 # How to calculate squares quickly

Let’s focus on calculating squares.
They can be used to speed up regular multiplication, like we discuss in the difference of squares thread.

The tens are easy, 10 squared = 100, 20 squared = 400, 30 squared = 900, etc.

The fives are easy:
35 squared is 3 times 4 times 100 and add 25 = 1225.
45 squared is 4 times 5 times 100 and add 25 = 2025.
55 squared is 5 times 6 times 100 and add 25 = 3025.
In other words take the first digit, multiply this by one more and concatenate ‘25’ to the result.

We see that we can easily do the tens and the fives, so all that is left to do are the numbers in between.
This is easier than it might sound. Let’s start with an example:

31 squared. If we start by calculating 30 squared = 900, then to get to 31, all we need to do is add 30 and then 31.
Alternatively we can double the number we just squared (2X30=60), add it to our result (add to 900 = 960) and add one (961).
The latter is what I do most of the time. I hope you see how easy this is.

Since these are the very numbers you already have in your mind when you do this calculation, this keeps the amount of digits to keep in your mind to a bare minimum and makes it a great technique for mental calculation.

Another example 36^2.
Start with 35^2=1225. Then double the 35 to get 70. This makes 1295, add one to get 1296.
Done.

We can go back.
34^2. Start with 35^2=1225, again double the 35 to get 70. Now instead of adding 70, we now subtract 70. 1225 - 70 = 1155, again add one to get 1156.

This is based on (a + b)^2 = a^2 + 2ab +b^2
We used b=1 in the previus examples and b=-1 for the example of going back.
If b = -1, when we go back one step we get a^a - 2a + 1
This is the reason we subtract 70 and add 1 in the previous example.

Tip: use small steps. For 36 squared we could start with 30 and use a step of 6 to get the following

intermediate results:
30^2 = 900
3X6X2 = 360. Add to 900 = 1260
6^2=36. Add to 1260 to get 1296.

Same result. However, I do 1225 + 70 + 1 a lot quicker than 900 + 360 + 36.

For 37 we can evaluate 35, then calculate 36, then move to 37.
This is great training the squares. Just start at a random number, square it and do the next, etc.

For 37, instead of first caclculating 36 and then 37, we can do 2 steps at a time.

We can use (a + b)^2 = a^2 + 2ab +b^2 with 35 for a and 2 for b.
37^2 =
(35 + 2)^2 =
35^2 + 2X2X35 +2^2 =
1225 + 4X35 +4 =
1225 +140 + 4 =
1365 + 4 = 1369.
In my head I start with 1225, then do 2 times 35 = 70, then double it to get 140, then add to 1225, then

If you cannot work with the formula, try this:
For one step forward add the number itself, then add the following number.
So for 36^2, if we know 35^2 we add first 35, then 36. So we now have added 35 twice and then 1.

For two steps forward, see this example.
For 37^2 if we know 35^2, we add first 35, then 36 (to go from 35 to 36), then 36, then 37 (to go from 36 to 37). So we add, 35, 36, 36, and 37.
If we add 35 four times, then there is 4 left to add.

It is a lot of text to write down, but once you get the hang of it, it goes really quick.

6 Likes

It’s a shame we only have a like button, because this post is 5 out of 5 stars!

The only way this could be better (five-and-a-half stars) is if you’d have called if squares with binomial formulas. It’s pretty clear that you are using the first and second formula when adding and subtracting, but even the numbers ending in 5 are done with the third:

\color{blue}{(a+b)^2=a^2+2ab+b^2}
\color{blue}{(a-b)^2=a^2-2ab+b^2}
\color{blue}{(a+b)*(a-b)=a^2-b^2}

For those interested, the nice thing here is how this is done in a mutually exclusive and collectively exhaustive fashion. If you are not into Vedic math and creatively picking the algorithm that you feel best suited… go with this method:

## Blocks of 10

As described, use your knowledge of single digit squares and add two zeros

## Split them in half

Use your knowledge of single digit multiplication tables for the left-hand-side and put 25 on the right

## Up and down

Worst case scenario is now stepping over 4 unknowns to get to the next known: 20 ? ? ? ? 25 ? ? ? ? 30, etc. However you can cut that again in half by seeing it as either 2 up or 2 down.

## One after the other

The first formula basically becomes the second formula if you plug in a negative number, so think of it this way rather than subtracting a positiv number. So again logically cut in half, there is only 1 up/down or 2 up/down now.

## Conclusion

The first two steps come from knowledge you already have. The next two is just looking at the problem from the right angle. Finally, if you do the double-double thinking

it really just comes down to doubling numbers and adding 1 or 4. The entire setup is very elegant!

This one should be way up in the Mental Math category

3 Likes

Again; excellent post!
I have nothing to add and thank you for the compliments!

Some nice calculation topics this week. When I finally get my memory system sorted it will be nice to come back to the suffering. I have a bunch of remedial math books to read this winter. Who knows maybe I will get over the hump this time round. There are so many pretty stones to play with it is hard to stick with any one very long.

Credit where credit is due…

Thanks, but just providing a commentary for people that may need help synthesizing the information… merely paraphrasing what you’ve already said.

I have been looking for this. Thank you!

Any Calendar calculation or Square root topic?

I have always been in love with this kind math.

Just search the forum for both…

Calendar is simply: century code + year code + month code + day and adjust for leap year if needed. I got an in-app tutorial here: https://itunes.apple.com/app/id1225480327

Square roots have shortcuts if you are talking about perfect squares. Otherwise search the forum (or google) for duplex method… I believe @Kinma wrote a post about it.

Do a search on this forum. We have talked about square roots and calendar calculations a lot.