Squaring Two-Digit Numbers (Redux)

If you want to square a two-digit number, you could just use a calculator, or you could use the traditional algorithm. I’m going to talk about a different method here, but not because I think this is a particularly useful method. The point of this discussion is to look at how numbers are interrelated; if you want to use it to square any two-digit number, fine, but that’s not the goal.

First, you’re going to need all the squares of one-digit numbers, so here they are:

\(x\)	1	2	3	4	5	6	7	8	9
\(x^2\)	1	4	9	16	25	36	49	64	81

I’ve seen a pair of “tricks” online about squaring multiples of five. Both of these just follow from basic patterns in mathematics.

With regards to the squares of multiples of five, there are two such cases: Multiples of ten, and the odd multiples of five (15, 25, 35, …).

For the multiples of ten (10, 20, 30…), take the first digit and square it, then append two zeroes on the end. For instance, \(30^2 = 900\) and \(60^2 = 3600\). In full, this gives us these:

\(x\)	10	20	30	40	50	60	70	80	90	100
\(x^2\)	100	400	900	1600	2500	3600	4900	6400	8100	10000

The mathematics behind this is that a multiple of ten is \(10k\), and \((10k)^2 =\) \(10^2 \cdot k^2 =\) \(k^2 \cdot 100\).

The second case involves odd multiples of five, and this is the one I most often see presented as a “trick” or a “hack”. Again, it just follows from the basic patterns in mathematics.

First, the process: Take the first digit, square it, and add the result to the original number; alternatively, take the first digit and multiply it by one larger. Then append 25 on the end. For instance, consider 35. \(3^2 + 3 = 12\), so \(35^2 = 1225\). Consider 65. \(6^2 + 6 = 42\), so \(65^2 = 4225\). In full, this gives us these:

\(x\)	15	25	35	45	55	65	75	85	95
\(x^2\)	225	625	1225	2025	3025	4225	5625	7225	9025

So why does this work? An odd multiple of five is \(10k + 5\), and \((10k+5)^2 =\) \((10k^2 + 2\cdot 10k\cdot 5 + 5^2 =\) \(100k^2 + 100k + 25 =\) \(100(k^2 + k) + 25 =\) \(100(k)(k+1) + 25\).

What about the rest of the two-digit numbers? All other integers are within two of a multiple of five.

We’re going to implement what’s called the Difference of Squares rule. The Difference of Squares rule says that, given two numbers \(a\) and \(b\), \(a^2 – b^2 = (a + b)(a – b)\).

If we know one of the squared values, we can calculate the other squared value. For instance, if \(a > b\), then \(a^2 = b^2 + (a + b)(a – b)\).

That’s a bunch of algebraic notation, so let’s look at some concrete examples.

Here’s the general algorithm:

1. Take a two digit number.
2. Find its closest multiple of five.
3. Add the two numbers, then multiply by the distance (1 or 2).
4. Add or subtract this (as appropriate) from the squared multiple of five.

For example (adding in step 4 because 60 < 61):

1. Consider 61.
2. Use 60.
3. \((60 + 61)\cdot 1 = 121\)
4. \(3600 + 121 = 3721\)

For example (subtracting in step 4 because 65 > 63):

1. Consider 63.
2. Use 65.
3. \((63+65)\cdot 2 = 128 \cdot 2 = 256\)
4. \(4225 – 256 = 3969\)

This algorithm is probably not much easier to do mentally than the traditional algorithm. As I said from the outset, the purpose of this is not to provide an easy trick for multiplying two-digit numbers; it’s to look at how numbers are interrelated and to explore the patterns inside.

For the purposes of mental mathematics, if one insists, I might recommend a modified version. Ignore the odd multiples of five; just use the multiples of ten:

1. Take a two digit number.
2. Find its next lower multiple of ten.
3. Add the two numbers, then multiply by the last digit (up to 9).
4. Add this from the squared multiple of ten.

This strikes me as less mentally cumbersome, but it relies in my perception more on brute force and misses a little bit of beauty. But here’s an example:

1. Consider 67.
2. Use 60.
3. \((60 + 67)\cdot 7 = 127 \cdot 7 = 889\)
4. \(3600 + 889 = 4489\)

Compare this to the approach above:

1. Consider 67.
2. Use 65.
3. \((65+67)\cdot 2 = 132 \cdot 2 = 264\)
4. \(4225 + 264 = 4489\)

Both methods return the same value. We could also use a “closest ten” algorithm for all but the multiples of five, which would mean we never multiply by more than four but we would then also need to subtract:

1. Consider 67.
2. Use 70.
3. \((70+67)\cdot 3 = 137 \cdot 3 = 411\)
4. \(4900 – 411 = 4489\)

And are any of these really easier than just multiplying 67 by 67 directly? I can’t answer that because that’s not even the goal of this article.

The purpose of this article is to explore how flexible approaches to mathematics will lead to the same results because numbers have internal consistency and structure.