Difference between revisions of "2014 USAJMO Problems/Problem 4"
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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: | ||
| − | Lemma: Define the function <math>f(p)</math> to equal the number of | + | Lemma: Define the function <math>f(p)</math> to equal the number of integers x less than <math>b^p</math> such that <math>S(x) \ge b^p</math>. If <math>b^{p+1} = S(y)</math> for some y, then <math>f(p+1) > f(p)</math>. |
Proof: Let <math>F(p)</math> be the set of integers x less than <math>b^p</math> such that <math>S(x) \ge b^p</math>. Then for every integer in <math>F(p)</math>, append the digit <math>(b-1)</math> to the front of it to create a valid integer in <math>F(p+1)</math>. Also, notice that <math>(b-1) \cdot b^p \le y < b^{p+1}</math>. Removing the digit <math>(b-1)</math> from the front of y creates a number that is not in <math>F(p)</math>. Hence, <math>F(p) \rightarrow F(p+1)</math>, but there exists an element of <math>F(p+1)</math> not corresponding with <math>F(p)</math>, so <math>f(p+1) > f(p)</math>. | Proof: Let <math>F(p)</math> be the set of integers x less than <math>b^p</math> such that <math>S(x) \ge b^p</math>. Then for every integer in <math>F(p)</math>, append the digit <math>(b-1)</math> to the front of it to create a valid integer in <math>F(p+1)</math>. Also, notice that <math>(b-1) \cdot b^p \le y < b^{p+1}</math>. Removing the digit <math>(b-1)</math> from the front of y creates a number that is not in <math>F(p)</math>. Hence, <math>F(p) \rightarrow F(p+1)</math>, but there exists an element of <math>F(p+1)</math> not corresponding with <math>F(p)</math>, so <math>f(p+1) > f(p)</math>. | ||
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. | 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. | ||
Revision as of 17:51, 18 August 2015
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 ![$(b^p, b^{p+1}]$](http://latex.artofproblemsolving.com/b/3/d/b3dc37dce0c363a7efe72ecddd69f687075e983e.png)
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 integers 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.
