# A very nice pattern: squaring numbers like 99, 999, 9999, etc

Quick! What’s 92?

I’m assuming (hoping) you rolled your eyes or thought “81” (thanks for playing along). Since that was easy enough, let’s make it more interesting… what about 992? or 9992? or even 999992?

By the end of this post, you will be able to rattle off the square of any number like this just as quickly as you think of the answer from the good ole times table. You’ll be the life of every party! You will simply dazzle others with your math magic! Yes I’m done.

It turns out that squaring numbers where the digits are all 9 follows a very predictable pattern. While there are patterns when multiplying any number by a number made up of only 9s, this pattern is much easier to remember and in my opinion, much more interesting.

## The pattern

Let’s just look at a few squared values to get an idea of what is going on.

 92 81 992 9801 9992 998001 99992 99980001 999992 9999800001

Based on this small sample, it seems you can square a number consisting of n 9s by writing (n-1) nines, 8, (n-1) zeros, and then 1. You can spot check some larger values and see the pattern continues. For example: 9999999999= 99999999980000000001.

Applying this, is someone randomly asked you “hey – what’s 992?”, you would mentally note that there are 2 digits, and so the answer would have one 9, 8, one zero, and then 1: 9801.

## Why does this work

As you know, simply showing a few examples is not a proof right? I’ve only shown that this pattern holds for these specific examples. A proof would show that it holds generally, without us having to type numbers into wolfram alpha to check for the rest of time. To understand the proof though, I will start by showing why it holds for one particular case: 99992.

Consider the definition of squaring. You know that 32 is just the product of 3 and 3. Naturally, the same is true here so: $9999^2 = 9999 \times 9999$

But what is multiplication really? It is repeated addition. With something like $3 \times 5$, we are saying “add 3 to itself 5 times”. Using this, you could write $3 \times 5$ as $3 + 3 + 3 + 3 + 3$ or, if you wanted, $3 \times 4 + 3$. Applying that here: $9999 \times 9999 = 9999 \times 9998 + 9999$

Now here is the “trick”. I’m going to add and subtract 9998. The way I will do that is by changing $9999 \times 9998$ to $10\,000 \times 9998$. This will mean I now have ten-thousand 9998s being added, which is too many, so I will subtract one of them from 9999. $9999 \times 9998 + 9999 = 10\,000 \times 9998 + 9999 - 9998$

Simplifying this, we get: $9999^2 = 10\,000 \times 9998 + 1$

And how do you multiply a value and 10,000? You write the value with four zeros following it. Here, you will then add 1. This will give us our pattern of three 9s, 8, three 0s, and 1.

## Generalized

The same technique will work for any number made up of all 9s. We just have to keep track of the number of digits along the way. Following the steps we used above: \begin{aligned}\underbrace{999 \cdots 9^2}_\text{n digits} &= 999 \cdots 9 \times 999 \cdots 9\\&= 999 \cdots 9 \times 999 \cdots 8 + 999 \cdots 9\\&= \underbrace{1000 \cdots 0}_\text{(n + 1) digits} \times 999 \cdots 8 + 999 \cdots 9 - 999 \cdots 8\\&= \underbrace{1000 \cdots 0}_\text{(n + 1) digits} \times 999 \cdots 8 + 1\\&= \underbrace{999 \cdots 8}_\text{n digits} \underbrace{000 \cdots 0}_\text{n digits} + 1\\&= \underbrace{999 \cdots 8}_\text{n digits} \underbrace{000 \cdots 1}_\text{n digits}\\&= \underbrace{999 \cdots 9}_\text{(n - 1) digits} 8 \underbrace{000 \cdots 0}_\text{(n - 1) digits} 1\\\end{aligned}

The number of digits is easy to keep track of since when we add 1, we know that we just went up to another digit (999 + 1 = 1000 which has one more digit than 999). I’m not sure this would be as simple when squaring similar types of numbers like 777, 888 etc.