A-REI Pairs of Whole Numbers

Alignments to Content Standards: A-REI.C.6


The sums of three whole numbers taken in pairs are 12, 17, and 19. What is the middle number?

IM Commentary

This task addresses A-REI.C.6, solving systems of linear equations exactly, and provides a simple example of a system with three equations and three unknown. Two (of many) methods for solving the system are presented. The first takes the given information to make three equations in three unknowns which can then be solved via algebraic manipulation to find the three numbers. The second solution is more clever, creating a single equation in three unknowns from the given information. This equation is then combined with the given information about the sums of pairs of numbers to deduce what the third number is. In reality, this solution is not simpler than the first: rather it sets up a slightly different set of equations which can be readily solved (the key being to take the sum of the three equations in the first solution). It provides a good opportunity for the instructor to show different methods for solving the same system of linear equations.

This task was adapted from problem #6 on the 2012 American Mathematics Competition (AMC) 12A Test. For the 2012 AMC 12A, which was taken by 72,238 students, the multiple choice answers for the problem had the following distribution:

Choice Answer Percentage of Answers
(A) 4 1.58
(B) 5 11.26
(C) 6 3.22
(D)* 7 66.15
(E) 8 3.90
Omit -- 13.84

Of the 72,238 students: 28,268, or 39%, were in 12th grade; 34,124 or 47%, were in 11th grade; 4,615, or 6%, were in 10th grade; and the remainder were below 10th grade.


Solution: 1 System of Equations

The three unknown numbers can be denoted by $x,y,$ and $z$. We are given the sums of any pair of these numbers as is indicated in the following system of equations:

\begin{align} x+y &= 12 \\ x+z &= 17 \\ y + z &= 19. \end{align}

so solve the system, we can start by subtracting the first equation from the second, giving $$ z - y = 5. $$ Adding this to the third equation in the list gives $$ 2z = 24 $$ so $z = 12$. We can now go back and plug this value of $z$ into the second and third of the original equations to find that $x = 5$ and $y = 7$. A quick check shows that these numbers do satisfy the three equations and the middle number is 7.

Note that the way we translated the three pieces of information into equations, it turned out that $x$ was the smallest number, $y$ the middle number, and $z$ the largest number. This is not necessary, however, as we could have written down the same system of equations with the right hand sides (12, 17, 19) in any order.

Solution: 2 Reasoning to produce special equations

Instead of three equations, each involving two unknons, there is another way to set up equations to find the solution. Suppose again that $x,y,$ and $z$ are the three whole numbers we are looking for and this time we arrange them so that $x \leq y \leq z$. We know that

\begin{align} 2(x+y+z) &=(x+y)+(x+z)+(y+z)\\ &=12+17+19\\ &=48 \end{align}

We know that $x + y$ is the smallest of the three sums followed by $x + z$ followed by $y + z$ because we assumed $x \leq y \leq z$. So we conclude that $$ x + y + z = 24 $$

Now we are given that $x + y = 12$. Subtracting this from the equation $x + y + z = 24$ gives us $z = 12$. Similarly from $x + z = 17$ we can subtract this from $x + y + z = 24$ to deduce that $y = 7$. Finally we get $x = 5$ by subtracting $y+ z = 19$ from $x + y + z = 24$.