TomĀ drew rectangles, like in Alison's explanation:
If we overlap the two rectangles we cut a piece off at the bottom and create a new piece at the right hand side. The difference between the two small rectangles is always two. This works for all numbers, not just $9, 10, 11, 12$.
Keone, from Sage Ridge School in Reno, started with four consecutive whole numbers:
The product of the first and last numbers is always $2$ less than the product of the middle two numbers.
Explanation: Suppose the first number is $x$. Then the second number is $x+1$, the third is $x+2$, and the fourth is $x+3$.
So the product of the first and fourth numbers is $x(x+3) = x^2 + 3x$.
Also, the product of the second and third numbers is $(x+1)(x+2) = x^2 + 3x + 2$.
So $(x+1)(x+2) = x(x+3) + 2$ for any chosen value of $x$.
When he considered five consecutive whole numbers he found that:
The product of the first and last numbers is always $3$ less than the product of the second and fourth numbers.
Explanation: Again, let the first number be $x$; then the second number is $x+1$, the third is $x+2$, the fourth is $x+3$, and the fifth is $x+4$.
So the product of the first and last numbers is $x(x+4) = x^2 + 4x$.
Also, the product of the second and fourth numbers is $(x+1)(x+3) = x^2 + 4x + 3$.
So $(x+1)(x+3) = x(x+4) + 3$ for any chosen value of $x$.
And with $n$ consecutive whole numbers he found that:
The product of the first and last numbers is always $n-2$ less than the product of the second and penultimate numbers.
Let us consider the general case where there are n consecutive whole numbers.
As before, let the first number be $x$; then the second number is x+1, the third is x+2, and so on.
The last number (the $n$th number) will be $x+n-1$. Thus, the penultimate (second-to-last) number will be $x+n-2$.
So the product of the first number and the last number will be $x(x+n-1) = x^2 + nx - x$
The product of the second and the penultimate numbers will be $(x+1)(x+n-2) = x^2 + nx - 2x + x + n - 2 = x^2 + nx - x + n - 2$
So $(x+1)(x+n-2) = x(x+n-1) + n - 2$;
that is, the product of the second and penultimate numbers will always exceed the product of the first and last numbers by exactly $n - 2$.
For example, if we take 6 numbers, the product of the $2$nd and $5$th numbers will be $4$ more than the product of the $1$st and last numbers.
We received similar findings from Aisling, from Grand Avenue Primary School:
Natasha, from the European School, generalised her findings in a similar way and then went on to check her conclusion:
We can generalise this problem by substituting particular numbers with letters.
Let the first number be $a$.
If there are $n$ numbers, where a is the first, then the last number is $(a+n-1)$.
The second number is $(a+1)$, the penultimate number $(a+n-2)$.
By multiplying the first and last numbers together, we get $a(a+n-1) = a^2 +an-a$
Multiplication of the second and penultimate numbers gives
$(a+1)(a+n-2) = a^2+an-a+n-2$
The difference therefore in the product of the first and last numbers and the product of the second and penultimate numbers is always $n-2$.
For added confirmation we can take a random example:
Consider the numbers $56, 57, 58, 59, 60, 61, 62$
where $n = 7$ and $a = 56$
Our general formula tells us that the difference in the product pairs should be $n-2$ (i.e. $5$).
When we do the calculation, we get
$$57 \times61 - 56 \times62 = 3477 - 3472 = 5$$
This result corresponds with the general one established above.
Well done to you all for such clear reasoning.