Difference between revisions of "2017 OIM Problems/Problem 1"
(Created page with "== Problem == For each positive integer <math>n</math>, let <math>S(n)</math> be the sum of its digits. We say that <math>n</math> has the property <math>E</math> if the terms...") |
|||
| Line 1: | Line 1: | ||
== Problem == | == Problem == | ||
| − | For each positive integer <math>n</math>, let <math>S(n)</math> be the sum of its digits. We say that <math>n</math> has the property <math>E</math> if the terms of the infinite sequence <math>n, S(n), S(S(n)), S(S(S(n))), \cdots </math>, are all even, and we say that <math>n</math> has property <math>O</math> if the terms of this sequence are all odd. Show that among all the positive integers <math>n</math> such that <math>1 \le n \le 2017</math> there are more who have property | + | For each positive integer <math>n</math>, let <math>S(n)</math> be the sum of its digits. We say that <math>n</math> has the property <math>E</math> if the terms of the infinite sequence <math>n, S(n), S(S(n)), S(S(S(n))), \cdots </math>, are all even, and we say that <math>n</math> has property <math>O</math> if the terms of this sequence are all odd. Show that among all the positive integers <math>n</math> such that <math>1 \le n \le 2017</math> there are more who have property <math>O</math> than those who have property <math>E</math>. |
~translated into English by Tomas Diaz. ~orders@tomasdiaz.com | ~translated into English by Tomas Diaz. ~orders@tomasdiaz.com | ||
Latest revision as of 13:36, 14 December 2023
Problem
For each positive integer 
, let 
be the sum of its digits. We say that has the property 
if the terms of the infinite sequence 
, are all even, and we say that has property 
if the terms of this sequence are all odd. Show that among all the positive integers such that 
there are more who have property than those who have property .
~translated into English by Tomas Diaz. ~orders@tomasdiaz.com
Solution
This problem needs a solution. If you have a solution for it, please help us out by adding it.
