1. operations

Gary performed the following three operations: 15 ÷ x, 21 ÷ x and 6 ÷ x. If he never got a remainder and x is an integer greater than 1, what is the value of x?

2. From these assumptions follows:

$ax = 15$
$bx = 21$
$cx = 6$

Where $x> 1, a,b,c$ are natural numbers.

Now we can write out the possible values of $x$
$x|6$ gives possible values $x= 2,3,6$
$x|21$ gives possible values $x = 3,7,21$
$x|15$ gives possible values $x = 3,5,15$

What is the only possible number $x$ in all 3 lists?

3. Originally Posted by sri340
Gary performed the following three operations: 15 ÷ x, 21 ÷ x and 6 ÷ x. If he never got a remainder and x is an integer greater than 1, what is the value of x?
Or $\frac{e}{f}=g\ \Rightarrow\ \frac{e}{g}=f$

Hence $x=\frac{15}{a}=\frac{21}{b}=\frac{6}{c}$

The prime factors of 6 are 1, 2 and 3.
15 and 21 are divisible by 3 but not by 2.