An elegant problem
- Do you think 2011 is elegant?
- What on earth do you mean?
- Add the sum of its digits squared.
- You mean, 2^2 + 0^2 + 1^1 + 1^1 = 4 + 0 + 1 + 1 = 6?
- Right on! Do the same with 6.
- You mean, 6^2 = 36?
- And with 36, and so on.
- 3^2 + 6^2 = 9 + 36 = 45. 4^2 + 5^2 = 16 + 25 = 41. 4^2 + 1^2 = 17.
- How long do I have to go on?
- To the bitter end!
- 1^2 + 7^1 = 1 + 49 = 50. 5^2 + 0^2 = 25. 2^2 + 5^2 = 29. 2^2 + 9^2 = 85. 8^2 + 5^2 = 89.
- I give up!
- That is OK! If the process ends in 1, the starting number, or for that matter any numbers in the sequence, is elegant.
- And you want me to find an elegant number?
- That would be nice, but I actually want you to find an infinite number of pairs of consecutive numbers that both are elegant.
- You must be out of your mind!
Problem source: El País.
November 15th, 2011 at 1:50 pm
There seems to be something wrong with the question. It’s only positive integers that can be elegant. So an infinite number of consecutive elegant numbers would mean all integers N and above, for some N.
Now, if there is any inelegant number at all, then it produces a never-ending sequence. By the definition of elegance, the terms of the sequence are all different and are all inelegant. They are all positive integers, therefore for any N there must be members of this sequence that are greater than N.
So unless every positive integer is elegant, there can’t be an infinite number of consecutive elegant integers. An infinite number of elegant integers, perhaps.
November 15th, 2011 at 2:54 pm
Sorry!
It should have read:
” I actually want you to find an infinite number of pairs of consecutive numbers that both are elegant.”
I have added “pairs of” in the post above.
November 22nd, 2011 at 3:28 pm
1111111 and 1111112 are both elegant.
So here are an infinite number of pairs of consecutive numbers that are both elegant:
10111111 and 10111112
100111111 and 100111112
1000111111 and 1000111112
etc.
Of course there are an infinite number of ways to insert the zeros; for example:
11111101 and 11111102
111111001 and 111111002
1111110001 and 1111110002
etc.
This leads to another question: suppose we call a number “cyclical” if, by taking the sum of squared digits, we eventually get into a cycle. So for example, 99 is cyclical because we have the sequence 99, 162, 41, 17, 50, 25, 29, 85, 89, 145, 42, 20, 4, 16, 37, 58, 89, 145, 42, … etc. The cycle repeats when we reach 89.
So the question is: Prove or disprove that all positive integers are either elegant or cyclical.