Digit Math Application: Proving the correctness of shortcut method to squaring numbers ending with 5


Please read Digit Math: Introduction before you continue.

The problem

Someday, somewhere I came to know that any number which ends with the digit 5 can be easily squared. The trick can be easily demonstrate using an example. Suppose we want to find the square of 25.

Trick is to take the number before 5 (which will be 2 here), add one to it (2 + 1 = 3) and then multiply them together (2 x 3 = 6). Now the final answer would be the product followed by the number 25, i.e. 625 in this case.

Now let’s try it out yo find \(215^2\).

$$
\begin{align}
215^2 &= (21 \times (21 + 1)) \omega 25\\
&= (21 \times 22) \omega 25\\
&= 462\omega25\\
&= 46225\tag{Answer}
\end{align}
$$

This always seemed to work out very well. The problem was, can I trust this trick? Will this always hold true? I didn’t have answers to those questions, until I proved it myself using Digit Math. Good news is that this trick will always hold true.

The proof

Let the number be \(x = a\omega 5\).  \(a\) can have any number of digits.

Case 1:

\(a\) has exactly one digit. So, \(]a[ = ]5[ = 1\).

$$
\begin{align}
\therefore (a\omega 5)^2 &= aa\omega (5a+5a) \omega 5.5\tag{Using Bimultiplication formula}\\
&= a^2 \omega 10a \omega \underline{2}5\\
&= a^2 \omega (a\omega0) \omega \underline{2}5\\
&= a^2 \omega (a\omega2) \omega 5\\
&= (a^2 + a) \omega 2 \omega 5\\
&= \big(a(a+1)\big) \omega 25\tag{Proved}
\end{align}
$$

Case 2:

\(a\) has more than one digits. So, \(]a[ > (]5[ = 1)\).

But to apply Bimultiplication formula \(a\) must have the same number of digits in \(5\), which is obviously not the case here. So, we will use one trick. We will pad \(5\) with some number of zeroes on the right, so that, \(]a[ = ]5\omega c[\), where, \(c\) is all zeroes and \(]c[ = ]a[ – 1\). So, if \(a = 123 \Rightarrow c = 00\).

$$
\begin{align}
\therefore (a\omega 5)^2 &= \big(a \omega ]5 \omega c[^{]a[}\big)^2\\
&= a^2 \omega \big(a(5\omega c) + a(5\omega c)\big) \omega (5\omega c)^2\tag{Using Bimultiplication}\\
&= a^2 \omega 2a(5\omega c) \omega (5\omega c)^2\\
&= a^2 \omega (10a \omega 2ac) \omega (5\omega c)^2\\
&= a^2 \omega (10a \omega c) \omega (5\omega c)^2\tag{Since, c is all zeroes}\\
&= a^2 \omega (a \omega 0 \omega c) \omega (5\omega c)^2\\
&= a^2 \omega (a \omega 0 \omega c) \omega (25 \omega 10c \omega c^2)\\
&= a^2 \omega (a \omega 0 \omega c) \omega (25 \omega c \omega c)\tag{1}
\end{align}
$$

Since each digit group must have \(]a[\) digits, so let us move one zero from the middle \(c\) in \(25 \omega c \omega c\) to the rightmost \(c\). So, now that group becomes \(25\omega d\omega e\), where \(]d[ = ]c[ – 1\) and \(]e[ = ]c[ + 1\).

$$
\begin{align}
\therefore (1) &= a^2 \omega \big( (a \omega 0 \omega c) + (25 \omega d)\big) \omega e\\
&= a^2 \omega \big( a \omega (0+25) \omega (c+d) \big) \omega e\\
&= a^2 \omega ( a \omega 25 \omega d ) \omega e\tag{2}
\end{align}
$$

Now,

$$
\begin{align}
]25 \omega d[ &= ]25[ + ]d[\\
&= 2 + (]c[ – 1)\\
&= 2 + \big((]a[ – 1) – 1\big)\\
&= ]a[
\end{align}
$$

So, in the group \(a \omega 25 \omega d\), \(a\) is excess.

$$
\begin{align}
\therefore (2) &= (a^2 + a) \omega (25 \omega d) \omega e\\
&= a(a+1) \omega 25 \omega d \omega e\\
&= a(a+1) \omega 25 \omega c \omega c\tag{Shifting a zero from e to d}\\
\end{align}
$$

So finally,

$$
\begin{align}
&(a\omega 5 \omega c)^2 = a(a+1) \omega 25 \omega c \omega c\\
&\Rightarrow \big((a\omega 5) \times 10^c\big)^2 = \big(a(a+1) \omega 25 \omega c\big) \times 10^c\\
&\Rightarrow (a\omega 5)^2 \times 10^{2c} = \big(a(a+1) \omega 25\big) \times 10^{2c}\\
&\Rightarrow (a\omega 5)^2 = a(a+1) \omega 25\tag{Proved}
\end{align}
$$

So, we see that this trick is applicable for all kinds of whole numbers that end with 5. 

Check out other applications of Digit Math

Link to list of other applications of Digit Math.

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