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# How many candies do you have?

By on 01/02/2016

You can perform this trick with any number of any kind of small items, say candies in a bag. You should have a sufficient supply of them. Besides that, you – the mathemagician – also need two volunteers for the trick; we shall call them Alexander and Barbara. If you are not sure that your volunteers are good in mental multiplication of small numbers, have a pen and paper or a calculator ready for them.

## First, prepare

First of all, you secretly choose two numbers (positive integers), add 1 to the first of them and multiply this number with the second and remember the result. For example, if you have chosen the numbers 6 and 2, you calculate (6+1)⋅2=14. Remember 14 and you are prepared to do the trick!

## Now it’s time to do the trick…

First you instruct Alexander to take any number of candies from the bag. You must turn away so that at no moment you see how many candies who of the two volunteers has. Now, Alexander is to show Barbara his candies and she should count them and then she should take 6 times as many from the bag. For example, if Alexander took 8 candies, she should take 48.

Now, tell Alexander to give Barbara 2 more from his candies (in the described example this leaves Alexander with 6 and Barbara with 50 candies). He may be grumbling now because Barbara has so many candies, but the next step will make Barbara grumble: tell her to give to Alexander from her candies 6 times as many as he still has. In our example this would mean Barbara giving 6⋅6 = 36 candies to Alexander.

Finally, its your turn to make a dramatical announcement that you know how many candies Barbara still has – the number is 14. Of course, if you chose some other two numbers than 6 (for the two multiplication instructions) and 2 (for the one subtraction), the number of candies Barbara will end up is the corresponding number you have calculated in the beginning.

## Finally, think about the maths

The maths behind this trick is, as often, simple algebra: the instructions are designed so that the number Barbara ends up with does not depend on the number of candies Alexander chooses in the beginning. If you closely follow the described procedure you should be able to work it out, but in case you prefer to have it all spelled out, here it is:

Denote your first number (the one used for multiplication) by m and the other by s; consequently the number to be remembered for the final disclosure is x = (m +1)n. Let n be the number of candies taken by Alexander in the first step. Then in the second step Barbara takes mn candies, in the third step she receives s candies from Alexander, so after this step he still has n – s candies and she mn+s. In the final step Barbara gives m(ns) candies to Alexander, so she ends up with mn+sm(ns) = s + ms = x candies, independently of the numbern of candies Alexander started with (while he ends up with n – s + m(ns) = (ns)(m + 1) candies, and since the number depends on n, the mathemagician cannot guess it without knowing that number).

Now, a problem for you: How many candies should you have in the bag in the beginning if you do not want to forbid Alexander taking more than 10 candies? Work out the solution for the suggested numbers 6 and 2 and also for the general case, and then generalise for other upper limits than 10.

Disclaimer: This trick is not mine in origin, but I cannot remember where I have picked it up.