Odd Perfect Number

A perfect number is a positive integer, the sum of whose factors(except the number itself) add upto the number.

Example: 28 = 14 + 7 + 4 + 2 + 1

There are about 48 known perfect numbers at the time of writing this article.But none of them are odd. Computers have checked upto $10^{300}$ but could never find one. So I tried my luck disproving their existence.I had a little experience proving this (pefect squares have odd number of factors). But rather than disproving their existence I ended up finding constraints on their existence.

Let $x$ be a odd perfect number.

Then $x$ can be denoted as $x = p^a.q^b.r^c ..….$ where $p, q, r.....$ are prime factors.

Theorem 1: One and only one of the prime factors have an odd power i.e. only one of $a, b, c....$ is odd and all others are even.

Proof: 

All the combination of factors of $x$ can be denoted by varying the powers of the prime factors.

The sum of all the factors of $x$ can then be written as

$s = (p^0 + p^1 + p^2 + ....p^a)(q^0 + ..... q^b)(r^0 + ...… r^c).....$

Balancing Chemical Equations [LifeHack]

Its not often that we come across chemical equations with coefficients like 48 and 50.But when we do most of us are meant to screw up.Thats exactly why a friend of mine asked me to come up with something.(Its hard enough just to memorise all those equations.)

I'll show you how to balance complicated chemical equations with huge coefficients easily and quickly.All you need to know is how to solve linear equations in variables.

The method might seem long but the equations that we get are simple and the method is much quicker than manually trying to balance the atoms.

Steps: 

1.Assign coefficients to each term in the equation (a,b,c.....)

2.Now balance each element i.e. solve for the coefficients.

(You can skip some steps by assigning coefficients directly as shown below)

Lets start with..

S₈ + HNO₃ = H₂SO₄ + NO₂ + H₂O

Only Perfect Squares Have Odd Number Of Factors

One day I came across a brain teaser,

There are 100 rooms in a hostel numbered 1,2...100.At first the hostel warden opens every room.Then he closes every 2nd room i.e. 2,4,6...100.In the next round he goes to every 3rd room 3,6,..99, if the room is already open he closes it and open closed rooms.This goes on and after the 100th round how many rooms will be open??

To solve this I tried using many methods.One of the methods was to find the number with odd number of factors. If the number had odd number of factors, it will stay opened after all the rounds been complete. Let's say 26. Factors are 1,2,13,26. It'll be opened in round 1, round 2, round 13 and round 26. After that it'll remain untouched. Now let's say a number $n$ has odd number of factors, then it'll remain open after $n$th round.

I tried other combinatorial methods.But all these seemed too long to yield an answer in a reasonable amount of time.After an hour or two one of my friends gave an answer 10.On asking how he arrived at the solution he said he tried upto 10 rounds and only the perfect squares 1, 4 & 9 remained open.

Frankly, I didn't like his answer.It was like mere guess work. And it was not yet conclusive. So the question was do all perfect squares actually have odd number of factors, do other numbers have odd number of factors as well?On factorizing random numbers only perfect squares seemed to have this property.

So we can vaguely say all perfect squares and only perfect squares have odd number of factors.

But I was never a fan of uncertainty, and the fact seemed too elementary to present itself in the form of a proof. After burning tons of glucose in my brain, the proof seemed pretty silly......