Trick for Squaring 40s (and 50s)
The Trick (I)
Here’s a really cute trick for squaring numbers in the \(40\)s.
$$ \begin{aligned} 40^2 &= 1600 \cr &= [16] \text{ and } [00] \cr 41^2 &= 1681 \cr &= [16] \text{ and } [81] \cr &= [16 + 0] \text{ and } [9^2] \cr 42^2 &= 1764 \cr &= [17] \text{ and } [64] \cr &= [16 + 1] \text{ and } [8^2] \cr &= [16 + \bold{1}] \text{ and } [(10-\bold{2})^2] \cr 43^2 &= 1849 \cr &= [18] \text{ and } [49] \cr &= [16 + \bold{2}] \text{ and } [(10-\bold{3})^2] \cr 44^2 &= 1936 \cr &= [19] \text{ and } [36] \cr &= [16 + \bold{3}] \text{ and } [(10-\bold{4})^2] \cr 45^2 &= 2025 \cr &= [20] \text{ and } [25] \cr &= [16 + \bold{4}] \text{ and } [(10-\bold{5})^2] \cr \end{aligned} $$
and so on! This goes all the way up to 50 and in fact, from 50 we get:
The Trick (II)
$$ \begin{aligned} 50^2 &= 2500 \cr &= [25] \text{ and } [00] \cr 51^2 &= 2601 \cr &= [26] \text{ and } [01] \cr &= [25 + 1] \text{ and } [1^2] \cr 52^2 &= 2704 \cr &= [27] \text{ and } [04] \cr &= [25 + \bold{2}] \text{ and } [\bold{2}^2] \cr 53^2 &= 2809 \cr &= [28] \text{ and } [09] \cr &= [25 + \bold{3}] \text{ and } [\bold{3}^2] \cr 54^2 &= 2916 \cr &= [29] \text{ and } [16] \cr &= [25 + \bold{4}] \text{ and } [\bold{4}^2] \cr 55^2 &= 3025 \cr &= [30] \text{ and } [25] \cr &= [25 + \bold{5}] \text{ and } [\bold{5}^2] \cr \end{aligned} $$
This goes on too! In fact, we can go “backwards” (into the 30s) and “forwards” (into the 60s) but the trick becomes less useful since you have to know stuff like \(13^2\) and what not, which isn’t nearly as fun/fast.
My bet is it has to do with the \(\bold{5}\) in the \(\bold{5}0\)s. Looking at just two examples (\(51\) and \(52\)) we can see that:
$$ \begin{aligned} 51^2 &= (50+1)^2 \cr &= (50^2 + 2*50 +1) \cr &= (50^2 + 2*50 +1) \cr &= (5^2*10^2 + 2*50 + 1) \cr &= (5*5*10*10 + 2*5*10 + 1) \cr &= (5*5*10*10 + 2*5*5*2 + 1) \cr &= (100(5*5 + 1) + 1) \cr &= (100(25 + \bold{1}) + \bold{1}) \cr \end{aligned} $$
and likewise with \(52^2\):
$$ \begin{aligned} 52^2 &= (50+2)^2 \cr &= (50^2 + 2*2*50 + 4) \cr &= (50^2 + 4*50 +4) \cr &= (5^2*10^2 + 4*50 + 4) \cr &= (5*5*10*10 + 2*2*5*10 + 4) \cr &= (5*5*10*10 + 2*2*5*5*2 + 4) \cr &= (100(5*5 + 2) + 4) \cr &= (100(25 + \bold{2}) + \bold{2^2}) \cr \end{aligned} $$
etc. At some point later on I’ll follow up with some justification for this.
Update: demonstrating why this works
Let’s generalize our numbers here, our numbers \(\bold{5}1, \bold{5}2, …, \bold{5}9, \) can be expressed as:
$$ 50 + n $$
(Note that any number can be expressed this way, including say 37, in this case \(\bold{n} = -13\). The formula we are about to derive will work for this case but it won’t be very useful).
Ok so our goal now is to come up with a representation of:
$$ (50 + n)^2 $$
that proves our assertion above. Let’s get started.
$$ \begin{aligned} (50 + n)^2 &= (50+n)(50 + n) \cr &= (50*50 + 50 * n + 50 * n + n^2) \cr \end{aligned} $$
Since \(50 = 2 * 25 = 2 * 5 * 5\), we substitute all occurrences of \(\bold{50}\):
$$ \begin{aligned} (50 + n)^2 &= (50*50 + 50 * n + 50 * n + n^2) \cr &= ((2 * 5 * 5) * (2 * 5 * 5) + 2 * (2 * 5 * 5) * n + n^2) \cr \end{aligned} $$
Let’s rearrange our number sentence a bit (note the bolds):
$$ \begin{aligned} (50 + n)^2 &= ((2 * 5 * 5) * (2 * 5 * 5) + 2 * (2 * 5 * 5) * n + n^2) \cr &= (\bold{2 * 2 * 5 * 5} * 5 * 5 + \bold{2 * 2 * 5 * 5} * n + n^2) \cr &= \bold{2 * 2 * 5 * 5}(5 * 5 + n) + n^2 \cr &= \bold{100}(25+ n) + n^2 \cr \end{aligned} $$
That last line: \(\bold{100}(25+ n) + n^2\) proves our assertion.
Let’s plug a few usecases in:
\(\bold{52}^2\)
$$ \begin{aligned} n &= 2 \cr (50 + n)^2 &= \bold{100}(25+ n) + n^2 \cr &= \bold{100}(25+ (2)) + (2)^2 \cr &= \bold{100}(27) + 4 \cr &= 2704 \end{aligned} $$
\(\bold{44}^2\)
Since \(50 + (-6) == 44\), here \(n == -6 \)
$$ \begin{aligned} n &= -6 \cr (50 + n)^2 &= \bold{100}(25+ n) + n^2 \cr &= \bold{100}(25+ (-6)) + (-6)^2 \cr &= \bold{100}(19) + 36 \cr &= 1936 \end{aligned} $$
\(\bold{24}^2\)
Since \(50 + (-26) == 24\), here \(n == -26 \)
$$ \begin{aligned} n &= -26 \cr (50 + n)^2 &= \bold{100}(25+ n) + n^2 \cr &= \bold{100}(25+ (-26)) + (-26)^2 \cr &= \bold{100}(-1) + 676 \cr &= -100 + 676 \cr &= 576 \cr \end{aligned} $$
^ This particular example shows us that while we can indeed use this formula for all numbers less than and greater than 50, it loses value as a mental math trick if we attempt to apply this for squares of numbers \( < 40 \) or \( > 50 \).