Why does (25)(25) = 2(2+1)(100)+25? An explanation.

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 give us: 625.

Now, let’s explore how our “trick” (more of an algorithm, really) works:

First, we take the number in the tens place.

In this case — 2.

Add 1 to the number in the tens place.

So 2+1, which is 3.

Multiply our first number (2) by the second number (3).

Ok, so 2 * 3 = 6.

Great! Almost there, hold 6 in the back of your head for a bit.

Take the number in the ones place (5) and square it.

5^2 = 5 * 5 = 25

Take the (6) from our previous computation and concatenate with (25). Result will be (625).

In computer science, concatenate means to smoosh together. For example, if we add 9 with 8, we get 17. But if we concatenate 9 and 8, we get 98. I use the term concatenate above in the same spirit.

To summarize then, the trick is to concatenate 5 * 5 (which is easy enough, just 25) with the product of the digit in tens place and its increment.

Cool! But here’s the million dollar question: WHY?!

Go ahead and try this out with any number that ends in — 5. It will always work. But why does it work???

I set out to find out, outlined below is my rationalization behind the mechanics of this rule.

Let’s begin by representing 25 slightly differently:

25 * 25 = 625
(20 + 5)(20 + 5) = 625

(NOTE: Remember that 25 * 25 = 625. Any and all operations we run on the left side of our equation should still evaluate to 625.)

This should look very similar to an elementary principle from middle school math:

(a + b)(a + b),
in this case, a = 20 and b = 5

Now, FOIL gives us a way to expand the number sentence above:

(a + b)(a + b) = 
a*a + ab + ab + b*b

And since our decomposition of 25 is essentially a + b, we can apply the same approach to our number sentence (with the caveat that we will not evaluate the products of the numbers, nor will we sum them):

(20 + 5)(20 + 5) = 625
20*20 + 20*5 + 5*20 + 5*5 = 625

It is natural to want to evaluate the number sentence right there (mainly because we are easily able to). But! To achieve what we seek (an explanation for why the algorithm we explored above works) we will abstain from evaluation and instead rely on strategically factoring.

(20 + 5)(20 + 5) = 625
20*20 + 20*5 + 5*20 + 5*5 = 625
--------- ^ FROM PREVIOUS BLOCK
20*20 + 20*5 + 5*20 + 5*5 = 625
2*10*20 + 2*10*5 + 5*2*10 + 5*5 = 625

Note the number sentence above once more, particularly realize that the first three terms all share a 2*10 multiplier. This means we can pull out the shared 2*10 terms from the first three numbers, like so:

(20 + 5)(20 + 5) = 625
20*20 + 20*5 + 5*20 + 5*5 = 625
2*10*20 + 2*10*5 + 5*2*10 + 5*5 = 625
--------- ^ FROM PREVIOUS BLOCK
2*10(20 + 5 + 5) + 5*5 = 625

It is interesting to note that just pulled out 2, which is the number from the tens place in 25. We will revisit the significance of that later on. For now, let’s trudge forward by adding the 5 + 5 term.

(20 + 5)(20 + 5) = 625
20*20 + 20*5 + 5*20 + 5*5 = 625
2*10*20 + 2*10*5 + 5*2*10 + 5*5 = 625
2*10(20 + 5 + 5) + 5*5 = 625
--------- ^ FROM PREVIOUS BLOCK
2*10(20 + 10) + 5*5 = 625

Notice now that our (20 + 10) term can be rewritten as follows:

(20 + 10) = 
(2 * 10 + 1 * 10) =
10(2 + 1)

Let’s substitute this back into our calculation:

(20 + 5)(20 + 5) = 625
20*20 + 20*5 + 5*20 + 5*5 = 625
2*10*20 + 2*10*5 + 5*2*10 + 5*5 = 625
2*10(20 + 5 + 5) + 5*5 = 625
2*10(20 + 10) + 5*5 = 625
--------- ^ FROM PREVIOUS BLOCK
2*10*10(2 + 1) + 5*5 = 625

Now, we are cooking. Let’s rearrange our last number sentence:

2*10*10(2 + 1) + 5*5 = 625
2*(2+1)*10*10 + 5*5 = // possible by commutative property of
                      // multiplication

That first term, 2*(2+1), is our holy grail. It reflects what our “trick” posited above: we can take the digit in the tens place (2), add one to it (2+1), multiply by the digit itself to arrive at 6 = 2*(2+1), the first component of our solution! Essentially, the factoring exercise shows us where the 6, or the 12 (if we try to compute 35 * 35), etc comes from.

Before making any conclusions, let’s try it with a few other numbers to validate the analysis above:

35 * 35 = 1225

35 * 35 = 1225
(30 + 5)(30 + 5) = 1225
30*30 + 30*5 + 5*30 + 5*5 = 1225
3*10*30 + 3*10*5 + 5*3*10 + 5*5 = 1225
3*10(30 + 5 + 5) + 5*5 = 1225
3*10(30 + 10) + 5*5 = 1225
---------
3*10*10(3 + 1) + 25 = 1225
3(3 + 1)*10*10 + 25 = 1225

75 * 75 = 5625

75 * 75 = 5625
(70 + 5)(70 + 5) = 5625
70*70 + 70*5 + 5*70 + 5*5 = 5625
7*10*70 + 7*10*5 + 5*7*10 + 5*5 = 5625
7*10(70 + 5 + 5) + 5*5 = 5625
7*10(70 + 10) + 5*5 = 5625
---------
7*10*10(7 + 1) + 25 = 5625
7(7 + 1)*10*10 + 25 = 5625

Ok, so it seems to be working. The real key to understanding why this always works is the number sentence below:

7*10(70 + 5 + 5) + 5*5 = 5625

Generally, when we expand any number that ends in — 5 in the manner that we have above, we get something like:

25 * 25 = (20 + 5)(20 + 5)
35 * 35 = (30 + 5)(30 + 5)
75 * 75 = (70 + 5)(70 + 5)

When the expanded sentence gets FOILed, we end up with an intermediary term that looks like this (bold for emphasis):

7*10(70 + 5 + 5) + 5*5 = 5625

That 5 + 5 term always shows up when squaring numbers that end in 5 (and if we were trying to compute 76 * 76, it would be 6 + 6, etc). And because 5 + 5 adds to 10 — always, we will always arrive at the last sentence below:

7*10(70 + 5 + 5) + 5*5 = 5625
7*10(70 + 10) + 5*5 = 5625
7*10*10(7 + 1) + 25 = 5625
7(7 + 1)*10*10 + 25 = 5625

And that’s why squaring any number ending in — 5 will result in a solution that can be computed by the algorithm above.

It’s just a happy coincidence! 🎉 🎈🎊 🙌

(This is also why this trick seems to only work with numbers that end in 5).

(PS: although I didn’t state explicitly, this trick should work for numbers > 100 as well but our p value would be > 10 which makes the “mental” part of this trick a bit more involved!)

I attempted an inductive proof of the concept above, written below with very little annotation.

For any number:
10p + q,
let p = any natural number
    q = 5
then,
(10p + 5)(10p + 5) = p(p+1)*100 + 25
Proof:
Let p = 1. Then:
(10p + 5)(10p + 5) = 
(10 + 5)(10 + 5) = 
10*10 + 5*10 + 5*10 + 25 = 
10(10 + 5 + 5) + 25 =
10(10 + 10) + 25 = 
1*(10 + 10)*10 + 25 =
1*(10*(1 + 1))*10 + 25 =
1*(1 + 1)*10*10 + 25 = p(p + 1)*10*10 + 25
Since p = 1
Let p = k. Then:
(10p + 5)(10p + 5) =
(10k + 5)(10k + 5) = 
10k * 10k + 5*10k + 5*10k + 25 =
10k(10k +5 + 5) + 25 = 
k(10k + 10)*10 + 25 = 
k(10(k+1))*10 + 25 =
k(k+1)*10*10 + 25 = p(p + 1)*10*10 + 25
Since p = 1
Let p = k+1. Then:
(10p + 5)(10p + 5) =
(10(k+1) + 5)(10(k+1) + 5) =
Let M = k + 1
(10M + 5)(10M + 5) =
10M*10M + 5*10M + 5*10M + 25 =
10M(10M +5 + 5) + 25 = 
M(10M + 10)*10 + 25 = 
M(10(M+1))*10 + 25 =
M(M+1)*10*10 + 25 = p(p + 1)*10*10 + 25
Since p = k + 1 = M