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# 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

# Digit Math Application: Proving the correctness of shortcut method to squaring numbers ending with 5

Please read Digit Math: Introduction before you continue.

# 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

# Digit Math Application: Proving that multiplying with 10**n puts n zeros at the end

Please read Digit Math: Introduction before you continue.

# The problem

This is a very fundamental concept that we were taught when we were in junior schools. Now, think of it, what it says. If you add \(x\) ten times then you will get \(x\omega0\). If you add \(x\) hundred times then you will get \(x\omega00\); and so on. How do we know that this will always be true? We know it since we have never seen one violation of it, but, anyway I will try to prove it know to rest the uneasy souls.

# The proof

Before I continue with this proof, I must make sure we understand another empirical rule.

When any number is multiplied by zero then the result is always zero. This is not hard to see why. When you multiply \(x\) by \(2\) then it is like adding \(x\) twice. When you multiply \(x\) by \(1\) then it is like adding \(x\) once. When you multiply \(x\) by \(0\) then it is like adding \(x\) zero times, which in other words is that we never added \(x\) in the first place so \(x\) never existed, so we had nothing, and that nothing is zero.

Back to proof.

Let the number \(a\omega b\) be multiplied by \(1\omega d\), where \(d\) is all all zeroes, \(n\) times. To use Bimultiplication forumula we need to make sure that \(b\) too has \(n\) digits. We can always partition a number such that \(b\) will always have \(n\) digits. For example if \(d = 000\) and \(a\omega b = 2\) then we make \(a = 000\) and \(b = 002\).

So,

$$

\begin{align}

& a \omega b \times 10^d\\

&= a \omega b \times 1 \omega d\\

&= a.1 \omega (a.d + b.1) \omega b.d\tag{By Bimultiplication}\\

&= a \omega (d + b) \omega d\tag{Since d is all zeroes}\\

&= a \omega b \omega d\tag{Since d is all zeroes}

\end{align}

$$

So, in the above equations we see that multiplying a number by \(10^d\) will give us the same number, but followed with \(d\) zeroes. **(Proved)**