Ok so for this one, no proof (at least not yet). But, we will demonstrate by example.

Squaring \( \bold{30} \)s

Let’s begin here.

So far, there’s no real pattern here. But! What if we decomposed our numbers a bit? eg:

Ok, bear with me here. The important parts are the middle and right columns. Suppose our number ( \( 30, 31, 32, …, 39 \)) is expressed as: \( 10a + n \) where \( a = 3 \) and \( n = 0,1,2,3…,9 \)

Then, it is clear that:

• the middle column can be expressed as: \( 2 * n * a \)
• the right colume can be expressed as: \( n^2 \)

If we look specifically at:

we notice that this works itself out quite nicely! \( 31^2 = 961 \) and our left, middle and right columns concatenate to this value as well. Great! But what about the rest…?

As it turns out, they also work provided we move any “extra” digit over. For instance, consider:

Here, middle is \( 12 \). If we “carry” the \(1\) from \(12\) over to the left column, we end up with:

As we can clearly see, this results in \( 1024 \) which is in fact \( 32^2 \). This principle applies across the board, for instance, consider:

Here, we first carry over the \( 3 \) in right over to middle.

Then, we do it again, carrying over \( 3 \) in middle to left:

As we can clearly see, this results in \( 1296 \) which is in fact \( 36^2 \).

Generically

Ok so let’s generalize this (we already kinda did). Given some number \( 10a + n \) where \( a, n \) are natural numbers:

And then ofc, we have to carry over any values in middle and right that are not digits.

Ex: \(\bold{78}^2\)

Here:

• \( a = 7 \)
• \( n = 8 \)

We first carry over from the right

Ok and now we do so again from middle:

Tada! Again, doesn’t work great for larger numbers but for squaring anything in the \( 20 \)s and \( 30 \)s, this is probably a very quick solution for solving mentally.

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