Category / Reports

by Nirupam Biswas

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.

by Nirupam Biswas

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)

Check out other applications of Digit Math

Link to list of other applications of Digit Math.

by Nirupam Biswas

Access Ext3/Ext2 file system on Mac OSX Lion (10.7)

On Mac if you want to access ext3/etx2 filesystems, which are used by Linux systems, you will find lots of links on net but all are pretty outdate and they don’t work for Lion. So, here is the updated version, which works. At least for me. 😉

You will need two softwares:-

  1. OSXFuse – Download link
  2. Fuse-ext2 – Download link

Download and install them in the sequence shown above.

Fuse-ext2 needs MacFuse to run, but this is no longer maintained and does not work on Lion. OSXFuse is the next generation MacFuse, but Fuse-ext2 is not meant to work with this. Fortunately OSXFuse includes “MacFUSE Compatibility Layer”. Just make sure to select this option when installing OSXFuse and you are good to go.

When both of them are installed, then try plugging in ext3 or etx2 partitioned disk and they should get automatically mounted, just like any other disk. Note, after installing them you may or may not need to restart your system.

PS. You will be able to read the disks but not write to it. As of now write option is not reliable.

by Nirupam Biswas

Can contact lens melt on the eye?

I believe no. This is yet another urban legend, spreading via mails and social networks. I decided to blog about it after seeing too many posts like – “A 21 year old guy had worn a pair of contact lenses during a barbecue party. After a few minutes, he started to scream for help and moved rapidly, jumping up and down….”.

A quick googling shows that contact lenses’ melting points are about 90°C. Some sites suggest that contact lenses can be boiled to disinfect them. Do you think your eyes and face can stand that much heat?

The bottom line is, before the contact lenses melt on your eyes, your face’s skin might have burnt out.

Something which people disregard is not wearing the contact lenses for too long. If you wear them for too long then the portion between your eyes and the lenses might dry up while rest of the eyes are wet. This could result in suction force that may prevent the lenses from being removed.

Disclaimer: I don’t wear contacts and have no experience about them. All the above information have been collected over net.