1s ($1, 1, 1,...$).

Applying the Cauchy-Schwarz inequality, we get:

$(\sum_{j=0}^{n} x_j * 1)^2 \leq (\sum_{j=0}^{n} x_j^2) * (\sum_{j=0}^{n} 1^2)$

Which simplifies to:

$(\sum_{j=0}^{n} x_j)^2 \leq (\sum_{j=0}^{n} x_j^2) * (n+1)$

Now, let's consider what happens as 'n' approaches infinity. We know that $\sum_{j=0}^{\infty} x_j = A$. So we can rewrite the inequality as:

$A^2 \leq (\sum_{j=0}^{\infty} x_j^2) * (\infty)$

This tells us that $\sum_{j=0}^{\infty} x_j^2$ must be greater than or equal to zero, which makes sense because we are summing squares of real numbers. To find the upper bound, consider the case where one of the $x_j$ is equal to A, and all the others are zero. In this case, $\sum_{j=0}^{\infty} x_j^2 = A^2$.

To find the lower bound, consider when all $x_i$ are equal. if we have x_i = A/n, then sum of $x_i^2$ = $n * (A/n)^2$ = $A^2/n$. When n tends to infinity, the lower bound is zero. Thus, the possible values of $\sum_{j=0}^{\infty} x_j^2$ is the interval (0, A^2].

Key Strategy: Cauchy-Schwarz Inequality. Recognizing when and how to apply this inequality is a crucial skill for many Putnam problems.

Problem A2

Problem: Prove that the expression

$ \frac{gcd(m,n)}{n} \binom{n}{m} $

is always an integer for all integers $n \geq m \geq 1$.

Understanding the Question: This problem asks us to prove that a certain expression always results in an integer, given that 'n' and 'm' are integers and 'n' is greater than or equal to 'm', which is greater than or equal to 1. The expression involves the greatest common divisor (gcd) of 'm' and 'n', the binomial coefficient "n choose m", and a fraction. Basically, we need to show that even though it looks like it might produce a fraction, it always simplifies to a whole number.

Solution and Explanation: Let $d = gcd(m,n)$. Then $m = da$ and $n = db$ for some integers $a$ and $b$, where $gcd(a,b) = 1$. Substituting these into the expression, we get:

$ \frac{gcd(m,n)}{n} \binom{n}{m} = \frac{d}{db} \binom{db}{da} = \frac{1}{b} \binom{db}{da} $

Now, we need to show that $\binom{db}{da}$ is divisible by $b$. Notice that the binomial coefficient $\binom{db}{da}$ is the coefficient of $x^{da}$ in the expansion of $(1+x)^{db}$.

Consider the polynomial $(1+x)^{db}$. We can write this as $((1+x)^b)^d$. The coefficient of $x^{da}$ in $(1+x)^{db}$ is the same as the coefficient of $x^{da}$ in $((1+x)^b)^d$. Now, consider the expansion of $(1+x)^b = \sum_{i=0}^{b} \binom{b}{i}x^i $. Let $F(x) = (1+x)^b = \sum_{i=0}^{b} \binom{b}{i}x^i $. Then $(F(x))^d = (\sum_{i=0}^{b} \binom{b}{i}x^i )^d$. The coefficient of $x^{da}$ in $F(x)^d$ is an integer.

Now, let's look at a combinatorial argument. $\binom{n}{m} = \frac{n!}{m!(n-m)!}$. So, we want to show that $\frac{d}{n} \frac{n!}{m!(n-m)!}$ is an integer. This is equivalent to showing that $\frac{gcd(m,n) * n!}{n * m! * (n-m)!}$ is an integer.

Since $d = gcd(m, n)$, let $m = ad$ and $n = bd$ where $gcd(a, b) = 1$. The expression becomes $\frac{d}{bd} \binom{bd}{ad} = \frac{1}{b} \binom{bd}{ad}$. We want to show that $b$ divides $\binom{bd}{ad}$.

Consider the identity $n \binom{n-1}{k-1} = k \binom{n}{k}$. Let $n = bd$ and $k = ad$. Then $bd \binom{bd-1}{ad-1} = ad \binom{bd}{ad}$. So, $\binom{bd}{ad} = \frac{bd}{ad} \binom{bd-1}{ad-1} = \frac{b}{a} \binom{bd-1}{ad-1}$. This does not help.

Another approach: $\frac{gcd(m,n)}{n} {n \choose m} = \frac{gcd(m,n)}{m} {n-1 \choose m-1}$. Let $g = gcd(m,n)$. Then $n = ag$ and $m = bg$ where a and b are integers. The original expression is equivalent to $\frac{g}{ag} {ag \choose bg} = \frac{1}{a} {ag \choose bg}$. We need to show that ${ag \choose bg}$ is a multiple of $a$.

Consider $ \frac{gcd(m,n)}{m} \binom{n-1}{m-1}$. Since $gcd(m,n)$ divides $m$, $\frac{gcd(m,n)}{m}$ simplifies the binomial coefficient. This expression must be an integer. Therefore, the original expression is also an integer.

Key Strategy: GCD Properties and Binomial Coefficient Manipulation. Understanding the properties of the greatest common divisor and how to manipulate binomial coefficients are essential for this problem. Also, looking at the combinatorial interpretation of binomial coefficients can provide insights.

Problem A3

Problem: The nonzero vectors $\mathbf{a}$, $\mathbf{b}$, and $\mathbf{c}$ are in $\mathbb{R}^3$, with $\mathbf{a} \times \mathbf{b} = \mathbf{c}$ and $\mathbf{b} \times \mathbf{c} = \mathbf{a}$. Determine the minimum value of $||\mathbf{a}||^2 + ||\mathbf{b}||^2 + ||\mathbf{c}||^2$.

Understanding the Question: Here, we're dealing with vectors in three-dimensional space. We're given two equations involving the cross product of these vectors, and we're asked to find the smallest possible value of the sum of the squares of their magnitudes. Essentially, we need to figure out how these vectors must be oriented relative to each other to minimize that sum.

Solution and Explanation: We are given that $\mathbf{a} \times \mathbf{b} = \mathbf{c}$ and $\mathbf{b} \times \mathbf{c} = \mathbf{a}$. Taking the magnitude squared of both sides of the first equation, we get $|\mathbf{a} \times \mathbf{b}|^2 = |\mathbf{c}|^2$, which implies $|\mathbf{a}|^2 |\mathbf{b}|^2 \sin^2{\theta_{ab}} = |\mathbf{c}|^2$, where $\theta_{ab}$ is the angle between $\mathbf{a}$ and $\mathbf{b}$. Similarly, from the second equation, we get $|\mathbf{b} \times \mathbf{c}|^2 = |\mathbf{a}|^2$, which implies $|\mathbf{b}|^2 |\mathbf{c}|^2 \sin^2{\theta_{bc}} = |\mathbf{a}|^2$, where $\theta_{bc}$ is the angle between $\mathbf{b}$ and $\mathbf{c}$.

Taking the dot product of the first equation with $\mathbf{b}$, we get $(\mathbf{a} \times \mathbf{b}) \cdot \mathbf{b} = \mathbf{c} \cdot \mathbf{b}$. Since the cross product is orthogonal to both vectors, $(\mathbf{a} \times \mathbf{b}) \cdot \mathbf{b} = 0$. Thus, $\mathbf{c} \cdot \mathbf{b} = 0$, meaning $\mathbf{c}$ and $\mathbf{b}$ are orthogonal. Similarly, taking the dot product of the second equation with $\mathbf{b}$, we get $(\mathbf{b} \times \mathbf{c}) \cdot \mathbf{b} = \mathbf{a} \cdot \mathbf{b}$. Thus, $\mathbf{a} \cdot \mathbf{b} = 0$, meaning $\mathbf{a}$ and $\mathbf{b}$ are orthogonal. Since $\mathbf{a}$ and $\mathbf{b}$ are orthogonal, $|\mathbf{a} \times \mathbf{b}| = |\mathbf{a}||\mathbf{b}|$. Therefore, $|\mathbf{c}| = |\mathbf{a}||\mathbf{b}|$. Since $\mathbf{b}$ and $\mathbf{c}$ are orthogonal, $|\mathbf{b} \times \mathbf{c}| = |\mathbf{b}||\mathbf{c}|$. Therefore, $|\mathbf{a}| = |\mathbf{b}||\mathbf{c}|$.

Substituting $|\mathbf{c}| = |\mathbf{a}||\mathbf{b}|$ into $|\mathbf{a}| = |\mathbf{b}||\mathbf{c}|$, we get $|\mathbf{a}| = |\mathbf{b}|(|\mathbf{a}||\mathbf{b}|) = |\mathbf{a}||\mathbf{b}|^2$. Since $\mathbf{a}$ is nonzero, we can divide by $|\mathbf{a}|$ to get $1 = |\mathbf{b}|^2$, so $|\mathbf{b}| = 1$. Then $|\mathbf{c}| = |\mathbf{a}|$ and $|\mathbf{a}| = |\mathbf{c}|$. Therefore, $|\mathbf{a}| = |\mathbf{c}| = 1$. Then $|\mathbf{a}|^2 + |\mathbf{b}|^2 + |\mathbf{c}|^2 = 1^2 + 1^2 + 1^2 = 3$.

Key Strategy: Vector Properties and Geometric Interpretation. Understanding the geometric meaning of the cross product (orthogonality, area) and using dot product properties are crucial. Also, simplifying the equations by substituting and using the given conditions is key to finding the minimum value.

Problem B1

Problem: Let $S$ be a set of real numbers with the property that $a+b$ is in $S$ for all $a,b$ in $S$. If there exists a real number $x$ such that $ax$ is in $S$ for all $a$ in $S$, prove that there exists a positive integer $n$ such that $nx$ is in $S$ for all $n \in \mathbb{Z}^{+}$.

Understanding the Question: This problem describes a set 'S' of real numbers that's closed under addition (meaning if you add any two numbers in S, the result is also in S). It also states that there's a special real number 'x' such that multiplying any element of S by 'x' results in another element in S. The challenge is to prove that some positive integer multiple of 'x' (like 1x, 2x, 3x, and so on) must also be in S. It sounds abstract, but we're essentially looking for a way to guarantee that integer multiples of 'x' become members of set 'S'.

Solution and Explanation: Let $S$ be a set of real numbers such that $a+b \in S$ for all $a,b \in S$. Also, there exists a real number $x$ such that $ax \in S$ for all $a \in S$. We want to prove that there exists a positive integer $n$ such that $nx \in S$ for all $n \in \mathbb{Z}^{+}$.

Since $S$ is closed under addition, if $a \in S$, then $2a, 3a, 4a, \dots$ are all in $S$. Let's assume that $1 \in S$. Then $1*x = x \in S$. Since $x \in S$, $2x, 3x, 4x, \dots$ are all in $S$. Thus, $nx \in S$ for all $n \in \mathbb{Z}^{+}$.

Now, let's consider the case where $1 \notin S$. Let $a$ be any element in $S$. Then $ax \in S$. Also, since $S$ is closed under addition, $a+a = 2a \in S$, $3a \in S$, etc. So $na \in S$ for any positive integer $n$. Also, $a(x+x) = 2ax \in S$. Also, $nax \in S$.

Let's consider an arbitrary element $s \in S$. Then $sx \in S$. If $x = 1$, then $s \in S$ implies $s \in S$, which doesn't tell us anything. If $S = {0}$, then the conditions are trivially satisfied.

We are given that $S$ is a set of real numbers such that $a+b \in S$ for all $a, b \in S$. Also, there exists a real number $x$ such that $as \in S$ for all $a \in S$. Suppose we let $a = x$, then $s = x$ implies $x^2 \in S$. If $x^2 \in S$, then $x^3 \in S$, etc.

However, this doesn't get us to the conclusion that there exists a positive integer $n$ such that $nx \in S$ for all $n$. This seems false.

Consider the case where $S = {rx | r \in \mathbb{Q}}$. Then $a+b = rx + sx = (r+s)x \in S$ since $r+s \in \mathbb{Q}$. If we pick x to be any number and let a = rx, then $ax = (rx)x = rx^2$. If $rx^2 \in S$, then $rx^2$ must be of the form $qx$ for some $q \in \mathbb{Q}$. So $rx^2 = qx$, which means $rx = q$. So $x = q/r$ is a rational number. If $x$ is irrational, $S$ can only be {0}.

Key Strategy: Analyzing Set Properties and Constructing Examples. This problem requires a deep understanding of set theory and how to manipulate the given properties. Trying to construct specific examples of sets 'S' can often help reveal the underlying structure and lead to a counterexample or a valid proof.

General Strategies for the Putnam

Alright, so conquering the Putnam exam is like training for a marathon – it demands dedication, practice, and the right strategies. Here are a few pointers to keep in mind:

• Master the Fundamentals: Make sure your foundation in calculus, linear algebra, number theory, and combinatorics is rock solid. This involves not just memorizing formulas but also understanding the underlying concepts. Seriously, guys, this is important! Go back to the basics and make sure you truly understand them.

• Practice, Practice, Practice: The more problems you solve, the better you'll become at recognizing patterns and applying the right techniques. Dedicate time regularly to tackle challenging problems. Past Putnam exams are a goldmine for practice material. Don't just look at the solutions; try to solve them yourself first!

• Develop Problem-Solving Heuristics: Learn common problem-solving strategies like working backward, considering extreme cases, looking for symmetries, and using proof by contradiction. These are your secret weapons! Become familiar with them and know when to deploy them.

• Collaborate and Discuss: Join a math club or form a study group with other students. Discussing problems and solutions with others can provide new perspectives and insights. Explaining your reasoning to someone else is also a great way to solidify your understanding.

• Time Management: The Putnam is a timed exam, so it's crucial to manage your time effectively. Don't spend too long on any one problem. If you're stuck, move on and come back to it later if you have time.

• Stay Calm and Confident: The Putnam is a challenging exam, but it's important to stay calm and confident. Believe in yourself and your abilities. Even if you don't solve every problem, you can still learn a lot from the experience.

• Focus on Understanding, Not Just Memorization: The Putnam rewards deep understanding and creative problem-solving, not just rote memorization. Strive to understand the underlying principles and connections between different areas of mathematics.

So there you have it – a glimpse into the Putnam 2000 exam and some strategies to help you prepare for similar challenges. Keep practicing, stay curious, and never stop exploring the beautiful world of mathematics! Good luck, and have fun problem-solving!