Least common factor of 17 and 6

Mathematics

2 answers:

marin [14]3 years ago

8 0

That will be 1, since 17 can only be factored by itself and 1

Even though 6 can be factored by 3 and 2, the least common factor between 17 and 6 will be 1.

MrMuchimi3 years ago

5 0

Hello there,

The correct answer is 2 the (LCM) or Least common factor of 17 and 6 is 2

If my answered helped you understand more greatly please mark me as brainliest thank you and have the best day ever!

You might be interested in

Suppose that, in addition to edge capacities, a flow network has vertex capacities. That is each vertex has a limit l./ on how m

storchak [24]

Answer:

See explanation and answer below.

Step-by-step explanation:

The tranformation

For this case we need to construct G' dividing making a division for each vertex v of G into 3 edges that on this case are $v_1, v_2 and l(v)$ .

We assume that the edges from the begin are the incoming edges of $v_1$ and all the outgoing edges from v are outgoing edges from $v_2$

We need to construct $G' = (V', E')$ with capacity function a' and we need to satisfy the follwoing:

For every $v \in V$ we create 2 vertices $v_1, v_2 \in V'$

Now we can add a new edge asscoiated to $v_1, v_2 \in E'$ with the condition $a' (v_1,v_2) = l(v)$

Now for each edges $(u,v)\in E$ we can create the following edge $( u_r, v_1) \in E'$ and the capacity is given by: $a' (u_r, v_1) = a (u,v)$

And for this case we can see this:

$|V'| = 2|V|, |E'|= |E| +|V|$

Now we assume that x is the flow who belongs to G respect vertex capabilities. We can create a flow function x' who belongs to G' with the following steps:

For every edge $(u,v) \in G$ we can assume that $x' (u_r ,v_1) = x(u,v)$

Then for each vertex $u \in V -t$ and we can define $x\(u_1,u_r) = \sum_{v \in V} x(u,v)$ and $x' (t_1,t_2) = \sum_{v \in V} x(v,t)$

And after see that the capacity constraint on this case would be satisfied since for every edge in G' on the form $(u_r, u_1)$ we have a corresponding edge in G because: