Understanding how This Works (shortcut for squaring $2$ digit numbers mentally)

Question

We all know that math is as much about finding the answer as it is about knowing how a method leads you to the answer.

In fact, not knowing the how has caused me to loose marks on several occasions at school.

So my question arises, how does the method for calculating the squares of $2$ digit numbers work? It seems very counter intuitive and doesn't make much sense from the surface.

The method: Take any $2$ digit number and round it to the nearest 10, take the difference and do the inverse operation, then multiply the answer of the round and the result of the inverse calculation. Then add the square of the difference.

• E.g. $54^{2} = 50*58 + 16$

Answer 1

Observe that

$$(10a+b)^2=100a^2+20ab+b^2=(10a)(10a+2b)+b^2\tag{1}$$

and

$$(10a-b)^2=100a^2-20ab+b^2=(10a)(10a-2b)+b^2\;.$$

If the second digit of the number is $b<5$, and the first digit is $a$, the number is $10a+b$, and rounding to the nearest $10$ leaves you with $10a$. In the process of rounding you subtract $b$; adding $2b$ gives you $10+2b$, and equation $(1)$ then shows that multiplying this by the rounded number and adding $b^2$ gives the correct result.

If the second digit is $d\ge 5$ and the first digit is $d$, let $b=10-d$ and $a=c+1$; then the number is $10c+d=10(a-1)+(10-b)=10a-b$. Rounding to the nearest $10$ gives you $10a$. In the process of rounding you add $b$, so subtracting $2b$ gives you $10a-2b$, and $(2)$ shows that multiplying these numbers together and adding $b^2$ gives the correct result.

Answer 2

Let the number whose square is to be calculated be $(xy)=10x+y$ whose square would be $(10x+y)^2 = 100x^2+y^2+20xy$

let after rounding it to nearest 10 the number becomes $(x(y-k))=10x+(y-k)$ then the inverse would evaluate to $(x(y+k))=10x+(y+k)$.

we are to evaluate $(x(y+k))(x(y-k))+k^2$ = $100x^2+10xy-10xk+10xy+y^2-yk+10xk+ky-k^2+k^2$ = $100x^2+y^2+20xy$ = $(10x+y)^2$ = $(xy)^2$.