In my earlier post Squaring 2 digit numbers – I , I introduced a method that simplified the problem of squaring two digit numbers.

It involved the addition of three numbers, two of which were easy to get and the third required a little mental computation, but would soon get easy after practice.

In this post, I write about a different method that is even faster.

**There is one prerequisite though. You need to memorize the first fifty squares.** I know. Seems ridiculous. But if you think about it, you already know the squares till 30. It doesn’t take long to get the next 20 in memory, trust me.

A two digit square would result in the following four digit answer:

_ _ _ _

What if I told you, once you get the first 50 squares down in memory, that there was a way to **instantly** get the last two digits from memory and the first two digits would require **light-weight computation. **

That is the point towards which mental math tries to converge. Coming up with algorithms where you fill in the digits of the answer with *minimal *computation is the goal.

**EXAMPLE:**

STEP 1: GET THE LAST TWO DIGITS

Immediately get the complement of 67, that is, 33. From memory (Remember, it requires that you know the squares till 50), recall that .

The last two digits of your answer are the same as the last two digits of the square of the complement.

Thus, so far you have:

_ _ 8 9

STEP 2: GET THE FIRST TWO DIGITS

The first two digits of your answer is the first digit squared plus a number. All this step would be doing is figuring out what that number is. So you already know your first two digits is 36 + .

Multiply the two digits of your number and double it. You get 6 x 7 x 2 = 840. You don’t need this entire number. You just need the first digit. If it’s 9 or above 9, your x is this first digit (or first two digits in case of a four digit number) plus 1. If it is below 9, your x is simply that first digit.

Here, we simply do 36 + 8 = 44.

Thus, our answer is :

4 4 8 9

**This is what went through my mind when I was doing the calculation:**

1.) “Okay, 67… complement 33..okay…square is 1089 so… _ _ 8 9”

2.) “Now, 67 is .. 6 x 7 is 42 .. times 2 is 8 hundred something… (below 9).. so 36 + 8 is 44.. okay .. so we have 4 4 8 9“.

**ANOTHER EXAMPLE (as it goes through my head):**

1.) “Okay, 79… complement 21…square is 441 so… _ _ 4 1”

2.) “Now, 79 is .. 7 x 9 is 63 .. times 2 is 12 hundred something… (above 9).. so 49 + 12 + 1 is 62.. okay .. so we have 6 2 4 1“.

P.S. I am providing a more rigorous mathematical formulation for retrieving the first two digits in case it’s unclear from the method above:

FIRST TWO DIGITS

Suppose the problem is . __Your first two digits are + x .__

__To compute x__, first multiply the two digits. do A x B = AB.

1.) If AB > 50, x = 2A + 1

2.) If AB < 50, do 2 x AB

a.) If the first digit of 2AB = 9, x = 9 + 1 = 10.

b.) Else, x = first digit of 2AB.

END