Different Products
Problem
Take any set of four consecutive whole numbers.
Find the product of the first and last numbers.
Find the product of the second and penultimate (in this case third!) numbers.
e.g. consider the set $5, 6, 7, 8$
$5 \times 8 = 40$
$6 \times 7 = 42$
Try a few more examples with sets of four consecutive numbers.
Do your results have anything in common?
Now take a set of five consecutive whole numbers.
Again, find the product of the first and last numbers, and the product of the second and penultimate numbers.
Try several examples.
Do your results have anything in common?
Can you predict the difference between the product of the first and last numbers, and the product of the second and penultimate numbers, when you have 6 consecutive whole numbers?
What if you have $7$, or $10$, or $20$, or $100$ consecutive whole numbers?
Select sets of consecutive numbers and test your predictions.
Can you predict what the difference will be when you have $n$ consecutive whole numbers?
Mathematicians aren't usually satisfied with a few examples to convince themselves that something is always true, and look to proofs to provide rigorous and convincing arguments and justifications.
Can you prove that when you have n consecutive numbers, the difference between the product of the first and last numbers, and the product of the second and penultimate numbers, will be $n-2$?
Below is a proof that has been scrambled up.
Can you rearrange it into its original order?
Click on student solutions to see some different proofs that students submitted.
Extension:
Can you prove that when you have $n$ consecutive even numbers, the difference between the product of the first and last numbers, and the product of the second and penultimate numbers, will be $4n-8$?
Student Solutions
Extension:
Can you prove that when you have $n$ consecutive even numbers, the difference between the product of the first and last numbers, and the product of the second and penultimate numbers, will be $4n-8$?
Sunhari from British School Muscat sent in this proof:
Let the n numbers be: $2(a), 2(a+1),…2(a+n-2), 2(a+n-1)$
First product: $2a(2a+2n-2) = 4a^2 + 4an - 4a$
Second product:
$\begin{split}(2a+2)(2a+2n-4)& = 4a^2 + 4an - 8a + 4a + 4n - 8\\&= 4a^2 + 4an - 4a + 4n - 8 \end{split}$
Difference: $4a^2 + 4an - 4a + 4n - 8 - 4a^2 - 4an + 4a= 4n-8$
Teachers' Resources
Why do this problem?
This problem, along with the rest of the problems in the Proof for All (st)ages feature, provides an excellent context for observing, conjecturing and thinking about proof, and for appreciating the power of algebra.
Possible approach
Key question
Is there a way to represent the products of pairs that will help to explain the patterns you noticed?
Possible support
This problem could also be approached purely numerically, as an exercise in developing fluency with multiplication tables while looking for pattern and structure.
Before embarking on this problem, students could take a look at I Like to Prove It!, which contains a collection of more accessible problems that also challenge students to develop convincing arguments.