Question

find the factor of am
$find the factors of am + an + al + bm + bl + bn$
​

Answer 1

Step-by-step explanation:

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The Stampede supercomputer

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The World's Largest Maths Problem Has Been Solved, And It Takes Up 200 TB

BEC CREW

31 MAY 2016

If you thought advanced maths in high school was bad, spare a thought for a trio of mathematicians whose solution to a single maths problem takes up 200 terabytes of basic text - even with the help of a supercomputer.

When you consider that just 1 terabyte can hold 337,920 copies of War and Peace - one of the longest novels ever written - you can start to get a sense of just how insane that is. The previous record-holder was reportedly a 13-gigabyte proof, published in 2014.

So what’s this ridiculous maths problem? It’s been named the Boolean Pythagorean Triples problem, and was first posed by California-based mathematician Ronald Graham back in the 1980s.

The problem centres around the Pythagorean formula a2 + b2 = c2, where a and b are the shorter sides of a triangle, and c is the hypotenuse, or longest side.

Certain sets of three positive integers known as Pythagorean triples can be inserted into the formula, such as 32 + 42 = 52, 52 + 122 = 132, and 82 + 152 = 172.

With this in mind, imagine that every integer is painted either red or blue.

Graham asked if it's possible to colour all the integers either red or blue, so that no set of Pythagorean triples - a, b and c - were all the same colour. He put $100 on the line for anyone who could solve the problem. (That should cover a 1 terabyte drive.)

Andrew Moseman over at Popular Mechanics explains why that $100 looks mighty meagre, given the task ahead:

"What makes it so hard is that one integer can be part of multiple Pythagorean triples. Take 5. So 3, 4, and 5 are a Pythagorean triple. But so are 5, 12, and 13. If 5 is blue in the first example, then it has to be blue in the second, meaning either 12 or 13 has to be red.

Carry this logic forward into much bigger numbers and you could see where this would start to get tricky. If 12 has to be red in that 5-12-13 triple, it might force changes down the line that would result in a monochrome triple somewhere."

${yxy {251 {y}^{2} }^{2} }^{?}$

Answer 2

am+an+al+bm+bl+bn

= regrouping the terms we get

= a(m+n+l) + b(m+l+n)

= as (m+n+l) and (m+l+n) are common factors we group them

= (a+b) (m+l+n)