a bit too easy for a chinese TST 3rd problem. We can simply choose $A=\left\{k+1,2k,8k^2+8k+2,252,5,1 \right\} $ $B=\left\{k,2k+2,8k^2+8k+1,251 \right\} $ for any $k>252$

For any positive integer $k>100$, we can take $A = \left\{ {1,5,252} \right\} \cup \left\{ {4k + 1,6k,12k - 1,8{k^2} + 3k - 1} \right\}$, and $B = \left\{ {251} \right\} \cup \left\{ {4k + 2,6k - 1,12k - 2,8{k^2} + 3k} \right\}$

yunxiu wrote: For any positive integer $k>100$, we can take $A = \left\{ {1,5,252} \right\} \cup \left\{ {4k + 1,6k,12k - 1,8{k^2} + 3k - 1} \right\}$, and $B = \left\{ {251} \right\} \cup \left\{ {4k + 2,6k - 1,12k - 2,8{k^2} + 3k} \right\}$ How can we get this “ugly” solution？ Because $\frac{{{x_n}}}{{{x_{n - 1}}}} = \frac{{2(2n - 1)}}{n}$, $\frac{{{x_a}}}{{{x_{a - 1}}}}\frac{{{x_b}}}{{{x_{b - 1}}}} = 4\left( {\frac{{2a - 1}}{a}} \right)\left( {\frac{{2b - 1}}{b}} \right)$, $\left( {\frac{{2a - 1}}{a}} \right)\left( {\frac{{2b - 1}}{b}} \right)$ with odd numerator and close to $4$, so we can suppose $\left( {\frac{{2a - 1}}{a}} \right)\left( {\frac{{2b - 1}}{b}} \right) = \frac{{8k - 1}}{{2k}}$, so $b = \frac{{2k(2a - 1)}}{{a - 4k}}$, take $a = 4k + 2$ or $a = 6k$ we get $b = k(8k + 3)$ or $b = 12k - 1$. So $\frac{{{x_{4k + 2}}}}{{{x_{4k + 1}}}}\frac{{{x_{8{k^2} + 3k}}}}{{{x_{8{k^2} + 3k - 1}}}} = \frac{{2(8k - 1)}}{k} = \frac{{{x_{6k}}}}{{{x_{6k - 1}}}}\frac{{{x_{12k - 1}}}}{{{x_{12k - 2}}}}$. For any positive $k \geqslant 100$,we can take $A = {A_0} \cup \left\{ {4k + 1,6k,12k - 1,8{k^2} + 3k - 1} \right\}$, $B = {B_0} \cup \left\{ {4k + 2,6k - 1,12k - 2,8{k^2} + 3k} \right\}$. We can suppose $\left( {\frac{{2a - 1}}{a}} \right)\left( {\frac{{2b - 1}}{b}} \right) = \frac{{4k - 1}}{k}$ to get an other solution. Because $b = \frac{{k(2a - 1)}}{{a - 2k}}$. Let $a = 3k$ or $a = 2k + 1$, we get $b = 6k - 1$ or $b = k(2a - 1) = k(4k + 1)$, hence $\frac{{{x_{3k}}}}{{{x_{3k - 1}}}}\frac{{{x_{6k - 1}}}}{{{x_{6k - 2}}}} = \frac{{4(4k - 1)}}{k} = \frac{{{x_{2k + 1}}}}{{{x_{2k}}}}\frac{{{x_{4{k^2} + k}}}}{{{x_{4{k^2} + k - 1}}}}$. So we can take $A = {A_0} \cup \left\{ {2k + 1,3k - 1,6k - 2,4{k^2} + k} \right\}$,$B = {B_0} \cup \left\{ {2k,3k,6k - 1,4{k^2} + k - 1} \right\}$.

A possible construction is $S=\{ 252, 5, 3c, 6c-1, 2c-1\}, T=\{ 251, 6c-2, 3c-1, 2c,1\}$ for $c>1000$. Note $\frac{x_i}{x_{i-1}}=\frac{2(2i-1)}{i}$. We verify this construction works. $\frac{x_{252}}{x_{251}}=2\cdot \frac{503}{252}$, so $x_{252} x_5 = 1006 x_{251}$ We now take care of the other terms. Note $$\frac{x_{6c-1}}{x_{6c-2}} \cdot \frac{x_{3c}}{x_{3c-1}}=4\cdot \frac{12c-3}{6c-1} \cdot \frac{6c-1}{3c} =4(4-\frac 1c) = 4\cdot (2\cdot (2-\frac{1}{2c}))=4\cdot \frac{x_{2c}}{x_{2c-1}}$$ Which gives left hand side another factor of 4. Finally the right hand side has a factor of 2, which gives a product of $\frac{1006\cdot 4}{2}=2012$. How do we motivate this solution? Dealing with the $\binom{2i}{i}$ terms are difficult, but $\frac{x_{i+1}}{x_i}$ can reduce this to algebraic manipulation, making the problem much easier. We will deal with the algebra first and the constants later. We can use telescoping to get a $4(4-\frac{3}{c})=2\frac{4c-3}{2c-1} \cdot 2\frac{2c-1}{c}=\frac{x_{2c-1}}{x_{2c-2}} \cdot \frac{x_c}{x_{c-1}}$ Now note $4-\frac{3}{c}=2(2-\frac{3}{2c})$ so the term returns. DONE. @veyron I do not think this is too easy for China TST 3. I find it harder than China 2012 TST 2 P6, and some say Quiz 1s are supposed to be relatively easy.

We first make an extraordinary claim. Claim: For every positive integer $t$, set $A_t:=\{ 10t,40t-2,8t-1\}$ and $B_t:=\{ 10t-1,40t-3,8t\}$. We have $$\frac{\prod\limits_{i\in A_{t}}{x_i}}{\prod\limits_{j\in B_{t}}{x_j}}=4$$Proof of the claim: Notice that $\frac{x_{2k}}{x_{2k-1}}=\frac{4k-1}{k}$, the claim is almost trivially true. $\blacksquare$ Now we can begin our construction of $A$ and $B$ as follow: Pick any $3$ distinct integers $t_{1}, t_{2} ,t_{3}$ larger than $1000$ such that the sets $A_{t_{1}},B_{t_{1}}, A_{t_{2}},B_{t_{2}}, A_{t_{3}},B_{t_{3}}$ have pairwise empty intersection. Now set $A=\{252,32,1\}\cup A_{t_{1}}\cup \dots\cup A_{t_{3}}$, $B=\{251,31\}\cup B_{t_{1}}\cup\dots\cup B_{t_{3}}$. It is not hard to see that such $A,B$ satisfies the condition that we are asking for, in addition there are obviously infinitely many such $A,B$.