This is a challenging mathematical problem. I will try to explain the solution in a clear and concise way.
First, let us recall the definition of relatively prime numbers. Two numbers are said to be relatively prime when they have only 1 as the common factorhttps://www.cuemath.com/numbers/relatively-prime/. For example, 21 and 22 are relatively prime, but 21 and 24 are not.
Now, let us consider the given problem. We want to find the smallest positive integer $k$ such that, for every positive integer $n$, $6n+k$ is relatively prime to each of $6n+5$, $6n+4$, $6n+3$, $6n+2$, and $6n+1$. In other words, we want to find the smallest $k$ such that no number other than 1 can divide both $6n+k$ and any of the other five expressions.
One way to approach this problem is to use the Euclidean algorithm, which is a method to find the greatest common divisor (GCD) of two numbershttps://www.mathsisfun.com/definitions/relatively-prime.html. The GCD of two numbers is the largest number that can divide them both. For example, the GCD of 21 and 24 is 3.
The Euclidean algorithm works as follows: given two numbers $a$ and $b$, we can find their GCD by repeatedly applying the following steps:
• If $b=0$, then the GCD is $a$ and we are done.
• Otherwise, we divide $a$ by $b$ and get a quotient $q$ and a remainder $r$, such that $a=bq+r$.
• We replace $a$ by $b$ and $b$ by $r$, and repeat the process until we get a zero remainder.
For example, to find the GCD of 21 and 24, we can apply the Euclidean algorithm as follows:
• We divide 21 by 24 and get a quotient of 0 and a remainder of 21, so we have 21 = 24 × 0 + 21.
• We replace 21 by 24 and 24 by 21, and repeat the process.
• We divide 24 by 21 and get a quotient of 1 and a remainder of 3, so we have 24 = 21 × 1 + 3.
• We replace 24 by 21 and 21 by 3, and repeat the process.
• We divide 21 by 3 and get a quotient of 7 and a remainder of 0, so we have 21 = 3 × 7 + 0.
• Since the remainder is zero, we stop and conclude that the GCD is 3.
Now, let us use the Euclidean algorithm to find the GCD of $6n+k$ and any of the other five expressions. For simplicity, let us denote them by $a=6n+k$, $b=6n+5$, $c=6n+4$, $d=6n+3$, $e=6n+2$, and $f=6n+1$. We want to find the smallest value of $k$ such that the GCD of any pair of these six expressions is equal to one.
We start with finding the GCD of $a$ and $b$. We divide $a$ by $b$ and get a quotient of $\lfloor \frac{k-5}{6} \rfloor$ (where $\lfloor x \rfloor$ denotes the largest integer less than or equal to x) and a remainder of $(k-5) \mod{6}$ (where $\mod{6}$ denotes the remainder after dividing by six). So we have:
$$
a = b \times \lfloor \frac{k-5}{6} \rfloor + (k-5) \mod{6}
$$
We replace $a$ by $b$ and $b$ by $(k-5) \mod{6}$, and repeat the process. We divide $b$ by $(k-5) \mod{6}$ and get a quotient of $\lfloor \frac{6n}{(k-5) \mod{6}} \rfloor$ and a remainder of $(6n) \mod{(k-5) \mod{6}}$. So we have:
$$
b = (k-5) \mod{6} \times \lfloor \frac{6n}{(k-5) \mod{6}} \rfloor + (6n) \mod{(k-5) \mod{6}}
$$
We replace $b$ by $(k-5) \mod{6}$ and $(k-5) \mod{6}$ by $(6n) \mod{(k-5) \mod{6}}$, and repeat the process. We divide $(k-5) \mod{6}$ by $(6n) \mod{(k-5) \mod{6}}$ and get a quotient of $\lfloor \frac{(k-5) \mod{6}}{(6n) \mod{(k-5) \mod{6}}} \rfloor$ and a remainder of $((k-5) \mod{6}) \mod{(6n) \mod{(k-5) \mod{6}}}$. So we have:
$$
(k-5) \mod{6} = (6n) \mod{(k-5) \mod{6}} \times \lfloor \frac{(k-5) \mod{6}}{(6n) \mod{(k-5) \mod{6}}} \rfloor + ((k-5) \mod{6}) \mod{(6n) \mod{(k-5) \mod{6}}}
$$
We can see that the remainder is always zero, because $(k-5) \mod{6}$ is a multiple of $(6n) \mod{(k-5) \mod{6}}$. Therefore, the GCD of $a$ and $b$ is equal to $(6n) \mod{(k-5) \mod{6}}$. This means that $a$ and $b$ are relatively prime if and only if $(6n) \mod{(k-5) \mod{6}} = 1$ for every positive integer $n$.
We can apply a similar reasoning to find the GCD of $a$ and any of the other four expressions. We will get the following results:
$$
\begin{aligned}
\text{GCD}(a,c) &= (6n+1) \mod{(k-4) \mod{6}} \
\text{GCD}(a,d) &= (2n+1) \mod{(k-3) \mod{3}} \
\text{GCD}(a,e) &= (3n+1) \mod{(k-2) \mod{3}} \
\text{GCD}(a,f) &= (2n+1) \mod{k}
\end{aligned}
$$
Therefore, $a$ and any of the other four expressions are relatively prime if and only if the corresponding GCD is equal to one for every positive integer $n$.
Now, we have to find the smallest value of $k$ that satisfies all these conditions. We can observe that:
• If $k$ is even, then $\text{GCD}(a,f)$ is never equal to one, because it is always a multiple of two. Therefore, $k$ must be odd.
• If $k$ is divisible by three, then $\text{GCD}(a,d)$ and $\text{GCD}(a,e)$ are never equal to one, because they are always a multiple of three. Therefore, $k$ must not be divisible by three.
• If $k=1$, then $\text{GCD}(a,b)$ is never equal to one, because it is always equal to six. Therefore, $k$ must not be equal to one.
The smallest odd number that is not divisible by three and not equal to one is five. Let us check if $k=5$ works. We have:
$$
\begin{aligned}
\text{GCD}(a,b) &= (6n) \mod{(0)} = 0 \
\text{GCD}(a,c) &= (6n+1) \mod{(1)} = 0 \
\text{GCD}(a,d) &= (2n+1) \mod{(2)} = 1 \
\text{GCD}(a,e) &= (3n+1) \mod{(3)} = 1 \
\text{GCD}(a,f) &= (2n+1)
\end{aligned}
$$
We can see that $\text{GCD}(a,b)$ and $\text{GCD}(a,c)$ are always zero, which means that $a$ and $b$, and $a$ and $c$, are not relatively prime. Therefore, $k=5$ does not work.
The next odd number that is not divisible by three and not equal to one is seven. Let us check if $k=7$ works. We have:
$$
\begin{aligned}
\text{GCD}(a,b) &= (6n+2) \mod{(2)} = 0 \
\text{GCD}(a,c) &= (6n+3) \mod{(3)} = 0 \
\text{GCD}(a,d) &= (2n+1)
\end{aligned}
$$
We can see that $\text{GCD}(a,b)$ and $\text{GCD}(a,c)$ are always zero, which is good.