0
$\begingroup$

I have these sequences :

$1\ 15\ 6\ 19\ 14\ 6\ 19\ 14$ ...

$2\ 20\ 20\ 20\ 20\ 20\ 20$ ...

$3\ 20\ 20\ 20\ 20\ 20\ 20$ ...

$4\ 6\ 19\ 14\ 6\ 19\ 14$ ...

$5\ 6\ 19\ 14\ 6\ 19\ 14$ ...

$6\ 19\ 14\ 6\ 19\ 14\ 6$ ...

$7\ 19\ 14\ 6\ 19\ 14\ 6$ ...

$8\ 5\ 6\ 19\ 14\ 6\ 19\ 14$ ...

$9\ 14\ 6\ 19\ 14\ 6\ 19$ ...

They are all built following the same path. Can you find it?

Bonus : (you'll probably find this when you'll solve the first part of this question)

You may notice that each line start with a natural number, and ends with a loop (which is 6, 19, 14, etc or 20, 20, 20, etc). Can you tell if we will find a loop for each natural number?

$\endgroup$
9
$\begingroup$

Each number in the sequences is

The position in the alphabet of the first letter of the number before it

The bonus has been covered here

$\endgroup$
  • $\begingroup$ The link seems to deal with something slightly different. I added an answer to explain, because I can't use spoiler tags here $\endgroup$ – bg6471 Jul 28 '16 at 1:23
1
$\begingroup$

Will got the answer already, but I'm not convinced about the link provided for the bonus. I'll have a go at the bonus. I don't want this marked as answer though, as it's only for the bonus marks :)

The linked question was "Any other cyclic number(s) that I may have missed (or a 'proof' that there are no others)" and the answer was "Every number must go to a number from 1 to 26, so these are the only numbers that can possibly be in cycles. It is simple to check that all 26 start with one of the letters efnost. t loops to itself, nfs form a cycle, and eo both go to f." -- This seems to deal with the numbers that are cycling, and ignoring the initial number as though it's outside of the cycle

But

In this question "You may notice that each line start with a natural number, and ends with a loop (which is 6, 19, 14, etc or 20, 20, 20, etc). Can you tell if we will find a loop for each natural number?" - The natural number being the number at the start of the line

So if we assume 1 to 26 work, why not try 27 (twenty-seven)
27 20 20 20 20 20 20 (seems fine)

Anything up to 40 starts with a T so it becomes 20 20 20 20
40 6 19 14 6 19 (as per case for 4)

Any number up to 99 starts with one of the numbers given in the question, sixty (6 19 14...) eighty (5 6 19 14...) etc

Any number from 100 to 999 will also, four hundred and one (4 6 19 etc)

Any number from then on will be expressed as some (1, 2, or 3) digit groupings of multiples of 1000 eg. two hundred and seventeen thousand, six hundred and ninety four is just T again (20 20 20 20)

It seems like given that we've proved 1 to 9 already, we've already proved that there is a loop for any natural number

$\endgroup$
  • 1
    $\begingroup$ Every number goes to another from 1 to 26, so in at most 27 numbers there must be a repeat, leading to a loop. What the numbers actually start with is not very relevant at all... $\endgroup$ – ffao Jul 28 '16 at 2:00
  • 1
    $\begingroup$ Every number immediately goes to something between 1 and 26. If 1 to 26 all loop, then every number must loop. $\endgroup$ – f'' Jul 28 '16 at 5:49

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.