# Digit Math Application: Proving that all numbers ending with 5 are divisible by 5

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.

# Check out other applications of Digit Math

Link to list of other applications of Digit Math.

Now that you have proven numbers ending in 5 are divisible by 5, does this also prove these division shortcuts?

$$

\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}

$$

So now we have to prove – that any no. ending with 0 when divided by 5 wil yield an even no., and, any no. ending with 5 when divided with 5 will yield an odd no.

We can see that the second one is already proven above. Notice the MSG (Most Significant Group) equation.

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.