# Difference between revisions of "2014 USAJMO Problems/Problem 4"

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==Solution== | ==Solution== | ||

Define <math>S(n) = n + s_b(n)</math>, and call a number ''unrepresentable'' if it cannot equal <math>S(n)</math> for a positive integer <math>n</math>. | Define <math>S(n) = n + s_b(n)</math>, and call a number ''unrepresentable'' if it cannot equal <math>S(n)</math> for a positive integer <math>n</math>. | ||

− | We claim that in the interval <math> | + | We claim that in the interval <math>[b^p, b^{p+1})</math> there exists an unrepresentable number, for every positive integer <math>p</math>. |

If <math>b^{p+1}</math> is unrepresentable, we're done. Otherwise, time for our lemma: | If <math>b^{p+1}</math> is unrepresentable, we're done. Otherwise, time for our lemma: |

## Revision as of 14:29, 14 August 2014

## Problem

Let be an integer, and let denote the sum of the digits of when it is written in base . Show that there are infinitely many positive integers that cannot be represented in the form , where is a positive integer.

## Solution

Define , and call a number *unrepresentable* if it cannot equal for a positive integer .
We claim that in the interval there exists an unrepresentable number, for every positive integer .

If is unrepresentable, we're done. Otherwise, time for our lemma:

Lemma: Define the function to equal the number of integer x less than such that . If for some y, then .

Proof: Let be the set of integers x less than such that . Then for every integer in , append the digit to the front of it to create a valid integer in . Also, notice that . Removing the digit from the front of y creates a number that is not in . Hence, , but there exists an element of not corresponding with , so .

Note that our lemma combined with the Pigeonhole Principle essentially proves the claim. Therefore, because there are infinitely many intervals containing an unrepresentable number, there are infinitely many unrepresentable numbers.