# Magic forcing grid – Maths Magic

What’s the effect?

A 4×4 square grid is created on a piece of paper with numbers from 1-16 (see top left hand picture). Spectators choose 4 numbers at random from the grid and the total always equals 34.

What you need?

• Pen and paper

What’s the method?

There was one crucial detail left out from the description of the effect. When the first number is chosen and circled, the remaining numbers on that row and column are crossed out (see the top right picture).

There is now only a choice of 9 numbers for the second spectator to choose from. Again the choice is circled and the row and column is crossed out. For the third choice there are only 4 numbers remaining. The final choice isn’t a choice at all as there’s only one number that hasn’t been circled or crossed out. The total of the 4 chosen numbers is 34. Every time! (see the bottom right picture for one possible example)This can be repeated with different choices to show that it always totals 34.

Classroom investigations

There are a number of extensions that can be made for classroom investigations.

The obvious extension is working out what the total T would be for different grid sizes (nxn). The table below gives the first six results.

There is a general formula that can be set as a task for students to find. The total T is the grid size n multiplied by the central number of the grid. The central number is obvious when odd numbered grid sizes are constructed. However it’s easy to calculate by adding the first and last numbers (n^2) and dividing by two. Hence the term in the brackets of the general equation.

A further investigation could explore the links between this trick and Latin Squares.

## 6 thoughts on “Magic forcing grid – Maths Magic”

• Yes. You’ve prompted me to think about the connection. Each combination of 4 choices on the magic forcing grid must be one of the options of a row or column on a magic square.

• The reason why each combination of n choices on the forcing grid equals the same as the row total on an n by n magic square is because each row of the magic square adds up to the same total, which is the sum of all the numbers on the grid divided by n.

Since there are n^2 numbers in the grid, the total of all the numbers is n^2(n^2 + 1)/2, and you divide this by n to get the total for each row or column in the magic square, which gives us n^2(n^2 + 1)/2n.

If you multiply this out you get
(n^4 + n^2)/2n
which simplifies to (n^3 + n)/2
or n(n^2 + 1)/2
which is the same as your formula for the result of the forcing grid 🙂

1. Reblogged this on Find the Factors and commented:
Mathematics is full of magic, sometimes more than we even expected. There appears to be a connection between this math trick and 4 x 4 magic squares. I will be interested to see if anyone can form a 4 x 4 magic square from the information provided by The Science Magician in this post. It seems like it would still be a lot of work. My daughter is getting married in 5 weeks, so I will resist the temptation to investigate it now.

2. I’ve just thought of another way of looking at all of this and deriving the formula for the magic total T for an nxn forcing grid…

Every selection of n numbers from an nxn forcing grid can be chosen like this:
One of any of the n from the top row, then one from any of the n-1 from the next row (so the column is different from the first choice), followed by any of the n-2 options from the next row, until you get to the last row that only has 1 option. This gives a total of n! different magic choices that all add up to the magic total T.

So the total of all the different choices of n numbers (a maths text book would probably call them n-selections or something like that) added together = n! x T
But every number in the forcing grid belongs to (n-1)! distinct n-selections (because after choosing any particular number as your first choice in an n-selection you have (n-1)! possible ways to complete your selection.)
That means that n! x T is actually (n-1)! times the total of all the numbers in the grid.
But we know that the total of all the numbers in the grid is n^2(n^2 + 1)/2
So we have
(n! x T)/(n-1)! = n^2(n^2 + 1)/2
but n!/(n-1)!=n so this gives us
nT = n^2(n^2 + 1)/2
and therefore
T = n^2(n^2 + 1)/2n

which is your formula… There’s loads of maths in this, isn’t there 🙂