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:

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

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

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

What is the only possible number $\displaystyle x$ 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 $\displaystyle \frac{e}{f}=g\ \Rightarrow\ \frac{e}{g}=f$

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

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