Question

Find a and b such that gcd(a, b) = 14, a > 2000, b > 2000, and the only prime divisors of a and b are 2 and 7.

Answer 1

Answer:

$a = 2^\alpha.14 \ ,\alpha\geq 8\\b = 7^\delta.14 \ , \delta\geq 3$

Step-by-step explanation:

As 2  and 7 are the only prime divisors of both $a$ and $b$ we know that both can be written as:

$a = 2^\alpha . 7^\beta \ , \alpha ,\beta\ \in \mathbb{N}_{0}\\b = 2^\gamma . 7^\delta \ , \gamma ,\delta\ \in \mathbb{N}_{0}$

Where $\mathbb{N}_{0}$ is the set of all natural numbers adding the zero (careful because this part is important as I'll explain next).

We also know that 14 divides both numbers and that is actually the greatest common divisor between them. So we can rewrite a and b as follows:

$a = 2^\alpha . 7^\beta.14 \ , \alpha ,\beta\ \in \mathbb{N}_{0}\\b = 2^\gamma . 7^\delta.14 \ , \gamma ,\delta\ \in \mathbb{N}_{0}$

Why do I write them like this? Because this way is easier to observe that if $\alpha$ and $\gamma$ were both greater than zero, then 28 would divide both hence 14 wouldn't be their g.c.d.. Likewise, if $\beta$ and $\delta$ were both greater than zero, then 98 would divide both and once again, 14 wouldn't be their g.c.d.

So either of them has to be equal to zero. And then we have that

$a = 2^\alpha.14 \\b = 7^\delta.14$

All we have left to do is find the possible values for $\alpha$ and $\delta$ so that $a>2000 \ , b>2000$ and that only happens if $\alpha\geq 8$ and $\delta\geq 3$