I think it is not very hard problem ! . The first I will prove following result : Lemma : Exist an arithmetic sequence of rational number $ \frac{a_i}{b_i},i=1,..,k$ such that $ \gcd(a_i,b_i)=1$ and $ b_j$ are pairwise distinct . Let $ p$ is a prime greater than $ k$ . Consider following sequence : $ \frac{1}{p},\frac{1}{p}+\frac{1}{k!},...,\frac{1}{p}+\frac{k-1}{k!}$ For all $ d=0,...,k-1$ we have : $ \frac{1}{p}+\frac{d}{k!}=\frac{\frac{k!}{d}+p}{p.\frac{k!}{d}}$ It is not hard to show that $ p+\frac{k!}{d}$ and $ p.\frac{k!}{d}$ is relative prime . Hence $ b_j=p.\frac{k!}{d}$ , and obvious they are different from each other . Now let consider $ p_1,p_2$ are two different prime number , such that each of them is greater than $ \max\{a_1,b_1,..,a_k,b_k\}$ .Because the sequence $ \{\frac{a_i}{b_i}\}$ is an arithmetic sequence so $ \{\frac{a_i}{b_i}+\frac{p_1}{p_2}=\frac{c_i}{d_i}\}$ is also an arithmetic sequence . Because each of $ p_1,p_2$ greater than $ a_i,b_i$ so we have $ \gcd(a_i.p_2+b_i.p_1,b_i.p_2)=1$ . Therefore we must have $ c_i=a_i.p_2+b_i.p_1$ and $ d_i=b_ip_2$ . We will show that all $ c_i,d_j$ are different from each other . i ) If $ c_i=c_j$ then we have $ a_ip_2+b_ip_1=a_jp_2+b_j p_1$ , it gives $ p_2|b_i-b_j$ , because $ p_2>max\{b_i,b_j\}$ and $ b_i,b_j$ are different so it can't hold . ii) $ c_i=d_j$ then we have $ a_ip_2+b_ip_1=p_2b_j$ ,it gives $ p_2|b_1$ , it also can't hold . iii ) $ d_i=d_j$ , this can't hold because $ b_i,b_j$ are different from each other. Thus this sequence satisfies condition.

ya... i agree it's an easy problem in fact, many ways to construct them. One is like how the official solution construct it, and another common one is $ \frac{a_i}{b_i} = \frac{k!+i}{k!}$ I believe that most candidates get this problem.

brianchung11 wrote: ya... i agree it's an easy problem in fact, many ways to construct them. One is like how the official solution construct it, and another common one is $ \frac {a_i}{b_i} = \frac {k! + i}{k!}$ I believe that most candidates get this problem. try $ k=3$ you will get $ \frac{7}{6}$, $ \frac{4}{3}$, $ \frac{3}{2}$ I think better get this one: $ \frac {a_i}{b_i} = \frac {(k!)^2 + i}{(k!)^2}$

Let $y_1,y_2, \ldots , y_{k-1}, x_1,x_2, \ldots , x_{k-1}$ be $2k-2$ distinct primes that are greater than $k$. By CRT, there exists $x$ so that $\gcd (x,k!)=1, x \equiv -i \pmod{x_iy_i}$ for all $1 \le i \le k-1$. Now, we pick $\frac{a_1}{b_1}= \frac{1}{x_1x_2 \cdots x_{k-1}}$ and $d= \frac{1}{xx_1x_2 \cdots x_{k-1}}$. Hence, it's not hard to show that $$\frac{a_{i+1}}{b_{i+1}}= \frac{(x+i)/x_i}{xx_1x_2 \cdots x_ix_{i+1} \cdots x_{k-1}} \; \quad \text{and} \quad \gcd \left( \frac{x+i}{x_i}, \frac{xx_1 \cdots x_{k-1}}{x_i} \right)=1.$$Thus, $a_{i+1}= \frac{x+i}{x_i}, b_{i+1}= \frac{xx_1 \cdots x_{k-1}}{x_i}$ and $\gcd (a_{i+1},b_{i+1})=1$. Note that $a_{i+1} \ne a_{j+1}$ for all $i \ne j$ because $y_i \mid a_{i+1}$ but $y_i \nmid a_{j+1}$ since $y_i \nmid x+j$. We also have $b_i \ne b_j$ and $a_i \ne b_j$ (since $y_{i-1} \mid a_{i}$ but $y_{i-1} \nmid b_j$). Thus, we've guaranteed that $a_1, \ldots , a_k,b_1, \ldots ,b_k$ are pairwise distinct. We are done.

$k\ge 2$ we consider the following sequence. $\frac{(k!) ^2+1}{k! },\frac{(k!)^2+2}{k! },.....\frac{(k!) ^2+k}{k!}$ $gcd(k!,(k!) ^2+i)=i$ ,$i\in\mathbb \{1,2,3,.....k\}$ $i|(k!) $ and $i|(k!) ^2+i$ for all $i$ $b_i=\frac{k!}{i}$ and $a_i =\frac{(k!)^2+i}{i}$ and are $Z$ $a_i=\frac{(k!) ^2+i}{k! } > a_j=\frac{(k!) ^2+j}{j}>b_i>b_j$ Which is true for $1\le i<j\le k$ Example of a sequence $k=2$ $\frac{5}{2},\frac{6}{2},......$

Nice brianchung11 wrote: Prove that for any positive integer $ k$, there exists an arithmetic sequence $ \frac{a_1}{b_1}, \frac{a_2}{b_2}, \frac{a_3}{b_3}, ... ,\frac{a_k}{b_k}$ of rational numbers, where $ a_i, b_i$ are relatively prime positive integers for each $ i = 1,2,...,k$ such that the positive integers $ a_1, b_1, a_2, b_2, ..., a_k, b_k$ are all distinct. Solution: Pick primes $P_1 , P{_2} , \dots , P_k$ such that $| P_i - P_j | >k$ and let $x$ be an integer such that : $$\begin{array} {c}x \equiv -1 \pmod {P_1} \\ x \equiv -2 \pmod {P_2} \\ \vdots \\ x \equiv -k \pmod {P_k} \end{array}$$By CRT the existence of such an $x$ is guaranteed. Now, let $M = \prod_{i = 1}^{k} $ and consider the rational numbers $$ \frac{x}{M} , \frac{x+1}{M} , \dots , \frac{x+k}{M}$$Clearly, this sequence of rationals is an arithmetic progression. Moreover, observe that each of the numerator and denominators are distinct as $P_i$ divides no other number than $x+i$.

I hope it's correct just put $a_1=k!.p$ and $b_1=q$ let $r$ the difference between the terms in the arithmetic sequence we will put $r=\frac{c}{d.n!}$ such that $p,q,c,d$ are different primes greater than $k!$ $\frac{a_2}{b_2}=\frac{(k!)^2pd+cq}{k!qd}$ $\frac{a_3}{b_3}=\frac{\frac{(k!)^2pd}{2}+cq}{\frac{k!qd}{2}}$ $\frac{a_4}{b_4}=\frac{\frac{(k!)^2pd}{3}+cq}{\frac{k!qd}{3}}$ in general $\frac{a_i}{b_i}=\frac{\frac{(k!)^2pd}{i-1}+cq}{\frac{k!qd}{i-1}}$ it's clear that $(a_i,b_i)=1$ if we have $a_i=b_j$ then $\frac{(k!)^2pd}{i}+cq=\frac{k!qd}{j}$ but since $q $doesn't divide $\frac{(k!)^2pd}{i}$ that's impossible and we're done

Consider $\frac{p(k! + 1)}{k!}, \frac{p(k!/2 + 1)}{k!/2}, \frac{p(k!/3 + 1)}{k!/3}, \dots, \frac{p(k!/k + 1)}{k!/k}$ for some prime $p > k$ so that all the $a$'s are greater than all the $b$'s and $\gcd(a_i, b_i) = 1$.

Solved with nprime06, E_6 Construction: Take sufficiently large $n$ and consider the sequence \[1+\frac{1}{n!}, 1+\frac{2}{n!}, \dots\]Proof of Construction: Note that for the denominators to be distinct, we need \[\gcd(a_1, b_1)\neq \gcd(a_1, b_1\cdot d+a_1) \neq \cdots \neq \gcd(a_1, d\cdot (k-1)b_k)\]which implies $b_i > b_i\cdot d+a_1$ by repeatedly applying the euclidean algorithm. For the numerators to be distinct, it is sufficient to let $n$ be sufficiently large.

If $k=1,2$ it is easy to verify an arithmetic sequence exists, so assume $k \geq 3$. Consider an increasing list of distinct primes $p_1, p_2, \dots p_k$, where all the $p_i>k$. Then consider the arithmetic sequence \[\frac{x+1}{n}, \frac{x+2}{n}, \dots, \frac{x+k}{n}.\]I will show that by simplifying the fractions, we produce the desired sequence $ \frac{a_1}{b_1}, \frac{a_2}{b_2}, \frac{a_3}{b_3}, ... ,\frac{a_k}{b_k}$. Let $n=p_1\dots p_2, \dots, p_k$. Let $x$ be the minimal number that satisfies the following system of congruences: $x\equiv -1 \pmod {p_1}, x \equiv -2 \pmod {p_2}, \dots, x \equiv -k \pmod{p_k}$. A solution exists due to CRT. Then, for each term $\frac{x+i}{n}$ of the arithmetic sequence, both the numerator and denominator are divisible by $p_i$. In fact, I claim that dividing the numerator and denominator of $\frac{x+i}{n}$ by $p_i$ results in the desired arithmetic sequence. I claim that the gcd of the numerator and denominator of $\frac{x+i}{n}$ is $p_i$. To see this, assume some other prime $p_j$ divides $x+i$, so $x+i \equiv 0 \pmod {p_j}$. But $x+j \equiv 0 \pmod {p_j}$, so this means that \[x+i \equiv x+j \pmod {p_j} \implies i \equiv j \pmod {p_j}, \]which is a contradiction since $i$ and $j$ are both distinct and less than $p_j$. This establishes that $\gcd(a_i, b_i)=1$. To see that no two $a_i$ are the same, we will proceed by contradiction. Assume $a_i=a_j$ and $i>j$. We also have $$a_ip_i-a_jp_j=x+i-(x+j) \implies a_i(p_i-p_j)=i-j,$$a contradiction since $p_i-p_j >i-j$. To see that no $a_i=b_j$, we will again proceed by contradiction. Since $b_j=n/p_j$, we have that $a_i=p_1p_2\dots p_{j-1}p_{j+1}\dots p_k.$ But, $b_i=p_1p_2\dots p_{i-1}p_{i+1}\dots p_k$, so for $k\geq 3$, we have that $\gcd(a_i, b_i) \neq 1, $contradiction. So, we've established that the arithmetic sequence we've constructed does indeed satisfy all the requirements.

Simplify the following fractions $\dfrac{(n+1)!+2}{n+1}, \dfrac{(n+1)!+3}{n+1}, \dfrac{(n+1)!+4}{n+1}, \cdots, \dfrac{(n+1)!+n+1}{n+1}$ This clearly satisfies the conditions of the problem.

brianchung11 wrote: Prove that for any positive integer $ k$, there exists an arithmetic sequence $ \frac{a_1}{b_1}, \frac{a_2}{b_2}, \frac{a_3}{b_3}, ... ,\frac{a_k}{b_k}$ of rational numbers, where $ a_i, b_i$ are relatively prime positive integers for each $ i = 1,2,...,k$ such that the positive integers $ a_1, b_1, a_2, b_2, ..., a_k, b_k$ are all distinct. Induct,Suppose that $$\frac{a_1}{b_i}=a \;\;\;\;\; \frac{a_{n-1}}{b_{n-1}}=a+kd$$then at the $n$th step choose $$a_n=\left[\prod_{j=1}^{n-1} a_j \prod_{j=1}^{n-1} b_j \right] \cdot[a+(k+1)d] \;\;\; b_j=\left[\prod_{j=1}^{n-1} a_j \prod_{j=1}^{n-1} b_j \right]$$which clearly works.

Let $$\frac{a_i}{b_i}=\frac{(k!)^{1000}+i}{(k!)^{1000}}=\frac{1+(k!)^{1000}/i}{(k!)^{1000}/i},$$which is fully simplied. The absolute difference between $(k!)^{1000}/i$ and $(k!)^{1000}/j$ for $i \neq j$ is at least $\tfrac{(k!)^{1000}}{k(k-1)}\gg 1$, so the difference between $a_i-1$ and $b_j$ is massive and thus $a_i \neq b_j$ for $i \neq j$. Since $a_i \neq b_i$ clearly, we are done. $\blacksquare$

Choose primes $p_1>p_2>\ldots>p_k>k$. By the Chinese Remainder Theorem, consider a positive integer solution $x$ to the system of congruences: \begin{align*} x+1&\equiv 0\pmod{p_1}\\ x+2&\equiv 0\pmod{p_2}\\ &\vdots\\ x+k&\equiv 0\pmod{p_k} \end{align*}Since $p_i>k$ for all $i$, we have that $p_i\mid x+j$ if and only if $i=j$. Observe that $\gcd(a_i,b_i)=1$ because $p_j\neq a_i$ for $j\neq i$. Now, let $a_i=M\frac{x+i}{p_i}$ and $b_i=\frac{1}{p_i}\prod_{r=1}^k p_r$ for some integer $M>p_1p_2\ldots p_k$. Since $p_i$'s are distinct $b_i$'s are distinct. Since $p_1>p_2>\ldots>p_k$ we have $a_1<a_2<\ldots<a_k$ so the $a_i$'s are pairwise distinct. Since $M>p_1\ldots p_n$ we know $a_i\geq M>b_j$ for all $i,j$ so the $a_1,b_1,\ldots,a_k,b_k$ are all distinct. It remains to check that the $\frac{a_i}{b_i}$'s form an arithmetic progression. But we have $$\frac{a_i}{b_i}=\frac{M\frac{x+i}{p_i}}{\frac{1}{p_i}\prod_{r=1}^k p_r}=M\frac{x+i}{p_1p_2\ldots p_k}$$so $\frac{a_i}{b_i}$'s indeed form an arithmetic progression with common difference $\frac{M}{p_1p_2\ldots p_k}$.

Consider the arithmetic sequence \[\frac{1000!+1}{500!},\frac{1000!+2}{500!},\dots,\frac{1000!+100}{500!}.\]I claim that when the fractions are simplified, we have a valid construction. In fact, \[\gcd(1000!+a,500!)=\gcd(a,500!)=a.\] Note that no numerator can match a denominator as the numerators all have to be greater than $\dfrac{1000!}{500!}$ and the denominators all have to be less than $500!$. Obviously, no two numerators and no two denominators can match(we already know what the simplified fraction looks like), so we are done.

Suppose that $n=p_1p_2\cdots p_k$ for distinct primes such that $p_i>k$ for all $i$. Then, let $m$ be the smallest positive integer such that: (1) $m\equiv -i\pmod {p_i}$ for all $1\leq i\leq k$, (2) $m\geq G_{10^{1428k^{1434}}}$, where $G$ denotes the Graham sequence. Then, let the $k$ fractions be $$\frac{m+1}{n},\frac{m+2}{n}\cdots +\frac{m+k}{n}.$$Note that for any $1\leq i\leq k$, $m+i$ is divisible by $p_i$ but not by any of the other $k-1$ primes. Thus, after reducing this is $$\frac{(m+1)/p_1}{n/p_1},\frac{(m+2)/p_2}{n/p_2}\cdots +\frac{(m+k)/p_k}{n/p_k}.$$The denominators are clearly all different, and the numerators are far larger than the denominators (e.g. by Bertrand's postulate), so we just have to show that the numerators are different from each other. However, this is clearly true as well, as $m$ is very large so the effect of the addition of a constant at most $k$ is dominated by the difference in the prime that it is being divided by. Thus, we are done.

Consider the sequence of numerators $1 + d, 1 + 2d, ..., 1 + kd$ such that $1 + d \equiv 0 \pmod p_1$ $1 + 2d \equiv 0 \pmod p_2$ \vdots We chose the $p_i$ such that they are all larger than $k$ (since d is not 0 mod any of the primes, this makes it so that only 1 of the terms is divisible by each $p_i$). By CRT, exists a solution for $d$. We then make the numerators $1 + d$, $1 + 2d$, etc. and the denominators for all of them $p_1 \cdot p_2 \dots p_k$ (then simplifying will produce k distinct denominators). Also, the numerators are distinct from each other because we can let $p_{a + 1} > 2 \cdot p_a$ so the numerators. always decrease as the primes increase. We then multiply the denominator by the smallest prime greater than all of the $a$ to make sure the numerators are all distinct from the denomonators.

\begin{remark} Here was a fakesolve that i would've walked away with if not for $\textbf{incompleteusern/yaoaops}$; i decided to include it because its funny

\end{remark} Ok now I got a hint, here's the correct attempt: Take the numbers $\frac{((10)^{10^{10^{10^{10}}}})!+i}{((1434)^{1434})!}\forall 1\le i\le n$, which works since numerators easily larger than denominators and obviously distinct since each numerator is divided by pairwise different numbers, and denominators are pairwise distinct since they're also divided by pairwise different numbers. (actually we'd need sufficiently larger numbers if n was even larger, but for the sake of generalizing i showed a funny case)

Written by abeot with ccarolyn4 Choose primes $p_1, p_2, \dots, p_k$, such that $p_1 > p_2 > \dots > p_k$. Let $N = p_1p_2\cdots p_k$. Then choose $x$ such that $x \equiv -i \bmod p_i$ for $i = 1, 2, \dots, k$. By CRT this is always possible. We show that the reduced sequence is valid. First, we examine the $b_i$. Note that $b_i = \frac{N}{p_i}$, so all the $b_i$ are distinct. Now, we just need to show that the $a_i$ are distinct. Suppose that $a_i = a_j$, where $i<j$. Then since $a_i = \frac{x+i}{p_i}$, so then we must have that $p_j(x+i) = p_i(x+j)$. This is impossible. Thus, the $a_i$ are all distinct. $\blacksquare$ Remark: Our motivation for this solution was trying to get values in the form $\frac{x+1}{N}, \frac{x+2}{N}, \dots$. However, there is one big problem; this means that all $b_i$ are the same; we fix this by making $N = p_1 p_2 \dots$ for some primes $p$ such that $x \equiv -i \pmod{p_i}$. This means that the top and bottom will simplify, and the rest is written in the proof above.

We claim that for "appropriate" positive integers $a,b$ and distinct primes $p_1,p_2,\ldots,p_k$, the following construction works: \[\frac{a}{p_1p_2\cdots p_k},\frac{b}{p_2p_3\cdots p_k},\frac{\frac{2bp_1-a}{p_2}}{p_1p_3\cdots p_k},\cdots,\frac{\frac{kbp_1-(k-1)a}{p_k}}{p_1p_2\cdots p_{k-1}}\]For this constuction to work, we need the primes and the numbers to follow: For each $1\leq i\leq k$, we have $p_i\nmid ab$, except $p_1\mid b$ is allowed. For each $1\leq i\leq k$, we have $p_i\mid ibp_1-(i-1)a$. For each $1\leq i\leq k$, we have $p_i\nmid bp_1-a$. For $1\leq i,j\leq k$ such that $i\ne j$, then $p_j\nmid ibp_1-(i-1)a$. This gives a sufficient condition to work with. The first three properties can be satisfied by Chinese Remainder Theorem. For satisfying the fourth property, note that the common difference of the arithmetic progression is $\frac{bp_1-a}{p_1p_2\cdots p_k}$. Consider the numerators obtained $\pmod{p_i}$ when the denominators are set $p_1p_2\cdots p_k$. Two numerators (suppose numerator $i_1$ and $i_2$) can be congruent $\pmod{p_i}$ only if $p_i\mid (i_2-i_1)(bp_1-a)$. But $p_i\nmid bp_1-a$, so $p_i\mid i_2-i_1$. However, $-k<i_2-i_1<k$, so we can avoid this issue if we set $p_i>k$ for each $1\leq i\leq k$. The construction is complete.

...? $\frac{(n!)^2+1}{n!}, \frac{(n!)^2+2}{n!}, \dots, \frac{(n!)^2+n}{n!}$.

Consider the following $n$ fractions \[\frac{n!+1}{n!} ,\frac{\left(\frac{n!}{2}\right)+1}{\left(\frac{n!}{2}\right)}, \frac{\left(\frac{n!}{3}\right)+1}{\left(\frac{n!}{3}\right)} ,\ldots ,\frac{\left(\frac{n!}{n}\right)+1}{\left(\frac{n!}{n}\right)}\]They all satisfies $\gcd(a_i,b_i)=1$, all numerators are distinct as well as all denominators

Consider the fractions \[\frac{k!+1}{k!}, \frac{k!+2}{k!}, \ldots, \frac{k!+k}{k!}.\] Since $\gcd(k!+i,k!)=i$ for each $i = 1,\ldots,k$, the denominators are pairwise distinct and the numerators are pairwise distinct. We then simply scale the denominator by an arbitrarily large prime to ensure all $2k$ numerators and denominators are distinct. $\blacksquare$

For $n=1$, $\frac{7}{6}$ works. For $n=2$, $\frac{7}{6}, \frac{4}{3}$ works For $n=3$, $\frac{13}{6}, \frac{7}{3}, \frac{5}{2}$ works Otherwise, $1+\frac{k}{n!}$ from $k=1$ to $n$ works. This is an arithmetic sequence because the common difference is $\frac{1}{n!}$ The denominator is $\frac{n!}{k}$ since $k$ divides $n!$. This value differs for values of $k$. The numerator is $1+\frac{n!}{k}$. This is different from the denominators since it is always odd but the denominators are even.

what? \[\frac{a_i}{b_i}=\frac{n!-i}{p\cdot n!}\]for a sufficiently large prime $p$ works. $\blacksquare$