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38 members have voted
6 is divisible by 1, 2 and 3. 1 + 2 + 3 = 6
28 is divisible by 1, 2, 4, 7 and 14. 1 + 2 + 4 + 7 + 14 = 28.
For any rational number a/b, a number is a/b-perfect if the sum of the proper divisors equals to the original number multiplied by (a/b). For example 10 is 4/5-perfect since 1 + 2 + 5 = 8 = (4/5)*10. The normal definition of a perfect number is when this fraction a/b = 1.
Given we are in the year 2022, what is the smallest 20/22-perfect number? The sum of the proper divisors is equal to (20/22)*(the number)?
Can you find a second such number? (I don't know).
Quote: ThatDonGuyI have a feeling this answer earns the QI Klaxon, but the "obvious" answer is...
If you start with the cards in order 13, 11, 9, 7, 5, 3, 1, 2, 4, 6, 8, 10, 12, it requires 12 reversals.
I don't think more than that is possible without getting into an infinite loop.
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Sorry, incorrect.
I'll let this ride another day although a solve would be surprising, shocking really.
Quote: GialmereQuote: ThatDonGuyI have a feeling this answer earns the QI Klaxon, but the "obvious" answer is...
If you start with the cards in order 13, 11, 9, 7, 5, 3, 1, 2, 4, 6, 8, 10, 12, it requires 12 reversals.
I don't think more than that is possible without getting into an infinite loop.
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Sorry, incorrect.For 13 cards, there are 13! = 6,227,020,800 decks you might need to consider. Computer assistance advised.
I'll let this ride another day although a solve would be surprising, shocking really.
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I discovered that even if you limit it to the numbers 1-7, you need 16
Quote: teliotHappy "Perfect Number Day" everyone! It's 6/28.
6 is divisible by 1, 2 and 3. 1 + 2 + 3 = 6
28 is divisible by 1, 2, 4, 7 and 14. 1 + 2 + 4 + 7 + 14 = 28.
For any rational number a/b, a number is a/b-perfect if the sum of the proper divisors equals to the original number multiplied by (a/b). For example 10 is 4/5-perfect since 1 + 2 + 5 = 8 = (4/5)*10. The normal definition of a perfect number is when this fraction a/b = 1.
Given we are in the year 2022, what is the smallest 20/22-perfect number? The sum of the proper divisors is equal to (20/22)*(the number)?
Can you find a second such number? (I don't know).
link to original post
While we're waiting, here are some Perfect Number facts:
Easy enough to prove: the sum of the integers from 1 to any prime number that is one less than a power of 2 (a "Mersenne prime") is a perfect number.
Examples: 1 + 2 + (2^2 - 1) = 6; 1 + 2 + ... + (2^3 - 1) = 28; 1 + 2 + ... + (2^5 - 1) = 496.
Slightly harder to prove: these are the only even perfect numbers.
Conjectured, but not proven (nor disproven) yet: there are no odd perfect numbers.
Quote: Gialmere...I'll let this ride another day although a solve would be surprising, shocking really...
Using brute force, I agree that with seven numbers - e.g. starting with 4762153 - it can take 16 turns. This one takes 6 turns to get the 7 into last position. I can't imagine doing 13 numbers (6Bn of them) in a similar fashion!
Quote: teliotHappy "Perfect Number Day" everyone! It's 6/28.
6 is divisible by 1, 2 and 3. 1 + 2 + 3 = 6
28 is divisible by 1, 2, 4, 7 and 14. 1 + 2 + 4 + 7 + 14 = 28.
For any rational number a/b, a number is a/b-perfect if the sum of the proper divisors equals to the original number multiplied by (a/b). For example 10 is 4/5-perfect since 1 + 2 + 5 = 8 = (4/5)*10. The normal definition of a perfect number is when this fraction a/b = 1.
Given we are in the year 2022, what is the smallest 20/22-perfect number? The sum of the proper divisors is equal to (20/22)*(the number)?
Can you find a second such number? (I don't know).
link to original post
Quote: Gialmere...I'll let this ride another day although a solve would be surprising, shocking really...
This shows where I essentially gave in! I couldn't see an obvious pattern emerging and could see myself putting together a tree almost by hand. While that may eventually show the largest route for 7 numbers, I couldn't see how to shorten the process for the next levels. I was looking for an answer based on loops as there was a puzzle previously on this topic and suspect this is similar or related logic.
However this method would only show there was a maximum while trying to move from level N to N-1. I'm pure guessing it's aboput N or N-1, which would give an upper bound but not an actual maximum. Saying it was less, than say 78 (12+...+1) or 91, doesn't really add any value, especially as we know it's 16 for 7 numbers!

I think you mean 13! permutations not 13! “decks”. If I understand the problem correctly there is only one deck of 13 cardsQuote: Gialmere
For 13 cards, there are 13! = 6,227,020,800 decks you might need to consider. Computer assistance advised.
Quote: teliotCan you find a second such number? (I don't know).
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At first, I had problems finding the first number, but then I discovered I was about 200 years before my time - I was looking for 18/22 numbers.
I have checked every number up to 4 million (so far), and have found no other numbers. That doesn't mean they don't exist; after all, there are no perfect numbers between 8128 and 33,550,336.