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$a$ is an integer such that:

$$a \mid \gcd(b_1,b_2,\ldots,b_z)$$

and $z$ can be very large.

Does the GCD approach $a$ as $z$ grows? If yes, what is the relation between $z$ and $a$?

Thanks...

Daniel Fischer
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orcking
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    It is impossible to answer this question without knowing something about the $b_i$. (Also, the word approach is inappropriate for a sequence of integers; either the GCD is eventually equal to $a$ for large enough $z$, or it isn't. But we can't tell which, without more information.) – TonyK Jul 02 '15 at 17:35
  • Thanks. the $bs$ are unique... not sure what other information you need ? – orcking Jul 02 '15 at 17:54
  • The GCD might approach any multiple of $a$ as $z$ grows. – NovaDenizen Jul 04 '15 at 23:02

2 Answers2

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Using the constant sequence $b_n=2$ for all $n$, and choosing $a=1$, we have $$\begin{align*} a&\mid \gcd(b_1) & 1 &\mid 2 & \checkmark\\\\ a&\mid \gcd(b_1,b_2) & 1 &\mid 2 & \checkmark\\\\ a&\mid \gcd(b_1,b_2,b_3) & 1 &\mid 2 & \checkmark\\\\ &&\Large\vdots \end{align*}$$ but $\gcd(b_1,\ldots,b_z)=2$ for any choice of $z$, and does not approach $a=1$.

Zev Chonoles
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I think that I'm about to answer a different question, but perhaps you had something similar in mind. Might be overkill, but here I go.

In the statement of your question, you seem to understand that by taking bigger and bigger samples of the $b_i$, all of them divisible by $a$, it must be the case that “in the limit, $\gcd(b_1,\dots,b_z)$ must approach $a$”.

In the following I'll only discuss the case $a=1$; I'm leaving the general case as an “exercise for the reader”.

It was made clear in comments and in @ZevChonoles answer that a straightforward interpretation of this does not make sense, specially since the integers are discrete (as opposed to the real numbers, where you may approach indefinitely some number). But we may ask

(1) which is the probability that $z$ integers taken “at random“ have $\gcd$ equal to $1$ (here the requirement that $a$ divide each $b_i$ trivializes).

Even defining how two integers are “taken at random” must be done carefully; in this case, we say that the probability that a natural number $n$ satisfies a property $P(n)$ is taken to be the limit of the quotient $$ \frac{\# \{n \in \mathbb{N}: n \leq N \text{ and } P(n)\}}{\# \{n \in \mathbb{N}: n \leq N \}} $$ as $N$ approaches infinity (whenever this limit exists).

Back to question (1), the case $z=2$ is a classical problem, and it has been addressed in other questions, for example this one. The answer is $\frac{1}{\zeta(2)} = \frac{6}{\pi^2}$, where $\zeta$ is Riemann zeta function. Now I'll rephrase your question in the following way:

(2) Take $z$ integers $b_1,\dots,b_z$ at random, all of them divisible by $a$. Then the limit of the probability that $\gcd(b_1,\dots,b_z) = a$ as $z\to\infty$, is $1$.

Recall I'm working under the assumption $a=1$. So we're interested in knowing the probability that $z$ integers are coprime. Now this Wikipedia article states a generalization of the result we quoted for $z=2$: This probability is $\frac{1}{\zeta(z)}$. Now, it can be seen that $\lim_{z\to\infty} \frac{1}{\zeta(z)}=1$. This shows that the interpretation of your question given by (2) is indeed true.

Community
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Pedro Sánchez Terraf
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  • Great Pedro... question # 2 only applies to $a=1$. My original question was intended to see if $a$ is unknown, and I have z integers that are divisible by $a$, could I solve for $a$ if $z\to\infty$. Thanks – orcking Jul 05 '15 at 03:56
  • @orcking I didn't think about it, but it is likely that the same argument will work for the general case. – Pedro Sánchez Terraf Jul 05 '15 at 13:11