Author 
Topic: A magical number? (Read 660 times) 

BMAD
Junior Member
Posts: 57


A magical number?
« on: Feb 24^{th}, 2015, 9:01am » 
Quote Modify

Write a list from 1 to 100. Pick two numbers at random Sum the two numbers and find the product of the two numbers Sum the above two numbers you found Erase the two chosen numbers from the list Add the final sum of the two numbers to the list Repeat until there is only one number. Repeat the above process multiple times. Is it the same number? Do the numbers center around a point? Bimodal? analyze. I recommend trying this for numbers 110 and 120 to see if anything special is happening ... The calculated numbers can be quite huge.

« Last Edit: Feb 24^{th}, 2015, 9:02am by BMAD » 
IP Logged 



towr
wu::riddles Moderator Uberpuzzler
Some people are average, some are just mean.
Gender:
Posts: 13614


Re: A magical number?
« Reply #1 on: Feb 24^{th}, 2015, 11:15am » 
Quote Modify

The operation is commutative and associative, so the answer is always the same for a given starting set of numbers It seems to be to sum of the products of each set in the powerset. But calculated much quicker than it sounds :p


IP Logged 
Wikipedia, Google, Mathworld, Integer sequence DB



pex
Uberpuzzler
Gender:
Posts: 880


Re: A magical number?
« Reply #2 on: Feb 24^{th}, 2015, 2:32pm » 
Quote Modify

Numerically, the result seems to be (n+1)!1. Given the form of the answer and towr's observation that the order in which the operations are performed doesn't matter, it looks like it shouldn't be too hard to produce a proof based on good old mathematical induction, but I don't have the time right now. Edit: Ok, the base case is obvious, and the induction step is essentially just (n!1) + n + n(n!1) = n!(n+1) +nn 1 = (n+1)!1. Neat little puzzle! Edit the Second: By the same reasoning, if the list of numbers does not have to be 1..n but can be anything, the result is product(x_{i}+1)1.

« Last Edit: Feb 25^{th}, 2015, 1:38pm by pex » 
IP Logged 



