# 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.

\begin{align} &a \omega 0 \times \frac 1 5 = 2 \times x &a \omega 5 \times \frac 1 5 = 2 \times x +1 \end{align}
It is not hard to that the first equation too is true. In the above proofs we notice that we get 1 in MSG as carry over from LSG is $$2\times 5 = 10$$. In case the last digit is 0, as in this case then we will not get the carry over 1.