Putnam 2000 A1: A Deep Dive

Hey guys, let's dive into a classic from the 2000 Putnam Competition, specifically problem A1. This problem, while seemingly straightforward, really tests your understanding of basic number theory and modular arithmetic. It's a fantastic warm-up for the tougher problems on the exam, and mastering it can give you a real confidence boost. So, grab your thinking caps, and let's unravel this gem together!

Understanding the Problem Statement

The Putnam 2000 A1 problem asks us to prove that for any integer $n > 0$, there exists a positive integer $m$ such that the decimal representation of $m$ consists only of digits 1 and 2, and $m$ is divisible by $n$. In simpler terms, we need to show that we can always find a number made up of only 1s and 2s that is a multiple of any given positive integer $n$. Sounds a bit wild, right? But trust me, there's a clever way to approach this. The core of the problem lies in the pigeonhole principle and constructing a sequence of numbers that will eventually lead us to our desired multiple. We're not looking for the smallest such $m$, just any such $m$. This is a crucial detail that simplifies the proof significantly.

The Pigeonhole Principle: Our Secret Weapon

Alright, let's get into the nitty-gritty. The pigeonhole principle is a mathematical concept that states if you have more pigeons than pigeonholes, then at least one pigeonhole must contain more than one pigeon. In our context, we'll be using this principle to guarantee the existence of a number with specific properties. Consider the sequence of numbers formed by concatenating the digit '1' repeatedly: 1, 11, 111, 1111, and so on. We can also consider numbers formed by only digits '1' and '2'. The key insight is to look at these numbers modulo $n$. That is, we examine the remainders when these numbers are divided by $n$. Since there are only $n$ possible remainders when dividing by $n$ (0, 1, 2, ..., $n-1$), if we can construct a sequence of numbers that is guaranteed to have at least $n+1$ terms, then by the pigeonhole principle, at least two of these numbers must have the same remainder when divided by $n$. This is where the magic happens, guys!

Let's consider the sequence of numbers $a_k$ where $a_k$ is the integer whose decimal representation consists of $k$ digits, all of which are '1'. So, $a_1 = 1$, $a_2 = 11$, $a_3 = 111$, and so on. Now, let's look at the remainders of these numbers when divided by $n$. We have a_1 mod n, a_2 mod n, a_3 mod n, ..., a_{n+1} mod n. This gives us $n+1$ remainders. Since there are only $n$ possible remainders (0 to $n-1$), by the pigeonhole principle, at least two of these remainders must be the same. Let's say a_i mod n = a_j mod n for some $1 The Putnam 2000 A1 problem is a beautiful demonstration of how abstract mathematical principles can be applied to construct concrete results. It challenges us to think about the properties of numbers formed by specific digits and to leverage powerful tools like the pigeonhole principle. The elegance of the solution lies in its simplicity and the fact that it guarantees the existence of such a number without explicitly constructing it in a trial-and-error fashion. The problem statement itself is a great starting point for exploration. It asks us to prove that for any positive integer $n$, we can find a number $m$ composed solely of the digits 1 and 2, such that $m$ is perfectly divisible by $n$. This means no remainder, a clean division. The implication here is that the set of numbers formed by only 1s and 2s is rich enough to contain a multiple of every positive integer. This is quite a statement, and the proof needs to be robust.

We're not asked to find the smallest such $m$, which would be a significantly harder problem, but just to prove existence. This is a common strategy in mathematics: proving something exists is often easier than finding it. The problem sets a very specific constraint on the digits of $m$: only '1' and '2'. This restriction is key to the solution. If any digits were allowed, the problem would be trivial (just consider $n$ itself, or $2n$, etc.). The constraint forces us to be clever.

Let's break down the core idea without giving away the whole game just yet. We're going to build a sequence of numbers. The properties of this sequence, when viewed through the lens of modular arithmetic, will reveal the existence of our target number $m$. The pigeonhole principle is going to be our guiding star here. It's a fundamental concept in combinatorics that helps us deduce facts about distributions. Imagine you have more pigeons than pigeonholes. What's bound to happen? At least one pigeonhole must have more than one pigeon. We'll be applying this very same logic, but with numbers and their remainders.

Consider the set of all positive integers whose decimal representation consists only of the digits 1 and 2. This set is infinite. We want to show that this infinite set