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Pair Products

Stage: 4 Challenge Level: Challenge Level:1
Kira, from the UK, started off by doing some calculations:

Say your numbers were 4, 5, 6, 7. When you times the outer numbers 4 and 7 it will equal 28, and when you times the two middle numbers 5 and 6 it will equal 30.

Let's try it again on four new consecutive numbers, such as 1, 2, 3, 4: when you times the outer numbers 1 and 4 it equals 4, and when you times the two middle numbers 2 and 3 it equals 6.

Conclusion: the product of the outer numbers will always be 2 less than the product of the numbers in the middle.

Well spotted! A student from Germany drew rectangles, like in Alison's explanation, to see why this was true. The class at Elgin Academy had the clever idea of testing this for some negative numbers too!

Sam, from Park Grove Primary, gave the following explanation:


If you choose the first number of the set to be n, then the outside multiplication will be n(n+3), which simplifies to $n^2+3n$.
The inside multiplication will be (n+1)(n+2), which simplifies to $n^2+3n+2$.
If you look back at the outside multiplication, you can see that it is 2 less than the inside multiplication, proving the answer.

Great! Nathan also gave a convincing explanation:

Call the number halfway in between the middle two x.
This is the second number in our list plus 0.5, or the third number minus 0.5, so if we multiply the second and third numbers together we will get $x^2-0.25$.
But x is also the first number plus 1.5, or the fourth number minus 1.5, so if we multiply them together we will get $x^2-2.25$.

Nice - here Nathan is making use of the identity: $(x-a)(x+a) = x^2 - a^2$. Well done!

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.

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.

Tom explained it a different way:

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 $9, 10, 11, 12$.

We received a slightly different summary of findings from Aisling, from Grand Avenue Primary School:

The product of the first and last numbers of a series of consecutive whole numbers, is the same as the product of the second and penultimate numbers of that series, minus the number of numbers separating the first and last numbers.

Well done to you all for such clear reasoning.