$x=\frac{a_1+a_2+a_3+....+a_{2n}}{2n}$

Clearly $x\not= a_i$ for any $i$. So we have the product of $2n$ distinct non-zero integers which clearly reaches it's minimum value when $\{ x-a_i \}=\{-n,\ldots -1,1\ldots n\}$. In fact if these sets are not equal then the equality cannot occur since their absolute values would differ. So $(x-a_1)+(x-a_2)\ldots +(x-a_{2n})=-n\ldots -1+1\ldots n=0$ which implies $x=\frac{a_1+a_2\ldots +a_{2n}}{2n}$ It's a well known problem, can you believe it appeared on a UK FST in 2002!!

AK1024 wrote: Clearly $x\not= a_i$ for any $i$. So we have the product of $2n$ distinct non-zero integers which clearly reaches it's minimum value when $\{ x-a_i \}=\{-n,\ldots -1,1\ldots n\}$. In fact if these sets are not equal then the equality cannot occur since their absolute values would differ. So $(x-a_1)+(x-a_2)\ldots +(x-a_{2n})=-n\ldots -1+1\ldots n=0$ which implies $x=\frac{a_1+a_2\ldots +a_{2n}}{2n}$ It's a well known problem, can you believe it appeared on a UK FST in 2002!! And in Singapore Math Olympiad Senior in 2017

Infinity_Integral wrote: AK1024 wrote: Clearly $x\not= a_i$ for any $i$. So we have the product of $2n$ distinct non-zero integers which clearly reaches it's minimum value when $\{ x-a_i \}=\{-n,\ldots -1,1\ldots n\}$. In fact if these sets are not equal then the equality cannot occur since their absolute values would differ. So $(x-a_1)+(x-a_2)\ldots +(x-a_{2n})=-n\ldots -1+1\ldots n=0$ which implies $x=\frac{a_1+a_2\ldots +a_{2n}}{2n}$ It's a well known problem, can you believe it appeared on a UK FST in 2002!! And in Singapore Math Olympiad Senior in 2017 Just realised, also in Singapore Math Olympiad Junior 2017 The SMO Junior is for sec 1 and sec 2 students and here they have ISL questions inside. At least better than the Vieta Jumping question in SMO Junior 2023

$|(x-a_1)....(x-a_{2n})| = (n!)^2$ $|x-a_i| \geq 1$ achievable for two $a_i $ since they can be whether $1 ,-1$ next since they are distinct the next possible minimum is 2 $|x-a_j| \geq 2$ . . . $|x-a_j| \geq n$ therefore $((x-a_1),...,x-a_{2n})= (1,2,3,4,....,n,-1,-2,-3,..,-n)$ $2nx-(a_1+....+a_{2n}) = 0$ $x =\frac{ a_1+....+a_{2n}}{2n}$