i don't understand!! you main $ a_k = \sum_{i\in S_k}a_i$ where $ S_k = \{i\in\{1,2,...,n\}: \ k|a_i\}$

aviateurpilot wrote: you main $ a_k = \sum_{i\in S_k}a_i$ where $ S_k = \{i\in\{1,2,...,n\}: \ k|a_i\}$ Yes....

Ahiles wrote: aviateurpilot wrote: you main $ a_k = \sum_{i\in S_k}a_i$ where $ S_k = \{i\in\{1,2,...,n\}: \ k|a_i\}$ Yes.... ok, and you $ a_1,a_2,...,a_n\in\mathbb{Z}$ ?? because $ a_1=\sum_{i=1}^{n}a_i$ gives $ \sum_{i=2}^{n}a_i=0$

aviateurpilot wrote: because $ a_1 = \sum_{i = 1}^{n}a_i$ gives $ \sum_{i = 2}^{n}a_i = 0$ $ a_1 = \sum_{i = 2}^{n}a_i$

ok. then $ a_k=\sum_{i\in S_k,i\neq k}a_i$ and $ a_k=0$ if $ S_k=\emptyset$ and for exemple $ n=2$ works: $ (a_1,a_2)=(2,2)$

Ahiles wrote: ... Find all possible values for $ n$. What about others ???

Let f(x)=(x-a1)...(x-an) We have f(x)=x^n-a1.x^(n-1)+...+(-1)^n.an

mto wrote: Let f(x)=(x-a1)...(x-an) We have f(x)=x^n-a1.x^(n-1)+...+(-1)^n.an I don't understand what you say. Do natural numbers conclude 0? Could we choose (2,2,0,0,0,0,0,0,0,……)？

Natural numbers certainly don't include 0.but I don't know for this problem....

Ahiles wrote: In a sequence of natural numbers $ a_1,a_2,...,a_n$ every number $ a_k$ represents sum of the multiples of the $ k$ from sequence. Find all possible values for $ n$. im am very bad in english, and i don't undersood the probleme at fist time, i asked $ Ahiles$ if $ a_k=\sum_{i\in S_k}$ where $ S_k=\{j\neq i: \ k|a_j\}$ and we say, that it's true . and for exemple $ n=2$ works where $ (a_1,a_2)=(2,2)$ but i think that the true si that $ a_k=\sum_{0\le i_1<i_2<...<i_k\le n}\prod_{j=1}^{k}a_{i_j}$ then it's obvious like say $ mto$ that $ \boxed{\prod_{i=1}^{n}(X-a_i)=X^n+\sum_{j=1}^{n}(-1)^{i}a_jX^{n-j}}$ but we will have $ a_1=\sum_{i=1}^na_i$ ===> $ \sum_{i=2}^{n}a_i=0$ ===> $ a_2=a_3=...=a_n=0$ . the probleme here is not complete, or is fals

Ahiles missunderstood the problem.The problem stated in the official TST was that $ a_{k}$ is the number of multiples of $ k$ from the sequence,for all $ k=1,..,n$.

Any solution please

Here is a solution where we simply use the meaning of double counting. Denote $D(x)$ the number of positive divisors of $x$ . Also, observe that for all $1\leq i\leq n$ we have $0\leq a_i\leq n$. Correction, there is no $i$ such that $a_i=0$, because $0$ itself is a multiple of $i$, hence a contradiction. However, double counting the sum of all $a_i$ , we find out that: $a_1 + a_2 + ... + a_n = D(a_1) + D(a_2) + ... + D(a_n)$ It is easy to check that for all $3\leq n$ we have : $D(n) \leq n-1$ and that $D(n)=n$ only for $n=1,2$ thus all $a_i$ must be either $1$ or $2$ but if $3 \leq n$ ,since $a_1$ is always equal to $n$ ,we reach a contradiction. So only $n=1,2$ is possible and you can easily check all possibilities for $n=1,2$ and find out that both $n=1$ and $n=2$ have at least one solution.