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Difference between revisions of "1978 USAMO Problems/Problem 3"
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== Solution ==
 
== Solution ==
 
''Lemma:'' If <math>g</math> is a good integer, then so is <math>2g + 2</math> and <math>2g + 9</math>.
 
''Lemma:'' If <math>g</math> is a good integer, then so is <math>2g + 2</math> and <math>2g + 9</math>.
 +
 
Proof: Let <math>g = a_1 + a_2 + \cdots + a_k</math> such that <cmath>\frac{1}{a_1}+\frac{1}{a_2}+\cdots+\frac{1}{a_k}=1,</cmath> where the <math>a_i</math> are positive integers. Then, notice that <cmath>1 = \frac{1}{2} + \frac{1}{2} + \frac{1}{2a_1}+\frac{1}{2a_2}+\cdots+\frac{1}{2a_k},</cmath> so <math>g_1 = 2 + 2a_1 + 2a_2 + ... + 2a_k = 2 + 2g</math> is also a good integer. Furthermore, <cmath>1 = \frac{1}{3} + \frac{1}{6} + \frac{1}{2} = \frac{1}{3} + \frac{1}{6} + \frac{1}{2a_1}+\frac{1}{2a_2}+\cdots+\frac{1}{2a_k},</cmath> so <math>g_2 = 3 + 6 + 2a_1 + 2a_2 + ... + a_k = 9 + 2g</math> is also good.
 
Proof: Let <math>g = a_1 + a_2 + \cdots + a_k</math> such that <cmath>\frac{1}{a_1}+\frac{1}{a_2}+\cdots+\frac{1}{a_k}=1,</cmath> where the <math>a_i</math> are positive integers. Then, notice that <cmath>1 = \frac{1}{2} + \frac{1}{2} + \frac{1}{2a_1}+\frac{1}{2a_2}+\cdots+\frac{1}{2a_k},</cmath> so <math>g_1 = 2 + 2a_1 + 2a_2 + ... + 2a_k = 2 + 2g</math> is also a good integer. Furthermore, <cmath>1 = \frac{1}{3} + \frac{1}{6} + \frac{1}{2} = \frac{1}{3} + \frac{1}{6} + \frac{1}{2a_1}+\frac{1}{2a_2}+\cdots+\frac{1}{2a_k},</cmath> so <math>g_2 = 3 + 6 + 2a_1 + 2a_2 + ... + a_k = 9 + 2g</math> is also good.
  

Revision as of 16:50, 14 August 2014

Problem

An integer $n$ will be called good if we can write

$n=a_1+a_2+\cdots+a_k$,

where $a_1,a_2, \ldots, a_k$ are positive integers (not necessarily distinct) satisfying

$\frac{1}{a_1}+\frac{1}{a_2}+\cdots+\frac{1}{a_k}=1$.

Given the information that the integers 33 through 73 are good, prove that every integer $\ge 33$ is good.

Hint

Is there an algorithm to deduce another good integer given a good integer? You might want to use the fact that $1 = \frac{1}{2} + \frac{1}{2}$.




Solution

Lemma: If $g$ is a good integer, then so is $2g + 2$ and $2g + 9$.

Proof: Let $g = a_1 + a_2 + \cdots + a_k$ such that \[\frac{1}{a_1}+\frac{1}{a_2}+\cdots+\frac{1}{a_k}=1,\] where the $a_i$ are positive integers. Then, notice that \[1 = \frac{1}{2} + \frac{1}{2} + \frac{1}{2a_1}+\frac{1}{2a_2}+\cdots+\frac{1}{2a_k},\] so $g_1 = 2 + 2a_1 + 2a_2 + ... + 2a_k = 2 + 2g$ is also a good integer. Furthermore, \[1 = \frac{1}{3} + \frac{1}{6} + \frac{1}{2} = \frac{1}{3} + \frac{1}{6} + \frac{1}{2a_1}+\frac{1}{2a_2}+\cdots+\frac{1}{2a_k},\] so $g_2 = 3 + 6 + 2a_1 + 2a_2 + ... + a_k = 9 + 2g$ is also good.

This completes the lemma. Now, we use induction to show that the integers from 33 to $n$ are all good, where $n \ge 33$.

Base case: Our given takes care of $33 \le n \le 73$. Inductive step: Assume the integers between 33 and $k > 73, inclusive, are all good. Now, we casework on parity of$k+1$.

If$ (Error compiling LaTeX. Unknown error_msg)k+1$is odd, then write$k+1 = 2g + 9$for some integer$g$.

It follows that because$ (Error compiling LaTeX. Unknown error_msg)k \ge 74$, we must have

<cmath>g = \frac{k}{2} - 4 \ge 37 - 4 = 33.</cmath>

Thus,$ (Error compiling LaTeX. Unknown error_msg)33 \le g < \frac{k}{2}$, and so$g$is good by the inductive hypothesis. Applying the lemma gives$2g + 9 = k+1$as a good integer.

If$ (Error compiling LaTeX. Unknown error_msg)k+1$is even, write$k+1 = 2g+2$for some integer$g$. Therefore, <cmath>g = \frac{k-1}{2} > \frac{73-1}{2} = 36 > 3,</cmath> so that$33 < g < \frac{k}{2}$, meaning that$g$is a good integer by the inductive hypothesis. From the lemma,$k+1$must be good as well.

This completes the inductive step and thus the induction. Therefore, all integers$ (Error compiling LaTeX. Unknown error_msg)n \ge 33$ are good integers.

See Also

1978 USAMO (Problems • Resources)
Preceded by
Problem 2 		Followed by
Problem 4
1 • 2 • 3 • 4 • 5
All USAMO Problems and Solutions

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