"Find the smallest positive integer \(k\) such that, for every positive integer \(n\), is relatively prime to each of \(6n+3\), \(6n+2\), and \(6n+1\)."

I'm not to do this. Please help. Thanks!

Guest Jun 9, 2023

#1**0 **

We have that 6n+1, 6n+2, and 6n+3 are all multiples of 3, so 6n+k must be odd. Therefore, k must be divisible by 2. If k is divisible by 4, then 6n+k is also divisible by 4, so k must be a multiple of 6.

Now, we need to show that if k is a multiple of 6, then 6n+k is relatively prime to each of 6n+3, 6n+2, and 6n+1.

Suppose k=6m for some positive integer m. Then \begin{align*} \gcd(6n+k, 6n+3) &= \gcd(6n+6m, 6n+3) \ &= \gcd(6n+3, 6m) \ &= \gcd(3, 6m) \ &= 3, \end{align*}unless m=0.

If m=0, then k=0, and in this case, 6n+k=6n is relatively prime to each of 6n+3, 6n+2, and 6n+1.

Similarly, \begin{align*} \gcd(6n+k, 6n+2) &= \gcd(6n+6m, 6n+2) \ &= \gcd(2, 6m) \ &= 2, \end{align*}unless m=0.

And \begin{align*} \gcd(6n+k, 6n+1) &= \gcd(6n+6m, 6n+1) \ &= \gcd(1, 6m) \ &= 1, \end{align*}unless m=0.

Therefore, if k is a multiple of 6, then 6n+k is relatively prime to each of 6n+3, 6n+2, and 6n+1.

Hence, the smallest positive integer k such that, for every positive integer n, 6n+k is relatively prime to each of 6n+3, 6n+2, and 6n+1 is 6

Guest Jun 9, 2023