A riddle for 2015 
How can one get $2015$ using $1,2,\dots,9$ in this order and only once, with the operations $+,-,\times,/$ ?

Solving this riddle with a computer (using python) turned out to be impossible for me due to the amount of memory needed.
Hence, I am looking for a mathematical way to find the/a solution (I am not just looking for the solution, but also for a technique to solve that kind of problems).
 A: A simple brute force Python program could be used to find one possible solution, although finding all possible solutions would require permuting parentheses as well.
target = 2015

def rec(seq, total, ops):
    if (seq == 10):
        if total == target:
            print ops
        return
    rec(seq+1, total*seq, ops + ['*'])
    rec(seq+1, total-seq, ops + ['-'])
    rec(seq+1, total+seq, ops + ['+'])
    rec(seq+1, float(total)/seq, ops + ['/'])

rec(1, 0, [])

This program returns:
['+', '+', '*', '*', '+', '*', '+', '*', '-']

which is a possible solution to the problem
$(((1+2)*3*4+5)*6+7)*8-9 = 2015$
A: One technique is to use a product of numbers to get close to the solution and then use the remaining numbers to correct: For example $7\times8\times9=504=\frac{1}{4}\times2016$. Hence we need to generate a factor of $-1$ to add, and a factor of $4$ to multiply. One solution using this idea is
$$-1+(2-3+4-5+6)\times7\times8\times9$$
Another technique is to attempt to use a factorization: $2015=31\times65$ for example. Noting that $-7+8\times9=65$, all we have left to do is generate a factor of $31$. A solution using this idea is
$$((1-2)\times(3-4)+5\times6)\times(-7+8\times9)$$
A: I did this: $2015+9=2024$
$2024\div 8=253$
$253-7=246$
$246\div 6=41$
$41-5=36$
$36\div 4=9$
$9\div 3=3$
$3-2=1$
$1-1=0$.
Then $2015=(((1+2)\cdot 3\cdot 4+5)\cdot 6+7)\cdot 8-9$
A: $$
\begin{align}
2015
&=-1+2016\\
&=-1+224\times9\\
&=-1+28\times8\times9\\
&=-1+4\times7\times8\times9\\
&=-1+2\times2\times7\times8\times9\\
&=-1+2\times(-3+4-5+6)\times7\times8\times9\\
\end{align}
$$
The "method" used above was to look for a number close to 2015 that has a number of factors. Use a combination of early integers to make up the difference and the rest to make up the factors. Here, $2015+1=2016=4\times7\times8\times9$. This reduces the problem to writing $4$ using $2,3,4,5,6$. A similar approach was used to break $4=2\times2$ where now the problem is reduced to writing $2$ using $3,4,5,6$.
A: A very easy approach can be
A) Find the prime factors of the number. That is,
$$5\times13\times31=2015$$
B) Express those prime factors as sum,difference,product and divison of numbers $1-9$, starting from the biggest prime factor to be easy. That is,
$$ (8-3)\times(7\cdot2-1)\times(9\cdot4-5)$$
Just one example. :)
