Inserting numbers to create a geometric progression Place three numbers in between 15, 31, 104 in such way, that they would be successive members of geometric progression. 
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There is no geometric progression that contains all of $15$, $31$, and $104$, let alone also the hypothetical "extra" numbers.
For suppose that $15=kr^a$, $31=kr^b$, and $104=kr^c$ where $a$, $b$, $c$ are integers ($r$ need not be an integer, and $a$, $b$, $c$ need not be consecutive).  Then $31=kr^ar^{b-a}=15r^{b-a}$.  Similarly, $104=31r^{c-b}$. 
Without loss of generality, we may assume that $r>1$.  So $b-a$ and $c-b$ are positive integers.
Let $b-a=m$ and $c-b=n$. Then
$$\frac{31}{15}=r^m \qquad \text{and}\qquad \frac{104}{31}=r^n.$$
Take the $n$-th power of $31/15$, and the $m$-th power of $104/31$.  Each is $r^{mn}$.  It follows that
$$\left(\frac{31}{15}\right)^n=\left(\frac{104}{31}\right)^m.$$
From this we conclude that
$$31^{m+n}=15^n \cdot 104^m.$$
This is impossible, since $5$ divides the right-hand side, but $5$ does not divide the left-hand side.
Comment:  Have we really answered the question?  It asks us to place $3$ numbers between $15$, $31$, $104$ "in such a way that they would be successive members of a geometric progression."  Who does "they" refer to?  Certainly not all $6$ numbers, since already as we have seen, $15$, $31$, and $104$ cannot all be members of a (single) geometric progression of any kind.
But maybe "they" refers to the interpolated numbers!  Then there are uncountably many solutions, and even several solutions where the interpolated numbers are all integers.  For example, we can use $16$, $32$, $64$. The numbers $15$ and $31$ could be a heavy-handed hint pointing to this answer. 
Or else we can use  $16$, $24$, $36$. Or else $16$, $40$, $100$.  Then there is $18$, $24$, $32$, or $18$, $30$, $50$, and so on.   
