we have $1\le 2500$, then there is subset $S$ of $A$ whose sum of elements is $1$ so, $card(S)=1$ => $S=\{a_{1}\}$ => $a_{1}=1$

aviateurpilot wrote: we have $1\le 2500$, then there is subset $S$ of $A$ whose sum of elements is $1$ so, $card(S)=1$ => $S=\{a_{1}\}$ => $a_{1}=1$ Are you sure ?

N.T.TUAN wrote: aviateurpilot wrote: we have $1\le 2500$, then there is subset $S$ of $A$ whose sum of elements is $1$ so, $card(S)=1$ => $S=\{a_{1}\}$ => $a_{1}=1$ Are you sure ? we have $a_{1}<a_{2}<..<a_{12}$, then $min(\{sum\ of\ elements \ of\ S/ \ S\in P(A)\})=a_{1}$ then there isn't subset $S\in P(A)$ whose sum of elements is n, if $n< a_{1}$ then, $a_{1}=0$

aviateurpilot wrote: we have $a_{1}<a_{2}<..<a_{12}$, then $min(\{sum\ of\ elements \ of\ S/ \ S\in P(A)\})=a_{1}$ then there isn't subset $S\in P(A)$ whose sum of elements is n, if $n< a_{1}$ then, $a_{1}=0$ Are you sure? again

I am null in English, wanted you to explain me the question ?

read, please! N.T.TUAN wrote: each positive integer $n \leq 2500$ there is a subset $S$ of $A$ whose sum of elements is $n$.

I agreee with aviateurpilot that $a_1=1$. Didn't you want to ask to $a_{12}$? Or didn't you want to say $a_1>a_2>\dots >a_{12}$?