0
$\begingroup$

Might be a silly question. But here it is: Say we have all the factors of a number, 84 for example. Those numbers would be 1,2,3,4,6,7,12,14,21,28,42,84. Is it possible for there to be a number that would evenly divide one of the factors of 84, without being able to evenly divide 84 itself? For the number 84 it doesn't seem to be possible, but does there exist a number where that is possible?

$\endgroup$
  • 2
    $\begingroup$ Welcome to Mathematics Stack Exchange. No; if $a$ divides $b$ and $b$ divides $c$ then $a$ divides $c$ $\endgroup$ – J. W. Tanner Oct 29 '19 at 22:07
  • $\begingroup$ Got it. Thanks. $\endgroup$ – comp1201 Oct 29 '19 at 22:08
1
$\begingroup$

It's not possible.

If $a$ divides $b$ (i.e., $b=ka$) and $b$ divides $c$ (i.e., $c=nb$), then $a$ divides $c$ (because $c=nka$).

$\endgroup$
  • $\begingroup$ divisibility is a transitive relation $\endgroup$ – J. W. Tanner Oct 29 '19 at 22:10
  • $\begingroup$ divisibility is also a partial order. $\endgroup$ – Don Thousand Oct 29 '19 at 22:20

Not the answer you're looking for? Browse other questions tagged or ask your own question.