Casework
Casework is the process of splitting up a problem into a finite number of cases and then determining mathematical results for each respective case. Once all cases are "solved," the solution to the underlying problem becomes clear. Casework is used in nearly all areas of mathematics.
Casework is not just useful in finite, clearly delineated situations. Casework can also be applied in problems which have a large number of possible outcomes or even infinite possible outcomes. The goal of casework in these problems is to define a pattern or formula rather than a specific solution.
When to use Casework
The most obvious application of casework is when a problem gives you a clear structure in which different things happen under different circumstances. An example of this would be solving absolute value equations.
Solve: \(|2x+3|=3-x\)
\(2x+3\) is either positive, negative, or zero. Depending on this, the absolute value will do different things.
Case 1: \(2x+3\) is positive or zero.
If this is the case, then the absolute value will do nothing. The equation becomes \(2x+3=3-x\). Solving this yields \(x=0\).
Case 2: \(2x+3\) is negative.
If this is the case, then the absolute value will negate the expression. The equation becomes \(-(2x+3)=3-x\). Solving this yields \(x=-6\).
Both of these solutions are confirmed by substitution. The solution set is \(\{-6,\ 0\}\).
Another use of casework is when there are a finite number of possibilities for something, but only one can be true. Casework allows one use the process of elimination to find the correct solution.
Let \(a\) and \(b\) be positive integers such that \(a^2-b^2=24\). What is the maximum value of \(a^2+b^2\)?
The left hand side of the equation can be factored:
\[(a+b)(a-b)=24\]
Given that \(a\) and \(b\) are integers, this means that \(a+b\) and \(a-b\) must be integer factors of \(24\). Given that \(a\) and \(b\) are positive, \(a+b>a-b>0\). This gives the following possibilities for \(a+b\) and \(a-b\):
\[\begin{array}{c|c} a+b & a-b \\ \hline 24 & 1 \\ 12 & 2 \\ 8 & 3 \\ 6 & 4 \end{array}\]
Let \(m=a+b\) and \(n=a-b\). Solving this system of equations yields:
\[\begin{align} a &= \frac{m+n}{2} \\ \\ b &= \frac{m-n}{2} \end{align}\]
This gives the values of \(a\) and \(b\):
\[\begin{array}{c|c|c|c} a+b & a-b & a & b \\ \hline 24 & 1 & \color{red}{12.5} & \color{red}{11.5} \\ 12 & 2 & 7 & 5\\ 8 & 3 & \color{red}{5.5} & \color{red}{2.5} \\ 6 & 4 & 5 & 1 \end{array}\]
The values in red are non-integers, so they cannot be possible. The solution will be the maximum value of \(a^2+b^2\). The two possible values for \(a^2+b^2\) are \(74\) and \(26\), so the maximum value is \(74\).
It is always possible to use casework, even if the problem does not suggest a clear separation of the problem into cases. If a problem solver invents a clever way to separate the problem, it can lead to an efficient solution.
The following problem can seem complicated, but there is a clever way to apply casework to find the solution. Consider an equation of the form \(a^b=1\). Consider what values of \(a\) and \(b\) would make this equation true. This problem can be split into cases depending on these possibilities.
The applications of casework are extensive. The rest of this page will explore the applications of casework in various areas of mathematics.
Algebra
The Rational Root Theorem is an application of casework. Applying this theorem involves listing out possible rational roots of a polynomial function, and using the process of elimination to find all the rational roots.
Rational Root Theorem
Let \(f(x)\) be a non-constant polynomial function with integer coefficients and non-zero constant coefficient. Let \(D\) be the set of all positive integer factors of the leading coefficient, and let \(N\) be the set of all positive integer factors of the constant coefficient. Let the set \(S\) be defined by:
\[S=\left\{x\ \left |\ x=\pm\frac{n}{d},\ n\in N,\ d\in D\right.\right\}\]
The set of all rational roots of \(f(x)\) is a subset of \(S\).
Find the rational roots of the function \(f(x)=2x^3-3x^2-18x+14\).
The positive integer factors of the leading coefficient, \(2\), are: \(\{1,2\}\).
The positive integer factors of the constant coefficient, \(-14\), are: \(\{1,2,7,14\}\).
The Rational Root Theorem gives the following set as possible rational roots: \(\left\{\pm 1, \pm 2, \pm 7, \pm 14, \pm \dfrac{7}{2}\right\}\)
The rational roots of the function will always be a subset of the above set (it could be an empty set). This is essentially a set of cases to test in order to find the solution.
Testing each possible root yields \(\dfrac{7}{2}\) as the only rational root.
As was mentioned earlier, casework is useful for solving absolute value equations. It is also useful for solving inequalities.
Find the solution set of \(\dfrac{x+3}{2x-5}\ge 0\)
The numerator and denominator present different cases for when the expression on the left hand side of the equation is positive or negative.
Case 1: \(x+3\ge 0 \cap 2x-5>0\)
If both the numerator and denominator are greater than 0, then the rational expression will be positive. Thus, the inequality will be satisfied. The inequality will also be satisfied when the numerator is \(0\), but the denominator cannot be \(0\). Solving these inequalities yields \(x>\frac{5}{2}\).
Case 2: \(x+3\le 0 \cap 2x-5<0\)
If both the numerator and denominator are less than 0, then the rational expression will be positive. Thus, the inequality will be satisfied. The inequality will also be satisfied when the numerator is \(0\), but the denominator cannot be \(0\). Solving these inequalities yields \(x\le -3\).
Either case will satisfy the inequality. Thus, the solution set is \(\left\{x\ \left |\ x\le -3 \cup x>\frac{5}{2}\right.\right\}\)
Casework is especially important when one must consider multiple possibilities. The following problem is a good example of this.
Number Theory
Casework can be used to solve Diophantine Equations. A Diophantine Equation is an equation in which the variables can only take integer values.
In American football, points can be scored in the following ways:
- Touchdown: 7 points
- Field Goal: 3 points
- Safety: 2 points
A team finished a game with 24 points. How many combinations of scoring methods could lead to this score? The order of the scoring methods does not matter.
Let \(x\) be the number of touchdowns, \(y\) be the number of field goals, and \(z\) be the number of safeties. The goal is to solve:
\[7x+3y+2z=24\]
Casework allows one to convert the above equation to a two-variable equation, which is much easier to solve as a Diophantine Equation.
Case 1: The team scored 0 touchdowns.
This gives \(3y+2z=24\). The team could have scored a maximum of \(8\) field goals with \(0\) safeties. Then, the field goals can be decreased by increments of \(2\) while the safeties are increased by increments of \(3\). Thus, the order pair solutions \((x,y,z)\) for this case are:
\[\begin{array}{ccccc} (0,8,0) & (0,6,3) & (0,4,6) & (0,2,9) & (0,0,12) \end{array}\]
Case 2: The team scored 1 touchdown.
This gives \(3y+2z=17\). The team could have scored a maximum of \(5\) field goals with \(1\) safety. Then, the field goals and safeties can be adjusted in same way as the previous case. The ordered pair solutions \((x,y,z)\) for this case are:
\[\begin{array}{ccc} (1,5,1) & (1,3,4) & (1,1,7) \end{array}\]
Case 3: The team scored 2 touchdowns.
This gives \(3y+2z=10\). The ordered pair solutions \((x,y,z)\) for this case are:
\[\begin{array}{cc} (2,2,2) & (2,0,5) \end{array}\]
Case 4: The team scored 3 touchdowns.
This gives \(3y+2z=3\). The only ordered pair solution \((x,y,z)\) for this case is:
\[(3,1,0)\]
In all, there are \(11\) possible combinations of scoring methods.
In basketball, a player can score points in 3 different ways: \(1\) point from the foul line, \(2\) points for shots close to the basket, or \(3\) points for shots far away from the basket. If a player scores \(30\) points in a game, how many different ways can this be achieved?
Details and assumptions
The order of the baskets doesn't matter. For example, if the player scores 9 3-pointers and then 3 1-pointers, this is the same as scoring 3 1-pointers and then 9 3-pointers.
Casework can often be used to eliminate or narrow a set of solutions by applying modular arithmetic techniques.
What is the largest 2-digit integer whose perfect square ends in \(29\)?
Let \(x\) be an integer whose perfect square ends in \(29\). Then,
\[x^2\equiv 29\pmod{100}\]
Then it is also true that,
\[x^2\equiv 9\pmod{10}\]
This gives \(x\equiv 3\pmod{10}\) or \(x\equiv 7\pmod{10}\).
Case 1: \(x\equiv 3\pmod{10}\)
Let \(x=10a+3\). Substituting this into the original congruence gives
\[\begin{align} 100a^2+60a+9 &\equiv 29\pmod{100} \\ 60a &\equiv 20\pmod{100} \\ 6a &\equiv 2\pmod{10} \\ a &\equiv 2,\ 7\pmod{10} \end{align}\]
Case 1 gives two possible two-digit solutions for \(x\): \(23\) and \(73\).
Case 2: \(x\equiv 7\pmod{10}\)
Let \(x=10a+7\). Substituting this into the original congruence gives
\[\begin{align} 100a^2+140a+49 &\equiv 29\pmod{100} \\ 40a &\equiv 80\pmod{100} \\ 4a &\equiv 8\pmod{10} \\ a &\equiv 2,\ 7\pmod{10} \end{align}\]
Case 2 gives two possible two-digit solutions for \(x\): \(27\) and \(77\).
The only possible two digit solutions are \(23\), \(27\), \(73\), and \(77\). The largest of these is \(77\).
In other problems in Number Theory, the applications of casework are less immediately apparent, but applying other techniques can lead to a situation in which casework is needed.
In the below problem, an application of Vieta's Formulas will get you part of the way to the solution. Then, it's necessary to think about the different cases for values which will lead to integer roots.
Combinatorics
The application of casework is common in combinatorics and probability problems. Splitting a counting or probability problem into parts can lead to an efficient computation.
In the below problem, consider splitting the colorings into cases depending on which colors are adjacent to each other on the cube.
If you wish to paint the 6 faces of a cube using 6 colors, with each face in one distinct color, how many different ways can you do it?
Clarification: Different implies that for two coloring combinations \(A\) and \(B\), you can't pick up \(A\) and reorient it to make it look exactly like \(B\).
Image credit: Puzzles.com
The below problem will seem familiar to those who have studied rectangular grid walks. However, there is an interesting twist in that it is also possible to move diagonally. Consider splitting this problem into cases that depend on how many times the person moves diagonally.
The protagonist of this story, Stanley, is standing peacefully at coordinates \( (0, \space 0) \) without any concern in his life whatsoever.
Now, Stanley will move at coordinates \( (3, \space 4) \).
If Stanley only goes either up, or right, or up-right (corresponding vectors in their respective order: \( \vec{u}(0, \space 1) \), \( \vec{r}(1, \space 0) \) , \( \vec{ur}(1, \space 1) \)), on how many ways can Stanley accomplish this tremendous stunt?
Logic
Casework is useful in truth-tellers and liars problems. It allows one to consider each case of which people are telling the truth and which people are lying.
Casework is also useful for K-level thinking problems. These problems often require you to simplify the problem into smaller cases, and then use the smaller cases to solve larger cases.
Five pirates of different ages found a treasure chest with 100 gold coins. In the middle of the ocean, with no land in sight, they decide to split the coins using this scheme:
The oldest pirate proposes how to share the coins, and ALL pirates (including the oldest) vote for or against it. If 50% or more of the pirates vote for it, then the coins will be shared that way. Otherwise, the pirate proposing the scheme will be thrown overboard, and the process is repeated with the pirates that remain.
The pirates are a bloodthirsty bunch with no loyalty. If a pirate would get the same number of coins if he voted for or against a proposal, he will vote against to watch the proposer be thrown overboard.
Assuming that the pirates are intelligent, rational, greedy, and do not wish to be thrown overboard, (and are rather good at math for pirates) what should the oldest pirate propose?