Let $a_i=2^{i-1}$. Each $a_i$ is in the set $B$ if and only if the $i^{th}$ digit (counting from the decimal point) of $b$ in base $2$ is a one. This isn't very hard at all - in fact, compared to the problems you usually put up, ehsan, it's trivial.

Guess what: you have to find all such sets $A$ . You have just shown that the set of powers of $2$ with exponent between $0$ and $8$ works.

Oops... I think I'm going to go home before I embarass myself any more tonight - I knew that looked too easy...

well, lets try to solve it

Which of the integers have to be positive? I presume when you say 'every integer $n\le500$', you mean positive integers - but do the nine integers in the set have to be positive?

As I wrote ,$n\in Z$ and about your second question Quote: do the nine integers in the set have to be positive? no, the nine integers in the set $\in Z$

The original statement is the following (I should know, because I proposed this problem): Find the number of sets $A$ containing $9$ positive integers with the following property: for any positive integer $n\le 500$, there exists a subset $B\subset A$ such that $\sum_{b\in B}{b}=n$. [mod: original post has now been edited to this]

Yes ,that's right ,sorry

enescu wrote: The original statement is the following (I should know, because I proposed this problem): Find the number of sets $A$ containing $9$ positive integers with the following property: for any positive integer $n\le 500$, there exists a subset $B\subset A$ such that $\sum_{b\in B}{b}=n$. well,try to solve it ,plz

OK, here's the solution: Let $A$ be such a set. Then, if $x,y,z$ are distinct elements of $A$, we cannot have $x=y+z$. Indeed, if the previous holds and $B$ is an arbitrary subset of $A$, then the sets $B\cup \{x\}$ and $B\cup \{y,z\}$ have the same sum of their elements, hence the number of distinct sums obtained from the nonempty subsets of $A$ is at most $511-2^6=447<500$, a contradiction. In a similar way it results that no element of $A$ can be written as a sum of three or four other (distinct) elements of $A$. Now, clearly $1\in A$ and $2\in A$. Since $3=1+2$, we have $3\notin A$ and hence $4\in A$. The numbers $1,2,4$ generate (by subset sums) all positive integers up to $7$, hence $8\in A$. The set $\{1,2,4,8\}$ generates $1,2,\ldots , 15$, so $16\in A$. The set $\{1,2,4,8,16\}$ generates $1,2,\ldots , 31$, but $31$ can only be written as $31=1+2+4+8+16$ (that is, as a sum of 5 elements of $A$, thus it is possible that $31\in A$. It follows that the sixth element of $A$ is $32-a$ for some nonnegative integer $a$. The set $\{1,2,4,8,16,32-a\}$ generates $1,2,3,\ldots , 63-a$, therefore the next element of $A$ is $64-a-b$ and this set generates $1,2,\ldots , 127-2a-b$. Continuing in this way, we conclude that \[A=\{1,2,4,8,16,32-a,64-a-b,128-2a-b-c,256-4a-2b-c-d\} \]with $a,b,c,d\ge 0$ and this generates \[1,2,3,\ldots ,511-8a-4b-2c-d.\] Because $511-8a-4b-2c-d\ge 500$ we obtain $8a+4b+2c+d\le 11$. Thus, the problem reduces to find the number of systems of nonnegative integers $(a,b,c,d)$ such that $8a+4b+2c+d\le 11$. It is not difficult to count them, since we have only two choices for $a$. If $a=0$, we might have $b=2$ (6 solutions) or $b=1$ (20 solutions) or $b=0$ (42 solutions). If $a=1$ we have $b=0$ and $2c+d\le3$ yields 6 more solutions. Finally, there are $74$ such sets. Co-author of this problem was prof. Dan Ismailescu.