# 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.

This site uses Akismet to reduce spam. Learn how your comment data is processed.