Squaring 2 digit numbers – II

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.






Immediately get the complement of 67, that is, 33. From memory (Remember, it requires that you know the squares till 50), recall that 33^2 = 1089.
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


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 + x.

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):

79 ^ 2

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:


Suppose the problem is (10A+B)^2 Your first two digits are A^2  + 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.



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