Third Generic Trick for Squaring All Numbers
TL;DR: The Trick.
\(\bold{56}^2\)
$$ \begin{aligned} (56)^2 &= (\bold{55} + 1)^2 \cr &= 3025 + \bold{2}(55) + 1 \cr &= 3025 + 110 + 1 \cr &= 3136 \end{aligned} $$
Note that:
- \(55^2\) can be computed using the ending in 5 trick
- the really important part here is the \(\bold{2}*55\) bit
- this applies generally but becomes less useful the further we get from a number ending in 5.
\(\bold{57}^2\)
$$ \begin{aligned} (57)^2 &= (\bold{55} + 2)^2 \cr &= 3025 + \bold{4}(55) + 4 \cr &= 3025 + 220 + 4 \cr &= 3249 \end{aligned} $$
Note that:
- since we are 2 away from \(55\) here, we multiply by 4.
\(\bold{58}^2\)
$$ \begin{aligned} (58)^2 &= (\bold{55} + 3)^2 \cr &= 3025 + \bold{6}(55) + 9 \cr &= 3025 + 330 + 9 \cr &= 3364 \end{aligned} $$
Note that:
- since we are 3 away from \(55\) here, we multiply by 9.
\(\bold{54}^2\)
$$ \begin{aligned} (54)^2 &= (\bold{55} - 1)^2 \cr &= 3025 - \bold{2}(55) + 1 \cr &= 3025 - 110 + 1 \cr &= 2916 \end{aligned} $$
Note that:
- since we are -1 away from \(55\) here, we multiply by -2.
Demonstrating why this works
Let’s start with a very simple example and then we can expound from there. Say we have a number that ends in \(6\), such as \(46\) or \(56\).
Applying to numbers ending in \( \bold{6} \) only.
We can express this number as:
$$ 10a + 6 $$
where \(a\) represents the digit in the tens place. Now, as we know, we have a shortcut for figuring out how to square numbers ending in 5. Let’s try to use that knowledge as a starting point here. Let’s rewrite our number as:
$$ \begin{aligned} 10a + 6 &= \bigg[\big( 10a + 5 \big) + 1 \bigg] \cr \end{aligned} $$
Ok! This is cool, because we can now express \( (10a + 6)^2 \) as:
$$ \begin{aligned} (10a + 6)^2 &= \bigg[\big( 10a + 5 \big) + 1 \bigg]^2 \cr &= \big( 10a + 5 \big)^2 + 2(1)\big( 10a + 5 \big) + 1 \cr \end{aligned} $$
Now remember, \( 10a + 5 \) is the number we know how to square quickly/easily. So, our problem now simplifies to:
- First, apply the trick for squaring the number ending in 5
- Then, double the number ending in 5
- Finally, add the result from (1) to the result from (2) and then add our third piece, the \( \bold{1} \) to the sum. This gives us our solution/trick for squaring all numbers ending in 6.
Generic solution
Instead of numbers ending in \(6\) let’s now consider any number \(n\) such that \( 0 < n < 10 \). Now we can express our number as:
$$ 10a + n $$
We still want to decompose our number, \( 10a + n \) into some form such that we have a number ending in 5. Let there be some number \(k\) such that \( n = 5 + k \). Then:
$$ \begin{aligned} 10a + n &= \bigg[\big( 10a + 5 \big) + k \bigg] \cr \end{aligned} $$
We’re in business! Note that if we were considering a number ending in 6, then \(k = 1\) as we saw in the previous section. Let’s simplify:
$$ \begin{aligned} (10a + n)^2 &= \bigg[\big( 10a + 5 \big) + k \bigg]^2 \cr &= \big( 10a + 5 \big)^2 + 2(k)\big( 10a + 5 \big) + k^2 \cr \end{aligned} $$
Tada! While it doesn’t look like much, this actually brings us to the examples we saw earlier. Let’s try one of our examples again, but now with our formula applied.
\(\bold{56}^2\)
$$ \begin{aligned} (10a + n)^2 &= \big( 10a + 5 \big)^2 + 2(k)\big( 10a + 5 \big) + k^2 \cr \end{aligned} $$
Here, \( a = 5 \), \( n = 6 \) and \( k = 1 \). Substitute:
$$ \begin{aligned} (10a + n)^2 &= \big( 10a + 5 \big)^2 + 2(k)\big( 10a + 5 \big) + k^2 \cr (10(5) + 6)^2 &= \big( 10(5) + 5 \big)^2 + 2(1)\big( 10(5) + 5 \big) + (1)^2 \cr &= \big( 55 \big)^2 + 2(1)\big( 55 \big) + (1)^2 \cr \end{aligned} $$
This may looks weird / difficult but it’s actually really easy to compute mentally since the ending in 5 trick is so simple. The real drawback to this solution is honestly the middle term, as our number ends in a digit further away from 5, the cognitive load for computing 2 times that number or 4 times that number starts to get more difficult.
Thankfully, there are better solutions to this (that I have stumbled on). More on those in subsequent posts.