Please read Digit Math: Introduction before you continue.

# 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.