Co-Prime Listing

The prime numbers are $2,3,5,...$

The positive two-digit integers which are divisible by $2$ are $10,12,14,...,96,98$

The positive two-digit integers which are divisible by $3$ are $12,15,18,...,96,99$

The positive two-digit integers which are divisible by $5$ are $10,15,20,...,90,95$

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Notice that the numbers in each list of multiples are in the form of arithmetic progression(A.M.)

The number of positive two-digit integers divisible by $2$ = $\cfrac{98-10}{2}+1=45$

The number of positive two-digit integers divisible by $3$ = $\cfrac{99-12}{3}+1=30$

The number of positive two-digit integers divisible by $5$ = $\cfrac{95-10}{5}+1=18$

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Since the number of positive two-digit integers divisible by $2$ is the most among all the primes, by Pigeonhole's Principle , the $\textbf{least}$ number of positive two-digit integers that I need to write down to guarantee that I could find a pair of $\textbf{co-prime}$ numbers $=45+1=\boxed{46}$

$\textbf{Note:}$

You may be asking why isn't the answer $2$ . The answer is not $2$ because in order to guarantee that I could find a pair of $\textbf{co-prime}$ numbers, I ought to consider the $\textbf{worst-case scenario}$ ! Here, the $\textbf{worst-case scenario}$ is I first choose all $45$ positive two-digit integers divisible by $2$ and go on to pick the next number which is not a multiple of $2$ , which guarantees me a pair of co-prime numbers(the number of positive two-digit integers divisible by $2$ is the most excluding $1$ which is not a prime).

And take note that $1$ cannot be chosen before the $45$ positive two-digit integers divisible by $2$ are chosen as this would make the task of finding co-prime pair easier and this will not be the $\textbf{worst-case scenario}$ we are looking for.