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Please read Digit Math: Introduction before you continue.

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The problem

It seems I have developed a fascination for the number 5, so here we go again. Here I would be using Digit Math to prove that all integers  which end with digit 5 are always divisible by 5.

The proof

Case 1:

Take a two digit number $x\omega 5$, i.e. $x$ has only one digit.

$x\omega 5$ is divisible by 5, if $x\omega 5 \times \frac 1 5$ (i.e. $x \omega 5 \times 0.2$) leave no remainder.

So, $x\omega 5 \times 2$ must end with one zero.

$$\begin{align} &x\omega5 \times 0\omega2\\ &= 0 \omega 2x \omega 10\tag{Using Bimultiplication}\\ &= (2x + 1) \omega 0\tag{1} \end{align}$$

Since (1) ends with zero so it is proved.

Case 2:

$x\omega 5$ is a number where $x$ has $n$ digits.

Multiplying $x\omega 5$ by $10^{(n-1)}$ to make both sides of $\omega$ equal in number of digits.

So,

$$\begin{align} &x\omega (5 \times 10^{n-1}) \times 0 \omega (2\times 10^{n-1})\\ &= 0 \omega (2x\times 10^{n-1})\omega (10\times 10^{n-1})\\ &= (2x\times 10^{n-1})\omega 10^n\\ &= (2x\times 10^{n-1} + 1) \omega ]0[^n\tag{Moving extra 1 to left} \end{align}$$

In the above equation it is easy to see that the result ends with $10$ after we divide the $10^{n-1}$. Therefore, this is not going to leave any remainder. (Proved)

Case 3:

$\frac 5 5 = 1$. (Proved)

Yeah, this is a trivial case, but for the sake of completeness.

Check out other applications of Digit Math

Link to list of other applications of Digit Math.