I was noticing today that you can calculate the square of numbers by taking the number you want to square and removing the ones digit and multiplying by the number you want to square plus the ones digit. Then you square the ones digit and place it back on the number to get the square
503^2 = 253009
and
50*506=25300
then tacking the 3^2 on the end gets 253009.
Tricks similar to that seem to work so far for larger numbers as well and with other 1s digits, sometimes slightly differently though. I assume it has something to do with base 10 but I am not sure on why this works or how you would write a proof of it.
Any help would be appreciated.

2. Originally Posted by Enkie
I was noticing today that you can calculate the square of numbers by taking the number you want to square and removing the ones digit and multiplying by the number you want to square plus the ones digit. Then you square the ones digit and place it back on the number to get the square
503^2 = 253009
and
50*506=25300
then tacking the 3^2 on the end gets 253009.
Tricks similar to that seem to work so far for larger numbers as well and with other 1s digits, sometimes slightly differently though. I assume it has something to do with base 10 but I am not sure on why this works or how you would write a proof of it.
Any help would be appreciated.
Let $n\in\mathbb{N}$ then $n=a_0+10a_1+\cdots+10^n a_n$ where $a_k\in\{0,\cdots,9\}$. You are claiming that $n^2=\left(10 a_1+\cdots +10^n a_n\right)\cdot\left(2a_0+10 a_1+\cdots 10^n a_n\right)+a_0^2$, or equivalently

$\left(a_0+10 a_1+\cdots+10^n a_n\right)^2-\left(10a_1+\cdots+10^n a_n\right)\cdot\left(2a_0+10a_1+\cdots+10^na_n\rig ht)-a_0^2$ $=\left(\left(a_0+10a_1+\cdots 10^n a_n\right)^2-a_0^2\right)-\left(10 a_1+\cdots+10^na_n\right)\cdot\left(2a_0+10a_1+\cd ots+10^n a_n\right)$ $=0$.

But,

$\left(a_0+10a_1+\cdots+10^na_n\right)^2-a_0^2=\left(a_0+10a_1+\cdots+10^na_n-a_0\right)$ $\cdot\left(a_0+10a_1+\cdots 10^na_n+a_0\right)$ $=\left(10a_1+\cdots+10^na_n\right)\cdot\left(2a_0+ 10a_1+\cdots+10^n a_n\right)$.

From where the conclusion immediately follows.

3. Hello, Enkie!

Another approach . . .

I was noticing today that you can calculate the square of a number by:
(1) taking the number you want to square and removing the ones digit,
(2) and multiplying by the number you want to square plus the ones digit,
(3) then square the ones digit and add it on to get the square.

Any integer $N> 9$ is of the form: . $N \:=\:10T + U$
. . where $U$ is the units digit and $T$ is the "rest of the number."

We note that: . $N^2 \:=\:(10T + U)^2 \:=\:100T^2 + 20TU + U^2$

(1) Remove the units digit and append a zero.
. . . We have: . $10T$

(2) Multiply by $N + U\!:\;\;10T(N + U) \;=\;10T(10T + U + U) \;=\;100T^2 + 20TU$

(3) Add $U^2\!:\;\;100T^2 + 20TU + U^2\quad \hdots$ and the result is $N^2$

4. Let us try with $1485$. We get :

$148 \times (1485 + 5) = 220520$

Now add on $5^2 = 25 \Rightarrow 22052025$

But $1485^2 = 2205225$.

I understand what you mean but you need to mention that when adding the square of the last digit, you actually overlap the previous result when the last digit is greater than three.