TAGS: math just for fun

Proving the Squaring Numbers Ending in 5 Trick

September 4, 2022

So here’s a neat trick. What’s a quick way to square numbers that end in — 5 in your head?

For example, what is $$25*25 = ?$$

(Plug into a calculator real quick). We expect our trick to help up compute the solution, 625, quickly.

Ok, let’s break down the approach.

Example: what is 25^2?.

First, let’s re-write our number (25) in a slightly different but equivalent manner:

$$ \begin{aligned} 25^2 &= (20 + 5)^2 \cr &= (\bold{2} * 10 + \bold{5} * 1)^2 \cr \end{aligned} $$

Here, we have decomposed the number 25, as we can see:

2 is in the tens place
5 is in the ones place

To quickly arrive at the square of 25, we:

1. Let t= the the number in the tens place, 2. Let p= t+1, or 3.

$$ \begin{aligned} \bold{Let: } \cr t &= 2 \cr p &= t + 1 \cr \bold{Then: } \cr p &= (2) + 1 \cr p &=3 \end{aligned} $$

2. Now, let k= **t * p**, or 6.

Next, we multiply t and p together:

$$ \begin{aligned} k &= t * p \cr &= 2 * 3 \cr &= 6 \cr \end{aligned} $$

3. Take k, or 6 and put it in front of 25, which gives us 625. Done!

For the last part, concatente k and 25 to arrive at our answer – 625.

Tada!

But WHY tho??

This method works - for all numbers ending in 5 that we would like to square. We will prove this below.

Suppose we have some number a that ends in 5. We may represent a as:

$$ \begin{aligned} \text{Let } \cr a &= (10 * \bold{t} + 1 * \bold{n}) \cr \text{Then } \cr a^2 &= (10 * \bold{t} + 1 * \bold{n}) * (10 * \bold{t} + 1 * \bold{n}) \cr &= (10t +n)(10t + n) \end{aligned} $$

NOTE: we assume here that t is an integer greater than or equal to 0.

Now, we can expand this number sentence by FOIL, but we will not immediately simplify the terms:

$$ \begin{aligned} a^2 &= (10t +n)(10t + n) \cr &= 10 * 10 * t * t + 10 * t * n + 10 * t * n + n * n \end{aligned} $$

Remember, we only care about numbers ending in –5, so n=5. Let’s substitue:

$$ \begin{aligned} n &= 5 \cr a^2 &= 10 * 10 * t * t + 10 * t * (5) + 10 * t * (5) + (5) * (5) \cr &= 10 * 10 * t * t + 10 * t * (5) + 10 * t * (5) + \bold{25} \cr \end{aligned} $$

As we can clearly see: the mystery of why the square of any number ends in 25 is (kinda) solved. Onwards!

We will now group our terms like so:

$$ \begin{aligned} a^2 &= \Bigg[ 10 * 10 * t * t + 10 * t * (5) + 10 * t * (5) \Bigg] + \bold{25} \cr &= \Bigg[ \bold{10} * 10 * t * t + \bold{10} * t * (5) + \bold{10} * t * (5) \Bigg] + \bold{25} \end{aligned} $$

Let’s pull out the 10:

$$ \begin{aligned} a^2 &= \bold{10}\Bigg[ 10 * t * t + t * (5) + t * (5) \Bigg] + \bold{25} \end{aligned} $$

Ok cool, next – we will factor the terms inside the large brackets:

$$ \begin{aligned} a^2 &= \bold{10}\Bigg[ 10 * t * t + t * (5) + t * (5) \Bigg] + \bold{25} \cr &= \bold{10}\Bigg[ \bold{(5)} * 2 * t * t + t * \bold{(5)} + t * \bold{(5)} \Bigg] + \bold{25} \cr &= \bold{10}\Bigg[ \bold{5}\bigg[ 2 * t * t + t + t \bigg] \Bigg] + \bold{25} \cr &= \bold{10}\Bigg[ \bold{5}\bigg[ 2 * t * t + 2 * t \bigg] \Bigg] + \bold{25} \cr &= \bold{10}\Bigg[ \bold{5} * \bold{2 * t}\bigg[ t + 1 \bigg] \Bigg] + \bold{25} \cr &= \bold{10}\Bigg[ \bold{10} * \bold{t}\bigg[ t + 1 \bigg] \Bigg] + \bold{25} \cr &= \bold{100}\Bigg[ \bold{t}\bigg[ t + 1 \bigg] \Bigg] + \bold{25} \cr \end{aligned} $$

Let’s look at that last representation once again:

$$ a^2 = \bold{100}\Bigg[ \bold{t}\bigg[ t + 1 \bigg] \Bigg] + \bold{25} $$

Recall that t is the number in our tens place and this result demonstrates that the square of a (where a ends with 5) is always equal to the number in the tens place, times the number in the tens place plus 1. Because this value, t(t+1) is multiplied by 100, we are guaranteed that this number will always end in 25.

Donezo!

Some quick examples.

Let’s put this to the test!

5^2 = 25

$$ \begin{aligned} a &= 5 \cr a &= (10 * \bold{t} + 1 * \bold{n}) \cr t &= 0 \cr n &= 5 \cr a^2 &= \bold{100}\Bigg[ \bold{t}\bigg[ t + 1 \bigg] \Bigg] + \bold{25} \cr 5^2 &= \bold{100}\Bigg[ \bold{0}\bigg[ 0 + 1 \bigg] \Bigg] + \bold{25} \cr 25 &= \bold{100}\Bigg[ 0 \Bigg] + \bold{25} \cr &= 0 + \bold{25} \cr &= 25 \end{aligned} $$

55^2 = 3025

$$ \begin{aligned} a &= 55 \cr a &= (10 * \bold{t} + 1 * \bold{n}) \cr t &= 5 \cr n &= 5 \cr a^2 &= \bold{100}\Bigg[ \bold{t}\bigg[ t + 1 \bigg] \Bigg] + \bold{25} \cr 55^2 &= \bold{100}\Bigg[ \bold{5}\bigg[ 5 + 1 \bigg] \Bigg] + \bold{25} \cr 55^2 &= \bold{100}\Bigg[ \bold{5}\bigg[ 6 \bigg] \Bigg] + \bold{25} \cr 3025 &= \bold{100}\Bigg[ 30 \Bigg] + \bold{25} \cr &= 3000 + \bold{25} \cr &= 3025 \end{aligned} $$

105^2 = 11025

$$ \begin{aligned} a &= 105 \cr a &= (10 * \bold{t} + 1 * \bold{n}) \cr t &= 10 \cr n &= 5 \cr a^2 &= \bold{100}\Bigg[ \bold{t}\bigg[ t + 1 \bigg] \Bigg] + \bold{25} \cr 105^2 &= \bold{100}\Bigg[ \bold{10}\bigg[ 10 + 1 \bigg] \Bigg] + \bold{25} \cr 105^2 &= \bold{100}\Bigg[ \bold{10}\bigg[ 11 \bigg] \Bigg] + \bold{25} \cr 11025 &= \bold{100}\Bigg[ 110 \Bigg] + \bold{25} \cr &= 11000 + \bold{25} \cr &= 11025 \end{aligned} $$

A beautiful outcome.

Honestly, this observation strikes me as exceedingly pretty. We are only able to end up with the t(t+1) result because n=5. With n=5, we are able to factor out our number sentence from above (reproduced below for convenience):

t(t+1) is only possiblebecause n was 5 . For instance, if n was 6, we would end up with:

$$ \begin{aligned} a^2 &= \bold{10}\Bigg[ 10 * t * t + t * (6) + t * (6) \Bigg] + \bold{36} \cr &= \bold{10}\Bigg[ \bold{(5)} * 2 * t * t + t * \bold{(6)} + t * \bold{(6)} \Bigg] + \bold{36} \cr &= \bold{10}\Bigg[ \bold{(5)} * 2 * t * t + \bold{12} * t \Bigg] + \bold{36} \cr \end{aligned} $$

With n=6 (as an example), we end up with terms that are cannot be easily simplified, which is why (AFAIK) no simple solution for squaring numbers that don’t end in 5 exists in the same way. (Heads up, I did find some interesting results for a more general case; will share that methodology in a future post).

In short:

It’s just a happy coincidence! 🎉 🎈🎊 🙌