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1992 USAMO Problems/Problem 3: Difference between revisions

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Done but not elegant...
Done but not elegant...


== Resources ==
== See Also ==


{{USAMO box|year=1992|num-b=2|num-a=4}}
{{USAMO box|year=1992|num-b=2|num-a=4}}
{{MAA Notice}}
{{MAA Notice}}
[[Category:Olympiad Combinatorics Problems]]

Latest revision as of 06:52, 19 July 2016

Problem

For a nonempty set $S$ of integers, let $\sigma(S)$ be the sum of the elements of $S$. Suppose that $A = \{a_1, a_2, \ldots, a_{11}\}$ is a set of positive integers with $a_1 < a_2 < \cdots < a_{11}$ and that, for each positive integer $n \le 1500$, there is a subset $S$ of $A$ for which $\sigma(S) = n$. What is the smallest possible value of $a_{10}$?

Solution

Let's a $n$-$p$ set be a set $Z$ such that $Z=\{a_1,a_2,\cdots,a_n\}$, where $\forall i<n$, $i\in \mathbb{Z}^+$, $a_i<a_{i+1}$, and for each $x\le p$, $x\in \mathbb{Z}^+$, $\exists Y\subseteq Z$, $\sigma(Y)=x$, $\nexists \sigma(Y)=p+1$.

(For Example $\{1,2\}$ is a $2$-$3$ set and $\{1,2,4,10\}$ is a $4$-$8$ set)


Furthermore, let call a $n$-$p$ set a $n$-$p$ good set if $a_n\le p$, and a $n$-$p$ bad set if $a_n\ge p+1$ (note that $\nexists \sigma(Y)=p+1$ for any $n$-$p$ set. Thus, we can ignore the case where $a_n=p+1$).


Furthermore, if you add any amount of elements to the end of a $n$-$p$ bad set to form another $n$-$p$ set (with a different $n$), it will stay as a $n$-$p$ bad set because $a_{n+x}>a_{n}>p+1$ for any positive integer $x$ and $\nexists \sigma(Y)=p+1$.


Lemma) If $Z$ is a $n$-$p$ set, $p\leq 2^n-1$.


For $n=1$, $p=0$ or $1$ because $a_1=1 \rightarrow p=1$ and $a_1\ne1\rightarrow p=0$.

Assume that the lemma is true for some $n$, then $2^n$ is not expressible with the $n$-$p$ set. Thus, when we add an element to the end to from a $n+1$-$r$ set, $a_{n+1}$ must be $\le p+1$ if we want $r>p$ because we need a way to express $p+1$. Since $p+1$ is not expressible by the first $n$ elements, $p+1+a_{n+1}$ is not expressible by these $n+1$ elements. Thus, the new set is a $n+1$-$r$ set, where $r\leq p+1+a_{n+1} \leq 2^{n+1}-1$

Lemma Proven


The answer to this question is $\max{(a_{10})}=248$.

The following set is a $11$-$1500$ set:

$\{1,2,4,8,16,32,64,128,247,248,750\}$

Note that the first 8 numbers are power of $2$ from $0$ to $7$, and realize that any $8$ or less digit binary number is basically sum of a combination of the first $8$ elements in the set. Thus, $\exists Y\subseteq\{1,2,4,8,16,32,64,128\}$, $\sigma(Y)=x \forall 1\le x\leq 255$.

$248\le\sigma(y)+a_9\le502$ which implies that $\exists A\subseteq\{1,2,4,8,16,32,64,128,247\}$, $\sigma(A)=x \forall 1\le x\leq 502$.

Similarly $\exists B\subseteq\{1,2,4,8,16,32,64,128,247,248\}$, $\sigma(A)=x \forall 1\le x\le750$ and $\exists C\subseteq\{1,2,4,8,16,32,64,128,247,248,750\}$, $\sigma(A)=x \forall 1\leq x\leq 1500$.


Thus, $\{1,2,4,8,16,32,64,128,247,248,750\}$ is a $11$-$1500$ set.



Now, let's assume for contradiction that $\exists a_{10}\leq 247$ such that ${a_1, a_2, \dots, a_{11}}$ is a $11$-$q$ set where $q\geq 1500$

${a_1, a_2, \dots a_8}$ is a $8$-$a$ set where $a\leq 255$ (lemma).

$max{(a_9)}=a_{10}-1\leq 246$

Let ${a_1, a_2, \dots, a_{10}}$ be a $10$-$b$ set where the first $8$ elements are the same as the previous set. Then, $256+a_9+a_{10}$ is not expressible as $\sigma(Y)$. Thus, $b\leq 255+a_9+a_{10}\leq 748$.

In order to create a $11$-$d$ set with $d>748$ and the first $10$ elements being the ones on the previous set, $a_{11}\leq 749$ because we need to make $749$ expressible as $\sigma(Y)$. Note that $b+1+a_{11}$ is not expressible, thus $d<b+1+a_{11}\leq 1498$.


Done but not elegant...

See Also

1992 USAMO (Problems • Resources)
Preceded by
Problem 2 		Followed by
Problem 4
1 • 2 • 3 • 4 • 5
All USAMO Problems and Solutions

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. Error creating thumbnail: Unable to save thumbnail to destination

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