# Mental arithmetics: the square of a two-digit number

Every once in a while I strain to the maximum my capacity to botch a simple explanation. This time around, I apply it to a method to calculate mentally the square of a two-digit number. Actually, I know two of these: one of them, I have known for years; the other one I learnt on the Internet some months ago following a link to a Wikipedia article on Vedic Mathematics (whose name, I've heard, is purely for marketing :)).

The first method I knew is based on the fact that (a+b)^{2}=a^{2}+b^{2}+2ab. If I call the tens part “a” and the units part “b”, it's just a matter of squaring two single-significative-digit numbers (easy), adding up the results (easy) and then multiplying each number by the other and doubling the result and adding it up to the previous result (less easy, but still feasible).

For example, 37 is 30+7, so 30^{2}=900, 7^{2}=49, 2·30·7=420, I add it all together and it gives 1369.

You can quickly see that the problem with this system is having to remember two or three intermediate results while you multiply three numbers mentally. We just can't fit that many numbers in our short-term memory :)

The system I learnt recently is harder to explain, but it's quicker to calculate because we don't have to hold as many intermediate results in our mind.

This method is based on (a+b)(a-b)=a^{2}-b^{2}. What we do is call the number we want to square “a”, and choose a number “b” which will make the product (a+b)(a-b) easy to calculate (most commonly, we'll choose the smallest number that will make a+b or a-b a multiple of 10). When we've chosen that number, we can calculate the product, add b^{2} to the result and what we get is a^{2}.

Let's see this with the previous example where we wanted to calculate 37 squared. Using 3 as the value of “b”, we can calculate (37+3)(37-3)=40·34=1360. To this we add 3^{2}, which is 9, and we get 1369.

Another example: 72^{2}. If I choose 2, I've got (72+2)(72-2)=74·70=5180. I add 2^{2}=4, and I get 5184.

As you can see, with this system it's only necessary, usually, to multiply a two-digit number with a single-digit one (zeroes don't count), square a single-digit number and add it to the previous result. I don't think there's any faster method to calculate the square of a two-digit number (if you don't count memorising them all, of course).