Mental Feats

Michael Curtis (who commented on a previous posting) has written several articles on mental feats involving memory and mathematics.

Coincidently, I too read about the Trachtenberg system (a method for performing mental arithmetic) many years ago.

Reading his site reminded me of a time when I'd relieve boredom during occasions such as high school assemblies by squaring 2-digit numbers in my head using the techniques I had read about.

Today, I'd still use Trachtenberg's method for numbers ending in 5, based on the equation (10 a + 5)^2 = 100 a (a+1) + 25. However I have since found faster methods for other numbers, which I haven't seen described on the web. They only require additions and subtractions, but one has more to memorize.

Let n be the 2-digit number to be squared. Then if n lies in the range:

1. 0-25: Memorize these answers.
2. 25-50: Work out how far n is from 50 and how far n is from 25. Then the answer is 100(n - 25) + (50 - n)^2.
3. 50-75: Compute 100((n-50) + 25) + (n-50)^2. (This is also Trachtenberg's method for squaring fifty-somethings.)
4. 75-100: Compute 100(100 - 2(100-n)) + (100-n)^2

While I'm at it, I'll record a method for finding square roots (of squares of 2-digit numbers):

1. Remove the last two digits of the square. Then the first digit of the answer is the largest digit whose square is less than this number.
2. The last digit of the square tells us what the last digit of the answer could be. If it is 0 or 5, then so is the last digit of the answer and we are done, otherwise:
3. Let the first digit of the answer is a. Compare the square with the square of 10a+5. If larger, then the last digit of the answer is between 6 and 9, and if smaller, it is between 1 and 4. Luckily, in base 10, the squares of 1 to 4 have distinct last digits. Also the squares of a and (10-a) end in the same digit, so it is now easy to determine the last digit of the answer.

The corresponding algorithm for cube roots is much simpler, because the cube of each digit has a distinct last digit (and similarly with other odd powers).