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If $a_{1},a_{2},\cdots,a_{n}(a_{i}\neq a_{j}),n\ge 3$ are positive integers,show that: we can always choose two positive integers among them, $a_{m},a_{k},m,k\in\{1,2,\cdots,n\}$.such that $$\dfrac{a_{m}+a_{k}}{3a_{p}}\notin N^{+},\forall p\in\{1,2,3,\cdots,n\}$$

This problem is from the Jiangxi province Mathematical Contest,2014 .(I failed This year's exam was very difficult)

my idea: if $n=3$,Assume that $a_{1}=1,a_{2}=2,a_{3}=3$,then we have $a_{p}=1$,then we can choose $a_{m}=2,a_{k}=3$,then it is clearly $$\dfrac{a_{m}+a_{k}}{3a_{p}}=\dfrac{5}{3}\notin N^{+}$$ if $a_{p}=2$,then we can choose $a_{m}=1,a_{k}=3$

if $a_{p}=3$ then we can choose $a_{m}=1,a_{k}=2$,

But for in general $n$,I can't prove it.

enter image description here

Thank you very much

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  • $\begingroup$ can you post you solution? Thank you $\endgroup$ – math110 Apr 12 '14 at 9:04
  • $\begingroup$ During a lot of time, I saw the expression "I fell" used by math110 and did not understand it. I’ve just realized that he means "I failed" $\endgroup$ – Ewan Delanoy Apr 12 '14 at 9:30
  • $\begingroup$ HaHa,I fell in chinese mean : 我感觉. $\endgroup$ – math110 Apr 12 '14 at 9:31
  • $\begingroup$ Must $a_m,a_k,a_p$ be distinct?The result then does not hold if you consider the list $1,3,6,9$ with $a_p=1$. $\endgroup$ – rah4927 Apr 12 '14 at 9:32
  • $\begingroup$ $a_{m},a_{k}$ be distinct.and $a_{p}$ is can equal $a_{m}$ or $a_{k}$, $\endgroup$ – math110 Apr 12 '14 at 9:36
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We can assume $a_1 < a_2 < a_3 < \ldots <a_n$ without loss. Since $n\geq 3$, we have $a_n \geq 3$.

Let us argue by induction on $m=a_n$. When $m=3$, we have $n=3, (a_1,a_2,a_3)=(1,2,3)$ and this case has already been treated by the OP.

Now, suppose that $m>3$ and that the result is true for $m-1$. Let $a_1 < a_2 < a_3 < \ldots <a_n$ be a sequence of positive integers. If, for some $i \neq j$, $a_i+a_j$ is not divisible by $3$, then we may take $m=i,k=j$ and we are done. So we can assume that all the $a_i+a_j(i \neq j)$ are divisible by $3$. So for any distinct $i,j,k$, we have that $a_j-a_i=(a_j+a_k)-(a_i+a_k)$ is divisible by $3$. So all the $a_i$ are congruent to a constant $a\in\lbrace 0,1,2\rbrace$ modulo $3$. Then $2a$ is divsible by $3$, so $a=0$. We see then that all the $a_i$ are divisible by $3$ ; write $a_i=3b_i$ where $b_i$ is a positive integer. By the induction hypothesis, there are indices $m\neq k$ such that $\frac{b_n+b_m}{3b_p} \not\in {\mathbb N}$ for any $p$. Since $\frac{a_n+a_m}{3a_p}=\frac{b_n+b_m}{3b_p}$, we are done.

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