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The gcd cannot be a solution as the greatest common divisor, is the largest value, that divides all integers in the set.
But, there can be positive integer values less than gcd, that are not divisor of any of the values in the set.
Say, for the set of numbers: [11, 22, 33, 44, 55]

have the value of the smallest positive integer, as 6.

The proof for this statement is that for a given set of n numbers:

a_1, a_2, a_3, ..., a_n

considering the integers having only 2 prime factors: let 2, 3; have:

gcd(a_1, a_2, ..., a_n) = min(2^a_1, 2^a_2, ..., 2^{n-1}).min(3^a_1, 3^a_2, ..., 3^{n-1})

The product of the given n numbers is:

2^{a_1+a_2+ ...+ a_n}.3^{a_1+a_2+ ...+ a_n}

So, there can be many positive integers that satisfy the given criteria, of not dividing any of the given numbers.
Such a number can be less than gcd, or greater than the gcd; but don't how to proceed.

jiten
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    If the gcd is 1 or 2, then such a non-divisor certainly cannot be smaller than the gcd. – Hagen von Eitzen Oct 11 '23 at 02:06
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    Even with gcd equal to 3 you can't have an integer with these properties. Take for example the set ${3,6}$ which has gcd=3 and $2|6$. – Fotis Oct 11 '23 at 02:09
  • @HagenvonEitzen So, no rule applies, in general; for the existence of any non-divisor, smaller than the gcd, vis-a-vis the gcd value, for a set of numbers? – jiten Oct 11 '23 at 02:53
  • @Fotis Should the question be modified to some minimum value of the gcd; as it seems that then it would be applicable. – jiten Oct 11 '23 at 08:46

1 Answers1

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Such a number need not to exist.

For example, let $n$ be any integer and $N$ be the set of all positive integers less than $n$.
Obviously $N$ is a finite set.

Now define $m=n\prod\limits_{k\in N}k$.

Then it easy to see that $\gcd(m,n)=n$ and every number smaller than $n$ is divisor of $m$.

Hence for the set $\{m,n\}$ no integer with the properties you want exists.

Fotis
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  • So, my question is applicable when particularly the above property is not there. But, want to know if this is a proof of failure of the hypothesis, in the OP, by showing a counter-example? – jiten Oct 11 '23 at 08:51
  • Yes pretty much, the existance of a counterexample is proof that the initial claim is false – Fotis Oct 11 '23 at 12:35