If Any Item Does Not Apply To The Property Being Documented, Enter "N A" F: Fill & Download for Free

You shall have no other gods before Me. n/aYou shall not make idols. n/aYou shall not take the name of the LORD your God in vain. n/aRemember the Sabbath day, to keep it holy. n/aHonor your father and your mother. I’m a decent human beingYou shall not murder. I’m a decent human beingYou shall not commit adultery. I’m a decent human beingYou shall not steal. I’m a decent human beingYou shall not bear false witness against your neighbor. I’m a decent human beingYou shall not covet. Thought crimes are stupid

A 10rs pen.Well, this happened last month. I was in hostel, preparing for something too clichéd to discuss here.So one day, Maths teacher came in and started the chapter "Set Theory". The discussion started and just all at random we fell on a formula.[math]n(A \cup B) = n(A) + n(B) - n(A \cap B) [/math]"How many of you know how this formula comes out?" He asked. All 7 of us started to look for each other's face.He took his pen out of his pocket and placed it on the table, "Any of you prove this formula, This pen is yours."Everybody started to write complicated mathematics on their copies, n(A) n(B) and I don't know what!Now because of reading this one guy, Ethan James, everyday I have started to believe that Venn Diagrams can do anything.Since I read that freaking guy, I developed a firm faith in Venn Diagrams.They can be fun, they can be helpful in roasting people, Venn diagrams can do Hell anything, then why not to try them in the way they are meant to be used?So, I adjusted the equation.[math]n(A \cup B) + n(A \cap B) = n(A) + n(B) [/math]I looked around. Guys were still were doing their heads in the complex mathematics.I Made a Venn Diagram of RHS.[math]n(A \cup B) + n(A \cap B) = [/math]WhateverAnd then LHS.[math]n(A) + n(B) = [/math]Whatever, but the same whatever as the last time[math]. [/math]BOOM!Ethan James Good. XD“Say Paneer!”

Here are two groups of number:A：1，5，10，18，23, 27 ；B：2，3，13，15，25，26These two sets of numbers share an amazing relationship: their sums are equal.[math]1 + 5 + 10 + 18 + 23 + 27 = 2 + 3 + 13 + 15 + 25 + 26 = 84[/math]“What the heck? How is this amazing? There are millions if not trillions of sets of numbers like those!” you might be screaming so painfully in your heart for having wasted 2 seconds of you life reading this ridiculous answer.Is that true? Let’s tweak those numbers a bit by squaring all of them:[math]1^2 + 5^2 + 10^2 + 18^2 + 23^2 + 27^2 = 2^2 + 3^2 + 13^2 + 15^2 + 25^2 + 26^2 = 1708[/math]Their sums are still the same! Weird, you might think, but it’s nowhere near amazing. Many numbers could do that too since there are altogether six in each set.Alright, how about cubing them?[math]1^3 + 5^3 + 10^3 + 18^3 + 23^3 + 27^3 = 2^3 + 3^3 + 13^3 + 15^3 + 25^3 + 26^3 = 46324[/math]And it doesn’t stop there.[math]1^4 + 5^4 + 10^4 + 18^4 + 23^4 + 27^4 = 2^4 + 3^4 + 13^4 + 15^4 + 25^4 + 26^4 = 916885[/math]Is that it? Nope.[math]1^5 + 5^5 + 10^5 + 18^5 + 23^5 + 27^5 = 2^5 + 3^5 + 13^5 + 15^5 + 25^5 + 26^5 = [/math][math]22777944[/math]AMAZING RIGHT? Nope.[math]1^6 + 5^6 + 10^6 + 18^6 + 23^6 + 27^6 = [/math][math]570484228[/math][math]2^6 + 3^6 + 13^6 + 15^6 + 25^6 + 26^6 = [/math][math]569274628[/math]When we raise the numbers to the power of six, their sums are different now.—————————————————————————————————————These numbers follow the patterns of identity discovered by Soviet mathematician Gelfond':[math]a^n + ([/math][math]a[/math][math] + b + 4c)^n + ([/math][math]a[/math][math] + 2b + c)^n + ([/math][math]a[/math][math] + 4b + 9c)^n + ([/math][math]a[/math][math] + 5b + 6c)^n + ([/math][math]a[/math][math] + 6b + 10c)^n = ([/math][math]a[/math][math] + b)^n + ([/math][math]a[/math][math] + c)^n + ([/math][math]a[/math][math] + 2b + 6c)^n + ([/math][math]a[/math][math] + 4b + 4c)^n + ([/math][math]a[/math][math] + 5b + 10c)^n + ([/math][math]a[/math][math] + 6b + 9c)^n [/math]where n= 1 , 2 , 3 , 4 , 5 (6 is not supported)The two sets of numbers that served as an example satisfy the identity where a=1, b=1 and c=2. If we were to change the value of a, b, and c, we can get literally infinite amount of sets that also satisfy the equation.NOW YOU GOTTA ADMIT IT’S FREAKING AMAZING.Follow me for more answers like this one!