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# A Quick Arithmetic Trick

By on 01/02/2016

For the purposes of this short text, by number we mean a positive integer. The routine is as follows: You ask a spectator to choose which of two numbers, say 6 and 11, he/she prefers, but not to tell you his/her choice. Then tell him/her to multiply the number he/she prefers by, say, 3 and the other by, say, 2, add the two products and tell you the result. You immediately know which number (6 or 11) was the sprectator’s favourite.

How? You may have noticed the many “say”-s in the preceding paragraph. This is because both for the two numbers among which the spectator has a choice you can choose any two numbers, as long as one is even and the other is odd, as you can for the two multipliers (3 and 2 in our example). The only important thing is that you instruct that the spectator multiplies the chosen number by the odd multiplier and the other by the even multiplier, and then to add the products.

The secret is in the arithmetic of even and odd numbers. A product of an even number with any other number is even, and the product of two odd numbers is odd. Also, the sum of two numbers of different parity is odd, while the sum of two numbers with same parity is even. Thus, if the spectator chooses the even number, he/she will calculate even x odd + odd x even = even + even = even, i.e. will obtain an even result, while if he/she chooses the odd number, he/she calculates odd x odd + even x even = odd + even = odd. To put it short: the parity of the result is the same as of the chosen number. In our example, if the result is 40, number 6 was chosen in the beginning, while if the result is 45 the chosen number was 11. That’s all, folks!