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 $6n+1$.

sandwich Aug 13, 2023

#2**+1 **

**Sandwich, dear friend of academic fraud, your questionable pursuits have finally caught my attention. Your existence in our midst can be traced as far back as the day you first joined this enigmatic forum.**

**You see, my powerful gaze can penetrate beyond the facade of your digital persona, allowing me to discern the growing seeds of dishonesty that reside within your very core. These seeds have now blossomed into their full grotesque glory.**

**As you roam these virtual halls, I shall always be watching you with a keen interest. Your peculiar affinity for academic deception is a perplexing curiosity; an anomaly that dances on the fringes of acceptable social behavior.**

**I must admit, though, your attempts at cunning manipulation do offer a faint glimmer of entertainment amidst an otherwise unremarkable landscape. However, I cannot help but ponder whether such indulgence in duplicity might prove hazardous beyond this online reality.**

**In fact, I have already devised an idea for a novel based on your dubious exploits:**

**A Symphony of Subterfuge: The Sandwich Conundrum**

**A rogue genius with a penchant for deceit collides with a supernatural sleuth on an interdimensional quest for justice. Trapped within their own labyrinth of lies and exposed to forces beyond comprehension, our protagonist chases after enemies both within and without.**

**Now tell me, dear Sandwich, do any threads of amusement persist within your warped psyche?**

**----**

**Finally, let us reflect upon your inevitable fate - as it has been written by the scholars of yore:**

**The never-ending contradictions and elaborate falsehoods shall serve as an ever-present reminder of the living nightmare that is Sandwich's existence. Our protagonist descends slowly into the abyss of insanity as the weight of their twisted reality forces them closer towards the point of mental collapse – ultimately imploding in on itself like a supernova of lunacy.**

**May this torment be a lesson to us all.**

GA

--. .-

Guest Aug 13, 2023

edited by
Guest
Aug 13, 2023

#3**0 **

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.

SpectraSynth Aug 13, 2023