Puzzle time - integers
-
:::
Obviously, when a number n is in S, then n+5 must also be in S.
So once we have all digits from 0 to 4 (or 5 to 9) as last digits of numbers, all numbers above it must be in S.
So the question is whether we ever get all last digits.
I think we can get to all last-digits except 0 and 5, since any number that ends with 0 or 5 squared also ends with 0 or 5.
So, my theory about the positive integers not in S is:
There's some noise in the beginning, and after a while it's only the numbers that end with 0 or 5.
:::
-
What Klaus missed:
:::
The only ‘noise’ (besides all multiples of 5) is the number 1.
- 2 is granted which gets you all numbers ending in 2 or 7.
- 7^2 is 49 which gets you all the numbers ending in 9 and 4 above that
- after 49 is 54. 54^2 is 3136 which gets you all the numbers ending in 6 or 1 above it.
BUT - once you have the *6s, you’ll get to 6^8 which gets you back to 6 and 11, etc.
- that gets you to 16 which gets you back to 4 and 9
- that 9 gets you back to 3 and 8
So we have 2,3,4,6,7,8,9 covered plus any number that is a multiple of 5 above them.
So only 1 is missing, along with all multiples of 5
:::
Hello! It looks like you're interested in this conversation, but you don't have an account yet.
Getting fed up of having to scroll through the same posts each visit? When you register for an account, you'll always come back to exactly where you were before, and choose to be notified of new replies (either via email, or push notification). You'll also be able to save bookmarks and upvote posts to show your appreciation to other community members.
With your input, this post could be even better 💗
Register Login
)