1992 USAMO Problems/Problem 3

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...

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

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