Once in a while, the newspaper talks about youngsters or adults who have the seemingly incredible ability to perform arithmetic computations in their head for large integers. Actually, this all boils down to a basic understanding of algebra and algorithmic thinking. Yes, we can all do it. Just to give you an example, suppose we wish to know 35×35. An algorithm for getting the right answer says:

“the last two digits are 25; and the digits in front are equal to the product of the first digit with itself added 1”

This gives the answer 1225, since the first digit is 3, and 3×4 = 12. To see why this algorithm is true, we first look at the simple case where there is only one digit in front of 5. The product of two integers with 5 as the last digit and a as the first digit can be written as:

a5 x a5 = (a x 10 + 5)(a x 10 + 5) = 100a^2 + 50a + 50a + 25 = 100a(a+1) + 25

The first term has two zero digits at the end, so that when 25 is added to it, it is clear that the last two digits are just 25. Finally, the digits in front are just a(a+1).

Now you should be able to figure out an algorithm for multiplying something like 85 x 35, and perhaps, later, more general things like 29 x 29, and finally 73 x 48. Have fun!

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## About Tsung Fei

A teacher, researcher in the bioinformatics division at the University of Malaya

The 25 thingy is an old trick lols. Well, algorithm is algorithm. But some people still does better and faster than others.

As for 29^2 the algorithm I use is a^2 – b^2 = (a+b) (a-b)

a^2 = (a+b)(a-b) + b^2

29^2 = (29+1)(29-1)+1^2

= 30(28) + 1

Easily 841.

You have a different algorithm?

I think once the algorithm is known, the excitement dies down … and one moves on to do other things.

My algorithm for doing 29 x 29 is this:

Take the square of the first digit (2^2 = 4), and add to it the product of the two digits in the integer divided by 5 (2×9 / 5 = 18 / 5 = 3.6) to get 7.6. Move the decimal two places behind to get 760. Finally add to 760 the square of the second digit (9^2 = 81). This give 760 + 81 = 841.

This algorithm can be generalised for arbitrary integers.

This is so cool. Never knew maths can be this fun. Still haven`t figured out how the algorithm came along though.

Long live algorithmic thinking!

Try this for squares where the last digit is 5: we can write (say) 25 as 2×10 + 5. So 25×25 is (2×10 + 5) (2×10 + 5). You should be able to figure out the rest … once you understand the specific, move to the general case of a5 x a5. This gives you the algorithm.

As you can see, these things involve only elementary algebra.