This section contains several examples of mathematics problems that can be solved using lists. These applications come for a variety of levels of math. Some of these problems can be solved in other ways on the calculator.
Contents
 1 Reducing fractions
 2 Checking for prime factors
 3 Graphing families of curves
 4 Solving systems of linear equations
 5 Calculating both solutions to quadratic equations
 6 Finding the polynomial formula for a sequence
 7 Solving two parts of a triangle problem at once
 8 Plot both halves of conic graphs
 9 The application problem that got me started on this list idea
Reducing fractionsEdit
Let's reduce to lowest terms by seeing which prime numbers divide into both the numerator and denominator. In a list we can divide tham both at the same time.
Enter the numerator and denominator of the fraction as a list
 {29302,75803}


 {29302 75803}


Divide by 2.
 Ans / 2


 {14651 37901.5}


As we might have guessed the top comes out even, but the bottom does not. So, multiply by 2 to get the original fraction back.
 Ans * 2


 {29302 75803}


Let's try dividing by 3.
 Ans / 3


 {9767.333 25267.667}


Neither one works. Multiply the 3 back in.
 Ans * 3


 {29302 75803}


Looking at the ones digit tells us not to bother with 5. Seven is the next prime.
 Ans / 7


 {4186 10829}


Both number came out even so the original fraction reduces by 7. Now we must see if there is another factor of 7 in both numbers. We can hit enter again to do the same operation again: dividing the previous answer by 7.
 Enter


 {598 1547}


Seven is a factor of both numbers again; what about a third time?
 Enter


 {85.429 221}


The denominator is dividible by 7 a third time, but not the numerator. We must put the 7 back in.
 Ans * 7


 {598 1547}


Next try 11.
 Ans / 11


 {54.364 140.636}


No good.
 Ans * 11


 {598 1547}


13 is next.
 Ans / 13


 {46 119}


They both reduce by 13.
Now since 13 time 13 is more than 100 and thus more than 46, if 13 divided into 46 evenly the answer would be smaller than 13. But we tried all the primes smaller than 13. This fraction is in lowest terms.
Checking for prime factorsEdit
We can check any three digit number to see if is a prime by dividing it by all the primes less that the square root of 1000. Create a list in L1 for these primes.
 {2,3,5,7,11,13,17,19,23,29,31} Sto> L1
It might be easier to create this list in the Editor.
Is 997 a prime number? Divide 997 by all those primes.
 997 / L1 Sto→ L2


 {498.5 332.333...


Since this contains lots of numbers with ten decimal digits, look at L2 in the list editor, and scroll down it. We see that none of the divisions come out as even integers, so 997 is prime.
Try 991
 991 / L1 Sto→ L2
Can you see that 991 is also a prime?
Try 989.
 989 / L1 Sto→ L2
Can you see in the editor that the factors of 989 are 23 × 43?
Graphing families of curvesEdit
A list of numbers can be used to display several related graphs.
For example put 0, 1, 2, and 3 into a list. We will use this list of numbers in several ways.
 {0,1,2,3,} STO> L1


 {0 1 2 3}


 Start with a set of linear graphs.
Press Y= and enter
 Y1=L1*X+1
 ZOOM 6(ZStandard)
Notice that the lines have different slopes. Each graph is steeper that the previous one. But the yintercept is 1 in each case.
 Next look at a family of quadratic curves.
Change to
 Y1=1+L1*X+0.25X^2
 GRAPH
This time each of these parabolas have the same shape, due to the second degree term, and the same yintercept, due to the constant (1). The list draws each graph in a different place on the axes.
 How will changing the cofficient on an exponential function affect the graph?
 Y1= L1*e^(X/4)
 GRAPH
Why do you only see three graphs? The list has affected the yintercept. What does multiplying by zero do to the rest of the formula?
 And now for some trigonometry.
Hit the MODE key and move down and over to Degree; hit ENTER.
 Y=L1*3*sin(36X)
 GRAPH
Once again there are only three visible graphs. (I guess zero wasn't the best number to put in our list.) The sine waves are similar with different amplitudes.
Solving systems of linear equationsEdit
One good way of solving a system of equations is called the elimination method. The plan is to multiply each equation by a different number. The multipliers are chosen so that one of the variables in each equation has the same coefficient, except for opposite sign. This allows us to add the equations together and eliminate that variable with opposite cofficients.
Example
 8x + 7y = 35
 6x + 5y = 27
Put the three numbers in each equation into a list.
 {8,7,35} STO> L1


 {8 7 35}


 {6,5,27} STO> L2


 {6 5 27}


35 is the smallest number that the ycoefficients are both multiples of; and 24 is the smallest common multiple of 8 and 6. Let's start with 24. We need a positive 24x in one equation and a negative 24x in the other.
We can multiply the first equation by 3:
 3 * L1 STO> L3


 {24 21 105}


and the second equation by 4:
 4 * L2 STO> L4


 {24 20 108}


Add the two new lists.
 L3 + L4


 {0 1 3}


This list represents 0x + 1y = 3 or y = 3.
Next let's fix the y coeficients to 35 and 35
 5 * L1 STO> L3


 {40 35 175}


 7 * L2 STO> L4


 {42 35 189}


And add
 L3 + L4


 {2 0 14}


Which represents 2x = 14, so divide the list by 2
 Ans / 2


 {1 0 7}


Solution: x = 7, y = 3
Calculating both solutions to quadratic equationsEdit
This process is begun by having a list containing a positive one and a negative 1.
 {1,1} STO> L6


 {1 1}


When this list is multiplied by a number, the result is a list with plus and minus the number.
 1.8 * L6


 {1.8 1.8}


Next, since the answers to a quadratic equation are often irrational, they need to be rounded off. So set the calculator to 3 decimal digits.
Press the MODE Key and move down and over to the 3 and press ENTER.
Example  Solve x2 + 5x + 3 = 0
Using the quadratic formula gives us:
Start with the calculation inside the square root:
 5 ^ 2  4 * 1 *3


 13.000


Next do the square root and get both positive and negative answers:
 L6 * √ (Ans)


 {3.606 3.606}


Add 5
 Ans  5


 {1.394 8.606}


And divide by 2
 Ans / 2


 {.697 4.303}


That list contains both of the answers.